Related
I've a vector of vectors say vector<vector<int> > items of different sizes like as follows
1,2,3
4,5
6,7,8
I want to create combinations in terms of Cartesian product of these vectors like
1,4,6
1,4,7
1,4,8
and so on till
3,5,8
How can I do that ? I've looked up several links and I've also listed them at the end of this post but I'm not able to interpret that as I'm not that familiar with the language. Could some body help me with this.
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
int main()
{
vector<vector<int> > items;
int k = 0;
for ( int i = 0; i < 5; i++ ) {
items.push_back ( vector<int>() );
for ( int j = 0; j < 5; j++ )
items[i].push_back ( k++ );
}
cartesian ( items ); // I want some function here to do this.
}
This program has equal length vectors and I put this so that it will be easier to understand my data structure. It will be very helpful even if somebody uses others answers from other links and integrate with this to get the result. Thank you very much
Couple of links I looked at
one
Two
Program from : program
First, I'll show you a recursive version.
// Cartesion product of vector of vectors
#include <vector>
#include <iostream>
#include <iterator>
// Types to hold vector-of-ints (Vi) and vector-of-vector-of-ints (Vvi)
typedef std::vector<int> Vi;
typedef std::vector<Vi> Vvi;
// Just for the sample -- populate the intput data set
Vvi build_input() {
Vvi vvi;
for(int i = 0; i < 3; i++) {
Vi vi;
for(int j = 0; j < 3; j++) {
vi.push_back(i*10+j);
}
vvi.push_back(vi);
}
return vvi;
}
// just for the sample -- print the data sets
std::ostream&
operator<<(std::ostream& os, const Vi& vi)
{
os << "(";
std::copy(vi.begin(), vi.end(), std::ostream_iterator<int>(os, ", "));
os << ")";
return os;
}
std::ostream&
operator<<(std::ostream& os, const Vvi& vvi)
{
os << "(\n";
for(Vvi::const_iterator it = vvi.begin();
it != vvi.end();
it++) {
os << " " << *it << "\n";
}
os << ")";
return os;
}
// recursive algorithm to to produce cart. prod.
// At any given moment, "me" points to some Vi in the middle of the
// input data set.
// for int i in *me:
// add i to current result
// recurse on next "me"
//
void cart_product(
Vvi& rvvi, // final result
Vi& rvi, // current result
Vvi::const_iterator me, // current input
Vvi::const_iterator end) // final input
{
if(me == end) {
// terminal condition of the recursion. We no longer have
// any input vectors to manipulate. Add the current result (rvi)
// to the total set of results (rvvvi).
rvvi.push_back(rvi);
return;
}
// need an easy name for my vector-of-ints
const Vi& mevi = *me;
for(Vi::const_iterator it = mevi.begin();
it != mevi.end();
it++) {
// final rvi will look like "a, b, c, ME, d, e, f"
// At the moment, rvi already has "a, b, c"
rvi.push_back(*it); // add ME
cart_product(rvvi, rvi, me+1, end); add "d, e, f"
rvi.pop_back(); // clean ME off for next round
}
}
// sample only, to drive the cart_product routine.
int main() {
Vvi input(build_input());
std::cout << input << "\n";
Vvi output;
Vi outputTemp;
cart_product(output, outputTemp, input.begin(), input.end());
std::cout << output << "\n";
}
Now, I'll show you the recursive iterative version that I shamelessly stole from #John :
The rest of the program is pretty much the same, only showing the cart_product function.
// Seems like you'd want a vector of iterators
// which iterate over your individual vector<int>s.
struct Digits {
Vi::const_iterator begin;
Vi::const_iterator end;
Vi::const_iterator me;
};
typedef std::vector<Digits> Vd;
void cart_product(
Vvi& out, // final result
Vvi& in) // final result
{
Vd vd;
// Start all of the iterators at the beginning.
for(Vvi::const_iterator it = in.begin();
it != in.end();
++it) {
Digits d = {(*it).begin(), (*it).end(), (*it).begin()};
vd.push_back(d);
}
while(1) {
// Construct your first product vector by pulling
// out the element of each vector via the iterator.
Vi result;
for(Vd::const_iterator it = vd.begin();
it != vd.end();
it++) {
result.push_back(*(it->me));
}
out.push_back(result);
// Increment the rightmost one, and repeat.
// When you reach the end, reset that one to the beginning and
// increment the next-to-last one. You can get the "next-to-last"
// iterator by pulling it out of the neighboring element in your
// vector of iterators.
for(Vd::iterator it = vd.begin(); ; ) {
// okay, I started at the left instead. sue me
++(it->me);
if(it->me == it->end) {
if(it+1 == vd.end()) {
// I'm the last digit, and I'm about to roll
return;
} else {
// cascade
it->me = it->begin;
++it;
}
} else {
// normal
break;
}
}
}
}
Here is a solution in C++11.
The indexing of the variable-sized arrays can be done eloquently with modular arithmetic.
The total number of lines in the output is the product of the sizes of the input vectors. That is:
N = v[0].size() * v[1].size() * v[2].size()
Therefore the main loop has n as the iteration variable, from 0 to N-1. In principle, each value of n encodes enough information to extract each of the indices of v for that iteration. This is done in a subloop using repeated modular arithmetic:
#include <cstdlib>
#include <iostream>
#include <numeric>
#include <vector>
using namespace std;
void cartesian( vector<vector<int> >& v ) {
auto product = []( long long a, vector<int>& b ) { return a*b.size(); };
const long long N = accumulate( v.begin(), v.end(), 1LL, product );
vector<int> u(v.size());
for( long long n=0 ; n<N ; ++n ) {
lldiv_t q { n, 0 };
for( long long i=v.size()-1 ; 0<=i ; --i ) {
q = div( q.quot, v[i].size() );
u[i] = v[i][q.rem];
}
// Do what you want here with u.
for( int x : u ) cout << x << ' ';
cout << '\n';
}
}
int main() {
vector<vector<int> > v { { 1, 2, 3 },
{ 4, 5 },
{ 6, 7, 8 } };
cartesian(v);
return 0;
}
Output:
1 4 6
1 4 7
1 4 8
...
3 5 8
Shorter code:
vector<vector<int>> cart_product (const vector<vector<int>>& v) {
vector<vector<int>> s = {{}};
for (const auto& u : v) {
vector<vector<int>> r;
for (const auto& x : s) {
for (const auto y : u) {
r.push_back(x);
r.back().push_back(y);
}
}
s = move(r);
}
return s;
}
Seems like you'd want a vector of iterators which iterate over your individual vector<int>s.
Start all of the iterators at the beginning. Construct your first product vector by pulling out the element of each vector via the iterator.
Increment the rightmost one, and repeat.
When you reach the end, reset that one to the beginning and increment the next-to-last one. You can get the "next-to-last" iterator by pulling it out of the neighboring element in your vector of iterators.
Continue cycling through until both the last and next-to-last iterators are at the end. Then, reset them both, increment the third-from-last iterator. In general, this can be cascaded.
It's like an odometer, but with each different digit being in a different base.
