remove copy struct member from std::vector - c++

I have a vector of struct member like below:
struct pnt
{
bool has;
int num;
};
std::vector<pnt> myvector;
let have a sample vector like:
myvector (num): 3 4 4 3 5 5 7 8 9 10 10
myvector (has): 1 1 0 1 0 1 0 0 0 1 0
What I want to do is to find duplicated members (in terms of having same int num) and remove the one with false bool member.
so that my vector become like this:
myvector (num): 3 4 3 5 7 8 9 10
myvector (has): 1 1 1 1 0 0 0 1
to do so, I write following function:
void removeDuplicatedPnt(pnt_vec& myvector)
{
std::vector<pnt>::iterator pnt_iter;
for( pnt_iter = myvector.begin(); pnt_iter != myvector.end(); ++pnt_iter)
{
if(pnt_iter->has)
{
if(pnt_iter->num == (pnt_iter+1)->num)
{
myvector.erase(pnt_iter+1);
}
if(pnt_iter == myvector.begin())
{
continue;
}
if(pnt_iter->num == (pnt_iter-1)->num)
{
myvector.erase(pnt_iter-1);
pnt_iter++;
}
}
}
}
I could also do it by sequential checking of members. but the real vector could be very long. so that is why first I went to find the member with true boolean then I checked the the next and previous member. The question is that how I can modify above code in terms of efficiency and robustness.
NOTE: I can only use C++03 (not C++11). I can also use boos (version 1.53), so feel free if think there is any useful function there. :)

You can use this algorithm:
Collect all nums where has is true in a set<int>
Go through your vector<pnt> again, and remove all entries where has is false and num is present in the set<int>
Here is a sample implementation:
struct filter {
set<int> seen;
bool operator()(const pnt& p) {
return !p.has && (seen.find(p.num) != seen.end());
}
};
...
filter f;
for (vector<pnt>::const_iterator i = v.begin() ; i != v.end() ; i++) {
if (i->has) {
f.seen.insert(i->num);
}
}
v.erase(remove_if(v.begin(), v.end(), f), v.end());
Demo.

You can use std::sort and std::unique with a custom comparison predicate:
[](const pnt& a, const pnt& b) { return a.num < b.num; }
Here's a convenient way to demo it using Boost Range to reduce on typing:
Update C++03 version Live On Coliru
#include <boost/range.hpp>
#include <boost/range/algorithm.hpp>
#include <iostream>
using namespace boost;
struct pnt {
int num;
bool has;
pnt(int num = 0, bool has = false) : num(num), has(has) {}
friend bool operator<(pnt const& a, pnt const& b) { return a.num<b.num; }
friend bool operator==(pnt const& a, pnt const& b) { return a.num==b.num; }
};
int main() {
std::vector<pnt> v { {10,0 },{10,1 },{9,0 },{8,0 },{7,0 },{5,1 },{5,0 },{3,1 },{4,0 },{4,1 },{3,1 } };
for (pnt p : boost::unique(boost::sort(v)))
std::cout << "{ num:" << p.num << ", has:" << p.has << "}\n";
}
This actually has a few subtler points that make it possible to e.g. do
it = std::find(v.begin(), v.end(), 3); // find a `pnt` with `num==