Here's my solution. Also iterative, but a little shorter than the above...
void xp(const vector<vector<int>*>& vecs, vector<vector<int>*> *result) {
vector<vector<int>*>* rslts;
for (int ii = 0; ii < vecs.size(); ++ii) {
const vector<int>& vec = *vecs[ii];
if (ii == 0) {
// vecs=[[1,2],...] ==> rslts=[[1],[2]]
rslts = new vector<vector<int>*>;
for (int jj = 0; jj < vec.size(); ++jj) {
vector<int>* v = new vector<int>;
v->push_back(vec[jj]);
rslts->push_back(v);
}
} else {
// vecs=[[1,2],[3,4],...] ==> rslts=[[1,3],[1,4],[2,3],[2,4]]
vector<vector<int>*>* tmp = new vector<vector<int>*>;
for (int jj = 0; jj < vec.size(); ++jj) { // vec[jj]=3 (first iter jj=0)
for (vector<vector<int>*>::const_iterator it = rslts->begin();
it != rslts->end(); ++it) {
vector<int>* v = new vector<int>(**it); // v=[1]
v->push_back(vec[jj]); // v=[1,3]
tmp->push_back(v); // tmp=[[1,3]]
}
}
for (int kk = 0; kk < rslts->size(); ++kk) {
delete (*rslts)[kk];
}
delete rslts;
rslts = tmp;
}
}
result->insert(result->end(), rslts->begin(), rslts->end());
delete rslts;
}
I derived it with some pain from a haskell version I wrote:
xp :: [[a]] -> [[a]]
xp [] = []
xp [l] = map (:[]) l
xp (h:t) = foldr (\x acc -> foldr (\l acc -> (x:l):acc) acc (xp t)) [] h
Since I needed the same functionality, I implemented an iterator which computes the Cartesian product on the fly, as needed, and iterates over it.
It can be used as follows.
#include <forward_list>
#include <iostream>
#include <vector>
#include "cartesian.hpp"
int main()
{
// Works with a vector of vectors
std::vector<std::vector<int>> test{{1,2,3}, {4,5,6}, {8,9}};
CartesianProduct<decltype(test)> cp(test);
for(auto const& val: cp) {
std::cout << val.at(0) << ", " << val.at(1) << ", " << val.at(2) << "\n";
}
// Also works with something much less, like a forward_list of forward_lists
std::forward_list<std::forward_list<std::string>> foo{{"boo", "far", "zab"}, {"zoo", "moo"}, {"yohoo", "bohoo", "whoot", "noo"}};
CartesianProduct<decltype(foo)> bar(foo);
for(auto const& val: bar) {
std::cout << val.at(0) << ", " << val.at(1) << ", " << val.at(2) << "\n";
}
}
The file cartesian.hpp looks like this.
#include <cassert>
#include <limits>
#include <stdexcept>
#include <vector>
#include <boost/iterator/iterator_facade.hpp>
//! Class iterating over the Cartesian product of a forward iterable container of forward iterable containers
template<typename T>
class CartesianProductIterator : public boost::iterator_facade<CartesianProductIterator<T>, std::vector<typename T::value_type::value_type> const, boost::forward_traversal_tag>
{
public:
//! Delete default constructor
CartesianProductIterator() = delete;
//! Constructor setting the underlying iterator and position
/*!
* \param[in] structure The underlying structure
* \param[in] pos The position the iterator should be initialized to. std::numeric_limits<std::size_t>::max()stands for the end, the position after the last element.
*/
explicit CartesianProductIterator(T const& structure, std::size_t pos);
private:
//! Give types more descriptive names
// \{
typedef T OuterContainer;
typedef typename T::value_type Container;
typedef typename T::value_type::value_type Content;
// \}
//! Grant access to boost::iterator_facade
friend class boost::iterator_core_access;
//! Increment iterator
void increment();
//! Check for equality
bool equal(CartesianProductIterator<T> const& other) const;
//! Dereference iterator
std::vector<Content> const& dereference() const;
//! The part we are iterating over
OuterContainer const& structure_;
//! The position in the Cartesian product
/*!
* For each element of structure_, give the position in it.
* The empty vector represents the end position.
* Note that this vector has a size equal to structure->size(), or is empty.
*/
std::vector<typename Container::const_iterator> position_;
//! The position just indexed by an integer
std::size_t absolutePosition_ = 0;
//! The begin iterators, saved for convenience and performance
std::vector<typename Container::const_iterator> cbegins_;
//! The end iterators, saved for convenience and performance
std::vector<typename Container::const_iterator> cends_;
//! Used for returning references
/*!
* We initialize with one empty element, so that we only need to add more elements in increment().
*/
mutable std::vector<std::vector<Content>> result_{std::vector<Content>()};
//! The size of the instance of OuterContainer
std::size_t size_ = 0;
};
template<typename T>
CartesianProductIterator<T>::CartesianProductIterator(OuterContainer const& structure, std::size_t pos) : structure_(structure)
{
for(auto & entry: structure_) {
cbegins_.push_back(entry.cbegin());
cends_.push_back(entry.cend());
++size_;
}
if(pos == std::numeric_limits<std::size_t>::max() || size_ == 0) {
absolutePosition_ = std::numeric_limits<std::size_t>::max();
return;
}
// Initialize with all cbegin() position
position_.reserve(size_);
for(std::size_t i = 0; i != size_; ++i) {
position_.push_back(cbegins_[i]);
if(cbegins_[i] == cends_[i]) {
// Empty member, so Cartesian product is empty
absolutePosition_ = std::numeric_limits<std::size_t>::max();
return;
}
}
// Increment to wanted position
for(std::size_t i = 0; i < pos; ++i) {
increment();
}
}
template<typename T>
void CartesianProductIterator<T>::increment()
{
if(absolutePosition_ == std::numeric_limits<std::size_t>::max()) {
return;
}
std::size_t pos = size_ - 1;
// Descend as far as necessary
while(++(position_[pos]) == cends_[pos] && pos != 0) {
--pos;
}
if(position_[pos] == cends_[pos]) {
assert(pos == 0);
absolutePosition_ = std::numeric_limits<std::size_t>::max();
return;
}
// Set all to begin behind pos
for(++pos; pos != size_; ++pos) {
position_[pos] = cbegins_[pos];
}
++absolutePosition_;
result_.emplace_back();
}
template<typename T>
std::vector<typename T::value_type::value_type> const& CartesianProductIterator<T>::dereference() const
{
if(absolutePosition_ == std::numeric_limits<std::size_t>::max()) {
throw new std::out_of_range("Out of bound dereference in CartesianProductIterator\n");
}
auto & result = result_[absolutePosition_];
if(result.empty()) {
result.reserve(size_);
for(auto & iterator: position_) {
result.push_back(*iterator);
}
}
return result;
}
template<typename T>
bool CartesianProductIterator<T>::equal(CartesianProductIterator<T> const& other) const
{
return absolutePosition_ == other.absolutePosition_ && structure_ == other.structure_;
}
//! Class that turns a forward iterable container of forward iterable containers into a forward iterable container which iterates over the Cartesian product of the forward iterable containers
template<typename T>
class CartesianProduct
{
public:
//! Constructor from type T
explicit CartesianProduct(T const& t) : t_(t) {}
//! Iterator to beginning of Cartesian product
CartesianProductIterator<T> begin() const { return CartesianProductIterator<T>(t_, 0); }
//! Iterator behind the last element of the Cartesian product
CartesianProductIterator<T> end() const { return CartesianProductIterator<T>(t_, std::numeric_limits<std::size_t>::max()); }
private:
T const& t_;
};
If someone has comments how to make it faster or better, I'd highly appreciate them.
I was just forced to implement this for a project I was working on and I came up with the code below. It can be stuck in a header and it's use is very simple but it returns all of the combinations you can get from a vector of vectors. The array that it returns only holds integers. This was a conscious decision because I just wanted the indices. In this way, I could index into each of the vector's vector and then perform the calculations I/anyone would need... best to avoid letting CartesianProduct hold "stuff" itself, it is a mathematical concept based around counting not a data structure. I'm fairly new to c++ but this was tested in a decryption algorithm pretty thoroughly. There is some light recursion but overall this is a simple implementation of a simple counting concept.
// Use of the CartesianProduct class is as follows. Give it the number
// of rows and the sizes of each of the rows. It will output all of the
// permutations of these numbers in their respective rows.
// 1. call cp.permutation() // need to check all 0s.
// 2. while cp.HasNext() // it knows the exit condition form its inputs.
// 3. cp.Increment() // Make the next permutation
// 4. cp.permutation() // get the next permutation
class CartesianProduct{
public:
CartesianProduct(int num_rows, vector<int> sizes_of_rows){
permutation_ = new int[num_rows];
num_rows_ = num_rows;
ZeroOutPermutation();
sizes_of_rows_ = sizes_of_rows;
num_max_permutations_ = 1;
for (int i = 0; i < num_rows; ++i){
num_max_permutations_ *= sizes_of_rows_[i];
}
}
~CartesianProduct(){
delete permutation_;
}
bool HasNext(){
if(num_permutations_processed_ != num_max_permutations_) {
return true;
} else {
return false;
}
}
void Increment(){
int row_to_increment = 0;
++num_permutations_processed_;
IncrementAndTest(row_to_increment);
}
int* permutation(){
return permutation_;
}
int num_permutations_processed(){
return num_permutations_processed_;
}
void PrintPermutation(){
cout << "( ";
for (int i = 0; i < num_rows_; ++i){
cout << permutation_[i] << ", ";
}
cout << " )" << endl;
}
private:
int num_permutations_processed_;
int *permutation_;
int num_rows_;
int num_max_permutations_;
vector<int> sizes_of_rows_;
// Because CartesianProduct is called first initially with it's values
// of 0 and because those values are valid and important output
// of the CartesianProduct we increment the number of permutations
// processed here when we populate the permutation_ array with 0s.
void ZeroOutPermutation(){
for (int i = 0; i < num_rows_; ++i){
permutation_[i] = 0;
}
num_permutations_processed_ = 1;
}
void IncrementAndTest(int row_to_increment){
permutation_[row_to_increment] += 1;
int max_index_of_row = sizes_of_rows_[row_to_increment] - 1;
if (permutation_[row_to_increment] > max_index_of_row){
permutation_[row_to_increment] = 0;
IncrementAndTest(row_to_increment + 1);
}
}
};
#include <iostream>
#include <vector>
void cartesian (std::vector<std::vector<int>> const& items) {
auto n = items.size();
auto next = [&](std::vector<int> & x) {
for ( int i = 0; i < n; ++ i )
if ( ++x[i] == items[i].size() ) x[i] = 0;
else return true;
return false;
};
auto print = [&](std::vector<int> const& x) {
for ( int i = 0; i < n; ++ i )
std::cout << items[i][x[i]] << ",";
std::cout << "\b \n";
};
std::vector<int> x(n);
do print(x); while (next(x)); // Shazam!
}
int main () {
std::vector<std::vector<int>>
items { { 1, 2, 3 }, { 4, 5 }, { 6, 7, 8 } };
cartesian(items);
return 0;
}
The idea behind this is as follows.
Let n := items.size().
Let m_i := items[i].size(), for all i in {0,1,...,n-1}.
Let M := {0,1,...,m_0-1} x {0,1,...,m_1-1} x ... x {0,1,...,m_{n-1}-1}.
We first solve the simpler problem of iterating through M. This is accomplished by the next lambda. The algorithm is simply the "carrying" routine grade schoolers use to add 1, albeit with a mixed radix number system.
We use this to solve the more general problem by transforming a tuple x in M to one of the desired tuples via the formula items[i][x[i]] for all i in {0,1,...,n-1}. We perform this transformation in the print lambda.
We then perform the iteration with do print(x); while (next(x));.
Now some comments on complexity, under the assumption that m_i > 1 for all i:
This algorithm requires O(n) space. Note that explicit construction of the Cartesian product takes O(m_0 m_1 m_2 ... m_{n-1}) >= O(2^n) space. So this is exponentially better on space than any algorithm which requires all tuples to be stored simultaneously in memory.
The next function takes amortized O(1) time (by a geometric series argument).
The print function takes O(n) time.
Hence, altogether, the algorithm has time complexity O(n|M|) and space complexity O(n) (not counting the cost of storing items).
An interesting thing to note is that if print is replaced with a function which inspects on average only O(1) coordinates per tuple rather than all of them, then time complexity falls to O(|M|), that is, it becomes linear time with respect to the size of the Cartesian product. In other words, avoiding the copy of the tuple each iterate can be meaningful in some situations.
This version supports no iterators or ranges, but it is a simple direct implementation that uses the multiplication operator to represent the Cartesian product, and a lambda to perform the action.
The interface is designed with the particular functionality I needed. I needed the flexibility to choose vectors over which to apply the Cartesian product in a way that did not obscure the code.
int main()
{
vector< vector<long> > v{ { 1, 2, 3 }, { 4, 5 }, { 6, 7, 8 } };
(Cartesian<long>(v[0]) * v[1] * v[2]).ForEach(
[](long p_Depth, long *p_LongList)
{
std::cout << p_LongList[0] << " " << p_LongList[1] << " " << p_LongList[2] << std::endl;
}
);
}
The implementation uses recursion up the class structure to implement the embedded for loops over each vector. The algorithm works directly on the input vectors, requiring no large temporary arrays. It is simple to understand and debug.
The use of std::function p_Action instead of void p_Action(long p_Depth, T *p_ParamList) for the lambda parameter would allow me to capture local variables, if I wanted to. In the above call, I don't.
But you knew that, didn't you. "function" is a template class which takes the type parameter of a function and makes it callable.
#include <vector>
#include <iostream>
#include <functional>
#include <string>
using namespace std;
template <class T>
class Cartesian
{
private:
vector<T> &m_Vector;
Cartesian<T> *m_Cartesian;
public:
Cartesian(vector<T> &p_Vector, Cartesian<T> *p_Cartesian=NULL)
: m_Vector(p_Vector), m_Cartesian(p_Cartesian)
{};
virtual ~Cartesian() {};
Cartesian<T> *Clone()
{
return new Cartesian<T>(m_Vector, m_Cartesian ? m_Cartesian->Clone() : NULL);
};
Cartesian<T> &operator *=(vector<T> &p_Vector)
{
if (m_Cartesian)
(*m_Cartesian) *= p_Vector;
else
m_Cartesian = new Cartesian(p_Vector);
return *this;
};
Cartesian<T> operator *(vector<T> &p_Vector)
{
return (*Clone()) *= p_Vector;
};
long Depth()
{
return m_Cartesian ? 1 + m_Cartesian->Depth() : 1;
};
void ForEach(function<void (long p_Depth, T *p_ParamList)> p_Action)
{
Loop(0, new T[Depth()], p_Action);
};
private:
void Loop(long p_Depth, T *p_ParamList, function<void (long p_Depth, T *p_ParamList)> p_Action)
{
for (T &element : m_Vector)
{
p_ParamList[p_Depth] = element;
if (m_Cartesian)
m_Cartesian->Loop(p_Depth + 1, p_ParamList, p_Action);
else
p_Action(Depth(), p_ParamList);
}
};
};
I'm trying to keep a vector of commands so that it keeps 10 most recent. I have a push_back and a pop_back, but how do I delete the oldest without shifting everything in a for loop? Is erase the only way to do this?
Use std::deque which is a vector-like container that's good at removal and insertion at both ends.
If you're amenable to using boost, I'd recommend looking at circular_buffer, which deals with this exact problem extremely efficiently (it avoids moving elements around unnecessarily, and instead just manipulates a couple of pointers):
// Create a circular buffer with a capacity for 3 integers.
boost::circular_buffer<int> cb(3);
// Insert threee elements into the buffer.
cb.push_back(1);
cb.push_back(2);
cb.push_back(3);
cb.push_back(4);
cb.push_back(5);
The last two ops simply overwrite the elements of the first two.
Write a wrapper around a vector to give yourself a circular buffer. Something like this:
include <vector>
/**
Circular vector wrapper
When the vector is full, old data is overwritten
*/
class cCircularVector
{
public:
// An iterator that points to the physical begining of the vector
typedef std::vector< short >::iterator iterator;
iterator begin() { return myVector.begin(); }
iterator end() { return myVector.end(); }
// The size ( capacity ) of the vector
int size() { return (int) myVector.size(); }
void clear() { myVector.clear(); next = 0; }
void resize( int s ) { myVector.resize( s ); }
// Constructor, specifying the capacity
cCircularVector( int capacity )
: next( 0 )
{
myVector.resize( capacity );
}
// Add new data, over-writing oldest if full
void push_back( short v )
{
myVector[ next] = v;
advance();
}
int getNext()
{
return next;
}
private:
std::vector< short > myVector;
int next;
void advance()
{
next++;
if( next == (int)myVector.size() )
next = 0;
}
};
What about something like this:
http://ideone.com/SLSNpc
Note: It's just a base, you still need to work a bit on it. The idea is that it's easy to use because it has it's own iterator, which will give you the output you want. As you can see the last value inserted is the one shown first, which I'm guessing is what you want.
#include <iostream>
#include <vector>
template<class T, size_t MaxSize>
class TopN
{
public:
void push_back(T v)
{
if (m_vector.size() < MaxSize)
m_vector.push_back(v);
else
m_vector[m_pos] = v;
if (++m_pos == MaxSize)
m_pos = 0;
}
class DummyIterator
{
public:
TopN &r; // a direct reference to our boss.
int p, m; // m: how many elements we can pull from vector, p: position of the cursor.
DummyIterator(TopN& t) : r(t), p(t.m_pos), m(t.m_vector.size()){}
operator bool() const { return (m > 0); }
T& operator *()
{
static T e = 0; // this could be removed
if (m <= 0) // if someone tries to extract data from an empty vector
return e; // instead of throwing an error, we return a dummy value
m--;
if (--p < 0)
p = MaxSize - 1;
return r.m_vector[p];
}
};
decltype(auto) begin() { return m_vector.begin(); }
decltype(auto) end() { return m_vector.end(); }
DummyIterator get_dummy_iterator()
{
return DummyIterator(*this);
}
private:
std::vector<T> m_vector;
int m_pos = 0;
};
template<typename T, size_t S>
void show(TopN<T,S>& t)
{
for (auto it = t.get_dummy_iterator(); it; )
std::cout << *it << '\t';
std::cout << std::endl;
};
int main(int argc, char* argv[])
{
TopN<int,10> top10;
for (int i = 1; i <= 10; i++)
top10.push_back(5 * i);
show(top10);
top10.push_back(60);
show(top10);
top10.push_back(65);
show(top10);
return 0;
}
I want to extract a sub-vector. Then modify its elements which affects the original vector. My sample code below:
#include <vector>
#include <iostream>
using namespace std;
void printvec(vector<int>& v){
for(int i = 0;i < v.size();i++) {cout << v[i] << " ";}
cout << endl;
}
int main(){
vector<int> original;
for(int i = 1;i <= 10;i++) original.push_back(i);
printvec(original);
vector<int> subvector(original.begin()+4, original.end()-2);
subvector[0]=0;
subvector[1]=0;
printvec(subvector);
printvec(original);
return 0;
}
In above code, subvector does not modify vector. Can some one point me to an elegant way to make a subvector which modifies original vector (hopefully without explicit use of pointers if possible).
If you don't want to use a pointer, you could create a slice class to forward the work to - which will just be a pair of iterators and whatever other operations you might need:
template <typename T>
class slice {
using iterator = typename T::iterator;
using reference = typename std::iterator_traits<iterator>::reference;
slice(iterator first, iterator last)
: first(first), last(last)
{ }
reference operator[](size_t idx)
{
return *std::next(first, idx);
}
iterator begin() const { return first; }
iterator end() const { return last; }
private:
iterator first, last;
};
With that, you can do your slicing thusly:
slice<vector<int>> subvector(original.begin()+4, original.end()-2);
subvector[0]=0; // changes original[4]
subvector[1]=0; // changes original[5]
If you change your printvec to take an arbitrary container and use a range-for to iterate over it, you can print the subvector too. It will contain:
0 0 7 8
The line:
vector<int> subvector(original.begin()+4, original.end()-2);
creates a new vector and copies the elements from original.begin()+4 to original.end()-2 to the new one. Even with pointers, there is no way I would call elegant to achieve what you want, because many change to the original vector (rezise / push_back) could potentially invalidate the pointers to its elements.
Depending on the exact functionality you want to implement, you can use a class like this:
#include <vector>
#include <iostream>
using namespace std;
template<class T>
class Subvector {
std::vector<T>*const vec;
size_t start;
size_t end;
public:
Subvector(std::vector<T>& vector, size_t start, size_t end) :
vec(&vector),
start(start),
end(end)
{}
size_t size() const { return end - start; }
T& operator[](size_t i) {
return (*vec)[start + i];
}
const T& operator[](size_t i) const {
return (*vec)[start + i];
}
};
template<class VEC>
void printvec(const VEC& v){
for (int i = 0; i < v.size(); i++) { cout << v[i] << " "; }
cout << endl;
}
int main(){
vector<int> original;
for (int i = 1; i <= 10; i++) original.push_back(i);
printvec(original);
Subvector<int> subvector(original,4, original.size() - 2);
subvector[0] = 0;
subvector[1] = 0;
printvec(subvector);
printvec(original);
return 0;
}
I've a vector of vectors say vector<vector<int> > items of different sizes like as follows
1,2,3
4,5
6,7,8
I want to create combinations in terms of Cartesian product of these vectors like
1,4,6
1,4,7
1,4,8
and so on till
3,5,8
How can I do that ? I've looked up several links and I've also listed them at the end of this post but I'm not able to interpret that as I'm not that familiar with the language. Could some body help me with this.
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
int main()
{
vector<vector<int> > items;
int k = 0;
for ( int i = 0; i < 5; i++ ) {
items.push_back ( vector<int>() );
for ( int j = 0; j < 5; j++ )
items[i].push_back ( k++ );
}
cartesian ( items ); // I want some function here to do this.
}
This program has equal length vectors and I put this so that it will be easier to understand my data structure. It will be very helpful even if somebody uses others answers from other links and integrate with this to get the result. Thank you very much
Couple of links I looked at
one
Two
Program from : program
First, I'll show you a recursive version.
// Cartesion product of vector of vectors
#include <vector>
#include <iostream>
#include <iterator>
// Types to hold vector-of-ints (Vi) and vector-of-vector-of-ints (Vvi)
typedef std::vector<int> Vi;
typedef std::vector<Vi> Vvi;
// Just for the sample -- populate the intput data set
Vvi build_input() {
Vvi vvi;
for(int i = 0; i < 3; i++) {
Vi vi;
for(int j = 0; j < 3; j++) {
vi.push_back(i*10+j);
}
vvi.push_back(vi);
}
return vvi;
}
// just for the sample -- print the data sets
std::ostream&
operator<<(std::ostream& os, const Vi& vi)
{
os << "(";
std::copy(vi.begin(), vi.end(), std::ostream_iterator<int>(os, ", "));
os << ")";
return os;
}
std::ostream&
operator<<(std::ostream& os, const Vvi& vvi)
{
os << "(\n";
for(Vvi::const_iterator it = vvi.begin();
it != vvi.end();
it++) {
os << " " << *it << "\n";
}
os << ")";
return os;
}
// recursive algorithm to to produce cart. prod.
// At any given moment, "me" points to some Vi in the middle of the
// input data set.
// for int i in *me:
// add i to current result
// recurse on next "me"
//
void cart_product(
Vvi& rvvi, // final result
Vi& rvi, // current result
Vvi::const_iterator me, // current input
Vvi::const_iterator end) // final input
{
if(me == end) {
// terminal condition of the recursion. We no longer have
// any input vectors to manipulate. Add the current result (rvi)
// to the total set of results (rvvvi).
rvvi.push_back(rvi);
return;
}
// need an easy name for my vector-of-ints
const Vi& mevi = *me;
for(Vi::const_iterator it = mevi.begin();
it != mevi.end();
it++) {
// final rvi will look like "a, b, c, ME, d, e, f"
// At the moment, rvi already has "a, b, c"
rvi.push_back(*it); // add ME
cart_product(rvvi, rvi, me+1, end); add "d, e, f"
rvi.pop_back(); // clean ME off for next round
}
}
// sample only, to drive the cart_product routine.
int main() {
Vvi input(build_input());
std::cout << input << "\n";
Vvi output;
Vi outputTemp;
cart_product(output, outputTemp, input.begin(), input.end());
std::cout << output << "\n";
}
Now, I'll show you the recursive iterative version that I shamelessly stole from #John :
The rest of the program is pretty much the same, only showing the cart_product function.
// Seems like you'd want a vector of iterators
// which iterate over your individual vector<int>s.
struct Digits {
Vi::const_iterator begin;
Vi::const_iterator end;
Vi::const_iterator me;
};
typedef std::vector<Digits> Vd;
void cart_product(
Vvi& out, // final result
Vvi& in) // final result
{
Vd vd;
// Start all of the iterators at the beginning.
for(Vvi::const_iterator it = in.begin();
it != in.end();
++it) {
Digits d = {(*it).begin(), (*it).end(), (*it).begin()};
vd.push_back(d);
}
while(1) {
// Construct your first product vector by pulling
// out the element of each vector via the iterator.
Vi result;
for(Vd::const_iterator it = vd.begin();
it != vd.end();
it++) {
result.push_back(*(it->me));
}
out.push_back(result);
// Increment the rightmost one, and repeat.
// When you reach the end, reset that one to the beginning and
// increment the next-to-last one. You can get the "next-to-last"
// iterator by pulling it out of the neighboring element in your
// vector of iterators.
for(Vd::iterator it = vd.begin(); ; ) {
// okay, I started at the left instead. sue me
++(it->me);
if(it->me == it->end) {
if(it+1 == vd.end()) {
// I'm the last digit, and I'm about to roll
return;
} else {
// cascade
it->me = it->begin;
++it;
}
} else {
// normal
break;
}
}
}
}
Here is a solution in C++11.
The indexing of the variable-sized arrays can be done eloquently with modular arithmetic.
The total number of lines in the output is the product of the sizes of the input vectors. That is:
N = v[0].size() * v[1].size() * v[2].size()
Therefore the main loop has n as the iteration variable, from 0 to N-1. In principle, each value of n encodes enough information to extract each of the indices of v for that iteration. This is done in a subloop using repeated modular arithmetic:
#include <cstdlib>
#include <iostream>
#include <numeric>
#include <vector>
using namespace std;
void cartesian( vector<vector<int> >& v ) {
auto product = []( long long a, vector<int>& b ) { return a*b.size(); };
const long long N = accumulate( v.begin(), v.end(), 1LL, product );
vector<int> u(v.size());
for( long long n=0 ; n<N ; ++n ) {
lldiv_t q { n, 0 };
for( long long i=v.size()-1 ; 0<=i ; --i ) {
q = div( q.quot, v[i].size() );
u[i] = v[i][q.rem];
}
// Do what you want here with u.
for( int x : u ) cout << x << ' ';
cout << '\n';
}
}
int main() {
vector<vector<int> > v { { 1, 2, 3 },
{ 4, 5 },
{ 6, 7, 8 } };
cartesian(v);
return 0;
}
Output:
1 4 6
1 4 7
1 4 8
...
3 5 8
Shorter code:
vector<vector<int>> cart_product (const vector<vector<int>>& v) {
vector<vector<int>> s = {{}};
for (const auto& u : v) {
vector<vector<int>> r;
for (const auto& x : s) {
for (const auto y : u) {
r.push_back(x);
r.back().push_back(y);
}
}
s = move(r);
}
return s;
}
Seems like you'd want a vector of iterators which iterate over your individual vector<int>s.
Start all of the iterators at the beginning. Construct your first product vector by pulling out the element of each vector via the iterator.
Increment the rightmost one, and repeat.
When you reach the end, reset that one to the beginning and increment the next-to-last one. You can get the "next-to-last" iterator by pulling it out of the neighboring element in your vector of iterators.
Continue cycling through until both the last and next-to-last iterators are at the end. Then, reset them both, increment the third-from-last iterator. In general, this can be cascaded.
It's like an odometer, but with each different digit being in a different base.
Here's my solution. Also iterative, but a little shorter than the above...
void xp(const vector<vector<int>*>& vecs, vector<vector<int>*> *result) {
vector<vector<int>*>* rslts;
for (int ii = 0; ii < vecs.size(); ++ii) {
const vector<int>& vec = *vecs[ii];
if (ii == 0) {
// vecs=[[1,2],...] ==> rslts=[[1],[2]]
rslts = new vector<vector<int>*>;
for (int jj = 0; jj < vec.size(); ++jj) {
vector<int>* v = new vector<int>;
v->push_back(vec[jj]);
rslts->push_back(v);
}
} else {
// vecs=[[1,2],[3,4],...] ==> rslts=[[1,3],[1,4],[2,3],[2,4]]
vector<vector<int>*>* tmp = new vector<vector<int>*>;
for (int jj = 0; jj < vec.size(); ++jj) { // vec[jj]=3 (first iter jj=0)
for (vector<vector<int>*>::const_iterator it = rslts->begin();
it != rslts->end(); ++it) {
vector<int>* v = new vector<int>(**it); // v=[1]
v->push_back(vec[jj]); // v=[1,3]
tmp->push_back(v); // tmp=[[1,3]]
}
}
for (int kk = 0; kk < rslts->size(); ++kk) {
delete (*rslts)[kk];
}
delete rslts;
rslts = tmp;
}
}
result->insert(result->end(), rslts->begin(), rslts->end());
delete rslts;
}
I derived it with some pain from a haskell version I wrote:
xp :: [[a]] -> [[a]]
xp [] = []
xp [l] = map (:[]) l
xp (h:t) = foldr (\x acc -> foldr (\l acc -> (x:l):acc) acc (xp t)) [] h
Since I needed the same functionality, I implemented an iterator which computes the Cartesian product on the fly, as needed, and iterates over it.
It can be used as follows.
#include <forward_list>
#include <iostream>
#include <vector>
#include "cartesian.hpp"
int main()
{
// Works with a vector of vectors
std::vector<std::vector<int>> test{{1,2,3}, {4,5,6}, {8,9}};
CartesianProduct<decltype(test)> cp(test);
for(auto const& val: cp) {
std::cout << val.at(0) << ", " << val.at(1) << ", " << val.at(2) << "\n";
}
// Also works with something much less, like a forward_list of forward_lists
std::forward_list<std::forward_list<std::string>> foo{{"boo", "far", "zab"}, {"zoo", "moo"}, {"yohoo", "bohoo", "whoot", "noo"}};
CartesianProduct<decltype(foo)> bar(foo);
for(auto const& val: bar) {
std::cout << val.at(0) << ", " << val.at(1) << ", " << val.at(2) << "\n";
}
}
The file cartesian.hpp looks like this.
#include <cassert>
#include <limits>
#include <stdexcept>
#include <vector>
#include <boost/iterator/iterator_facade.hpp>
//! Class iterating over the Cartesian product of a forward iterable container of forward iterable containers
template<typename T>
class CartesianProductIterator : public boost::iterator_facade<CartesianProductIterator<T>, std::vector<typename T::value_type::value_type> const, boost::forward_traversal_tag>
{
public:
//! Delete default constructor
CartesianProductIterator() = delete;
//! Constructor setting the underlying iterator and position
/*!
* \param[in] structure The underlying structure
* \param[in] pos The position the iterator should be initialized to. std::numeric_limits<std::size_t>::max()stands for the end, the position after the last element.
*/
explicit CartesianProductIterator(T const& structure, std::size_t pos);
private:
//! Give types more descriptive names
// \{
typedef T OuterContainer;
typedef typename T::value_type Container;
typedef typename T::value_type::value_type Content;
// \}
//! Grant access to boost::iterator_facade
friend class boost::iterator_core_access;
//! Increment iterator
void increment();
//! Check for equality
bool equal(CartesianProductIterator<T> const& other) const;
//! Dereference iterator
std::vector<Content> const& dereference() const;
//! The part we are iterating over
OuterContainer const& structure_;
//! The position in the Cartesian product
/*!
* For each element of structure_, give the position in it.
* The empty vector represents the end position.
* Note that this vector has a size equal to structure->size(), or is empty.
*/
std::vector<typename Container::const_iterator> position_;
//! The position just indexed by an integer
std::size_t absolutePosition_ = 0;
//! The begin iterators, saved for convenience and performance
std::vector<typename Container::const_iterator> cbegins_;
//! The end iterators, saved for convenience and performance
std::vector<typename Container::const_iterator> cends_;
//! Used for returning references
/*!
* We initialize with one empty element, so that we only need to add more elements in increment().
*/
mutable std::vector<std::vector<Content>> result_{std::vector<Content>()};
//! The size of the instance of OuterContainer
std::size_t size_ = 0;
};
template<typename T>
CartesianProductIterator<T>::CartesianProductIterator(OuterContainer const& structure, std::size_t pos) : structure_(structure)
{
for(auto & entry: structure_) {
cbegins_.push_back(entry.cbegin());
cends_.push_back(entry.cend());
++size_;
}
if(pos == std::numeric_limits<std::size_t>::max() || size_ == 0) {
absolutePosition_ = std::numeric_limits<std::size_t>::max();
return;
}
// Initialize with all cbegin() position
position_.reserve(size_);
for(std::size_t i = 0; i != size_; ++i) {
position_.push_back(cbegins_[i]);
if(cbegins_[i] == cends_[i]) {
// Empty member, so Cartesian product is empty
absolutePosition_ = std::numeric_limits<std::size_t>::max();
return;
}
}
// Increment to wanted position
for(std::size_t i = 0; i < pos; ++i) {
increment();
}
}
template<typename T>
void CartesianProductIterator<T>::increment()
{
if(absolutePosition_ == std::numeric_limits<std::size_t>::max()) {
return;
}
std::size_t pos = size_ - 1;
// Descend as far as necessary
while(++(position_[pos]) == cends_[pos] && pos != 0) {
--pos;
}
if(position_[pos] == cends_[pos]) {
assert(pos == 0);
absolutePosition_ = std::numeric_limits<std::size_t>::max();
return;
}
// Set all to begin behind pos
for(++pos; pos != size_; ++pos) {
position_[pos] = cbegins_[pos];
}
++absolutePosition_;
result_.emplace_back();
}
template<typename T>
std::vector<typename T::value_type::value_type> const& CartesianProductIterator<T>::dereference() const
{
if(absolutePosition_ == std::numeric_limits<std::size_t>::max()) {
throw new std::out_of_range("Out of bound dereference in CartesianProductIterator\n");
}
auto & result = result_[absolutePosition_];
if(result.empty()) {
result.reserve(size_);
for(auto & iterator: position_) {
result.push_back(*iterator);
}
}
return result;
}
template<typename T>
bool CartesianProductIterator<T>::equal(CartesianProductIterator<T> const& other) const
{
return absolutePosition_ == other.absolutePosition_ && structure_ == other.structure_;
}
//! Class that turns a forward iterable container of forward iterable containers into a forward iterable container which iterates over the Cartesian product of the forward iterable containers
template<typename T>
class CartesianProduct
{
public:
//! Constructor from type T
explicit CartesianProduct(T const& t) : t_(t) {}
//! Iterator to beginning of Cartesian product
CartesianProductIterator<T> begin() const { return CartesianProductIterator<T>(t_, 0); }
//! Iterator behind the last element of the Cartesian product
CartesianProductIterator<T> end() const { return CartesianProductIterator<T>(t_, std::numeric_limits<std::size_t>::max()); }
private:
T const& t_;
};
If someone has comments how to make it faster or better, I'd highly appreciate them.
I was just forced to implement this for a project I was working on and I came up with the code below. It can be stuck in a header and it's use is very simple but it returns all of the combinations you can get from a vector of vectors. The array that it returns only holds integers. This was a conscious decision because I just wanted the indices. In this way, I could index into each of the vector's vector and then perform the calculations I/anyone would need... best to avoid letting CartesianProduct hold "stuff" itself, it is a mathematical concept based around counting not a data structure. I'm fairly new to c++ but this was tested in a decryption algorithm pretty thoroughly. There is some light recursion but overall this is a simple implementation of a simple counting concept.
// Use of the CartesianProduct class is as follows. Give it the number
// of rows and the sizes of each of the rows. It will output all of the
// permutations of these numbers in their respective rows.
// 1. call cp.permutation() // need to check all 0s.
// 2. while cp.HasNext() // it knows the exit condition form its inputs.
// 3. cp.Increment() // Make the next permutation
// 4. cp.permutation() // get the next permutation
class CartesianProduct{
public:
CartesianProduct(int num_rows, vector<int> sizes_of_rows){
permutation_ = new int[num_rows];
num_rows_ = num_rows;
ZeroOutPermutation();
sizes_of_rows_ = sizes_of_rows;
num_max_permutations_ = 1;
for (int i = 0; i < num_rows; ++i){
num_max_permutations_ *= sizes_of_rows_[i];
}
}
~CartesianProduct(){
delete permutation_;
}
bool HasNext(){
if(num_permutations_processed_ != num_max_permutations_) {
return true;
} else {
return false;
}
}
void Increment(){
int row_to_increment = 0;
++num_permutations_processed_;
IncrementAndTest(row_to_increment);
}
int* permutation(){
return permutation_;
}
int num_permutations_processed(){
return num_permutations_processed_;
}
void PrintPermutation(){
cout << "( ";
for (int i = 0; i < num_rows_; ++i){
cout << permutation_[i] << ", ";
}
cout << " )" << endl;
}
private:
int num_permutations_processed_;
int *permutation_;
int num_rows_;
int num_max_permutations_;
vector<int> sizes_of_rows_;
// Because CartesianProduct is called first initially with it's values
// of 0 and because those values are valid and important output
// of the CartesianProduct we increment the number of permutations
// processed here when we populate the permutation_ array with 0s.
void ZeroOutPermutation(){
for (int i = 0; i < num_rows_; ++i){
permutation_[i] = 0;
}
num_permutations_processed_ = 1;
}
void IncrementAndTest(int row_to_increment){
permutation_[row_to_increment] += 1;
int max_index_of_row = sizes_of_rows_[row_to_increment] - 1;
if (permutation_[row_to_increment] > max_index_of_row){
permutation_[row_to_increment] = 0;
IncrementAndTest(row_to_increment + 1);
}
}
};
#include <iostream>
#include <vector>
void cartesian (std::vector<std::vector<int>> const& items) {
auto n = items.size();
auto next = [&](std::vector<int> & x) {
for ( int i = 0; i < n; ++ i )
if ( ++x[i] == items[i].size() ) x[i] = 0;
else return true;
return false;
};
auto print = [&](std::vector<int> const& x) {
for ( int i = 0; i < n; ++ i )
std::cout << items[i][x[i]] << ",";
std::cout << "\b \n";
};
std::vector<int> x(n);
do print(x); while (next(x)); // Shazam!
}
int main () {
std::vector<std::vector<int>>
items { { 1, 2, 3 }, { 4, 5 }, { 6, 7, 8 } };
cartesian(items);
return 0;
}
The idea behind this is as follows.
Let n := items.size().
Let m_i := items[i].size(), for all i in {0,1,...,n-1}.
Let M := {0,1,...,m_0-1} x {0,1,...,m_1-1} x ... x {0,1,...,m_{n-1}-1}.
We first solve the simpler problem of iterating through M. This is accomplished by the next lambda. The algorithm is simply the "carrying" routine grade schoolers use to add 1, albeit with a mixed radix number system.
We use this to solve the more general problem by transforming a tuple x in M to one of the desired tuples via the formula items[i][x[i]] for all i in {0,1,...,n-1}. We perform this transformation in the print lambda.
We then perform the iteration with do print(x); while (next(x));.
Now some comments on complexity, under the assumption that m_i > 1 for all i:
This algorithm requires O(n) space. Note that explicit construction of the Cartesian product takes O(m_0 m_1 m_2 ... m_{n-1}) >= O(2^n) space. So this is exponentially better on space than any algorithm which requires all tuples to be stored simultaneously in memory.
The next function takes amortized O(1) time (by a geometric series argument).
The print function takes O(n) time.
Hence, altogether, the algorithm has time complexity O(n|M|) and space complexity O(n) (not counting the cost of storing items).
An interesting thing to note is that if print is replaced with a function which inspects on average only O(1) coordinates per tuple rather than all of them, then time complexity falls to O(|M|), that is, it becomes linear time with respect to the size of the Cartesian product. In other words, avoiding the copy of the tuple each iterate can be meaningful in some situations.
This version supports no iterators or ranges, but it is a simple direct implementation that uses the multiplication operator to represent the Cartesian product, and a lambda to perform the action.
The interface is designed with the particular functionality I needed. I needed the flexibility to choose vectors over which to apply the Cartesian product in a way that did not obscure the code.
int main()
{
vector< vector<long> > v{ { 1, 2, 3 }, { 4, 5 }, { 6, 7, 8 } };
(Cartesian<long>(v[0]) * v[1] * v[2]).ForEach(
[](long p_Depth, long *p_LongList)
{
std::cout << p_LongList[0] << " " << p_LongList[1] << " " << p_LongList[2] << std::endl;
}
);
}
The implementation uses recursion up the class structure to implement the embedded for loops over each vector. The algorithm works directly on the input vectors, requiring no large temporary arrays. It is simple to understand and debug.
The use of std::function p_Action instead of void p_Action(long p_Depth, T *p_ParamList) for the lambda parameter would allow me to capture local variables, if I wanted to. In the above call, I don't.
But you knew that, didn't you. "function" is a template class which takes the type parameter of a function and makes it callable.
#include <vector>
#include <iostream>
#include <functional>
#include <string>
using namespace std;
template <class T>
class Cartesian
{
private:
vector<T> &m_Vector;
Cartesian<T> *m_Cartesian;
public:
Cartesian(vector<T> &p_Vector, Cartesian<T> *p_Cartesian=NULL)
: m_Vector(p_Vector), m_Cartesian(p_Cartesian)
{};
virtual ~Cartesian() {};
Cartesian<T> *Clone()
{
return new Cartesian<T>(m_Vector, m_Cartesian ? m_Cartesian->Clone() : NULL);
};
Cartesian<T> &operator *=(vector<T> &p_Vector)
{
if (m_Cartesian)
(*m_Cartesian) *= p_Vector;
else
m_Cartesian = new Cartesian(p_Vector);
return *this;
};
Cartesian<T> operator *(vector<T> &p_Vector)
{
return (*Clone()) *= p_Vector;
};
long Depth()
{
return m_Cartesian ? 1 + m_Cartesian->Depth() : 1;
};
void ForEach(function<void (long p_Depth, T *p_ParamList)> p_Action)
{
Loop(0, new T[Depth()], p_Action);
};
private:
void Loop(long p_Depth, T *p_ParamList, function<void (long p_Depth, T *p_ParamList)> p_Action)
{
for (T &element : m_Vector)
{
p_ParamList[p_Depth] = element;
if (m_Cartesian)
m_Cartesian->Loop(p_Depth + 1, p_ParamList, p_Action);
else
p_Action(Depth(), p_ParamList);
}
};
};
How to create iterator/s for 2d vector (a vector of vectors)?
Although your question is not very clear, I'm going to assume you mean a 2D vector to mean a vector of vectors:
vector< vector<int> > vvi;
Then you need to use two iterators to traverse it, the first the iterator of the "rows", the second the iterators of the "columns" in that "row":
//assuming you have a "2D" vector vvi (vector of vector of int's)
vector< vector<int> >::iterator row;
vector<int>::iterator col;
for (row = vvi.begin(); row != vvi.end(); row++) {
for (col = row->begin(); col != row->end(); col++) {
// do stuff ...
}
}
You can use range for statement to iterate all the elements in a two-dimensional vector.
vector< vector<int> > vec;
And let's presume you have already push_back a lot of elements into vec;
for(auto& row:vec){
for(auto& col:row){
//do something using the element col
}
}
Another way to interpret this question is that you want a 1D iterator over a vector<vector<>> for example to feed it to for_each() or some other algorithm.
You can do that like this:
#include <iostream>
#include <iterator>
#include <vector>
#include <algorithm>
// An iterator over a vector of vectors.
template<typename T>
class vv_iterator : public std::iterator<std::bidirectional_iterator_tag, T>{
public:
static vv_iterator<T> begin(std::vector<std::vector<T>>& vv) {
return vv_iterator(&vv, 0, 0);
}
static vv_iterator<T> end(std::vector<std::vector<T>>& vv) {
return vv_iterator(&vv, vv.size(), 0);
}
vv_iterator() = default;
// ++prefix operator
vv_iterator& operator++()
{
// If we haven't reached the end of this sub-vector.
if (idxInner + 1 < (*vv)[idxOuter].size())
{
// Go to the next element.
++idxInner;
}
else
{
// Otherwise skip to the next sub-vector, and keep skipping over empty
// ones until we reach a non-empty one or the end.
do
{
++idxOuter;
} while (idxOuter < (*vv).size() && (*vv)[idxOuter].empty());
// Go to the start of this vector.
idxInner = 0;
}
return *this;
}
// --prefix operator
vv_iterator& operator--()
{
// If we haven't reached the start of this sub-vector.
if (idxInner > 0)
{
// Go to the previous element.
--idxInner;
}
else
{
// Otherwise skip to the previous sub-vector, and keep skipping over empty
// ones until we reach a non-empty one.
do
{
--idxOuter;
} while ((*vv)[idxOuter].empty());
// Go to the end of this vector.
idxInner = (*vv)[idxOuter].size() - 1;
}
return *this;
}
// postfix++ operator
vv_iterator operator++(int)
{
T retval = *this;
++(*this);
return retval;
}
// postfix-- operator
vv_iterator operator--(int)
{
T retval = *this;
--(*this);
return retval;
}
bool operator==(const vv_iterator& other) const
{
return other.vv == vv && other.idxOuter == idxOuter && other.idxInner == idxInner;
}
bool operator!=(const vv_iterator &other) const
{
return !(*this == other);
}
const T& operator*() const
{
return *this;
}
T& operator*()
{
return (*vv)[idxOuter][idxInner];
}
const T& operator->() const
{
return *this;
}
T& operator->()
{
return *this;
}
private:
vv_iterator(std::vector<std::vector<T>>* _vv,
std::size_t _idxOuter,
std::size_t _idxInner)
: vv(_vv), idxOuter(_idxOuter), idxInner(_idxInner) {}
std::vector<std::vector<int>>* vv = nullptr;
std::size_t idxOuter = 0;
std::size_t idxInner = 0;
};
int main()
{
std::vector<std::vector<int>> a = {{3, 5, 2, 6}, {-1, -4, -3, -5}, {100}, {-100}};
std::reverse(vv_iterator<int>::begin(a), vv_iterator<int>::end(a));
for (const auto& v : a)
{
std::cout << "{ ";
for (auto i : v)
std::cout << i << " ";
std::cout << "}\n";
}
}
Prints:
{ -100 100 -5 -3 }
{ -4 -1 6 2 }
{ 5 }
{ 3 }
Note this won't work with std::sort() because that requires a random access iterator. You could make it a random access iterator but you'd have to scan the vector at the start so you can map from flat index to idxOuter and idxInner in constant time. Not totally trivial but not hard either.
Suppose you have a vector like this:-
vector <vector<int>> vect{{1,2,3},{4,5,6},{7,8,9}};
Now to use iterators with 2D vectors :-
for(auto i = vect.begin() ; i<vect.end() ; i++)
{
for(auto j = i->begin() ; j<i->end() ; j++)
cout << *j <<" ";
cout <<"\n";
//similarly you can do other things
}
Also other shorter way is
for(auto i : vect)
{
for(auto j : i)
cout << j <<" ";
cout << "\n";
//similarly you can do other things also.
}
Please note the way of calling variables is different in both the cases.
You can use auto keyword for such cases:
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
int main() {
// your code goes here
vector<vector<int>>v;
for(int i=0;i<5;i++)
{
vector<int> x={1,2,3,4,5};
v.push_back(x);
}
cout<<"-------------------------------------------"<<endl;
cout<<"Print without iterator"<<endl;
cout<<"-------------------------------------------"<<endl;
for(int i=0;i<5;i++)
{
vector<int> y=v[i];
for(int j=0;j<y.size();j++)
{
cout<<y[j]<<" ";
}
cout<<endl;
}
cout<<"-------------------------------------------"<<endl;
cout<<"Print with iterator"<<endl;
cout<<"-------------------------------------------"<<endl;
for(auto iterator=v.begin();iterator!=v.end();iterator++)
{
vector<int> y=*iterator;
for(auto itr=y.begin();itr!=y.end();itr++)
{
cout<<*itr<<" ";
}
cout<<endl;
}
return 0;
}
Since it's 2020, I'll post an updated and easy method. Works for c++11 and above as of writing.
See the following example, where elements (here: tuples of <string, size_t>) of 2D vector (vector of vector) is iterated to compare with another value (string query) and the function then returns first element with match, or indicate "Not found".
tuple<string, size_t> find_serial_data( vector <vector <tuple <string, size_t>>> &serial,
string query)
{
for (auto& i : serial)
{
for (auto& j : i)
{
if ( get<0>(j).compare(query) == 0) return j;
}
}
cout << "\n Not found";
return make_tuple( "", 0);
}
Here is one example without the tuple thing:
string find_serial_data( vector <vector <string> > &serials,
string query)
{
for (auto& i : serials)
{
for (auto& j : i)
{
if ( j.compare(query) == 0) return j;
}
}
cout << "\n Not found";
return "";
}
Assuming you mean an STL iterator, and a custom container that implements a generic 2D array of objects, this is impossible. STL iterators support only increment and decrement (i.e. "next" an "previous") operations, where motion through a 2D set requires four such primitives (e.g. left/right/up/down, etc...). The metaphors don't match.
What are you trying to do?
Assuming you mean a vector of vectors, and you have std::vector in mind, there's no built in way to do it, as iterators only support increment and decrement operations to move forward and backwards.
A 2D vector is a matrix, and so you'd need two iterator types: a row iterator and a column iterator. Row iterators would move "up" and "down" the matrix, whereas column iterators would move "left" and "right".
You have to implement these iterator classes yourself, which is not necessarily a trivial thing to do. Unless, of course, you simply want to iterate over each slot in the matrix, in which case a double for loop using index variables i and j will work just fine. Depending on your needs (your post is a bit lacking in content here), you may want to use boost::numeric::ublas::matrix, which is a matrix class from the Boost linear algebra library. This matrix class has built-in row and column iterators, which make it generally easy to iterate over a matrix.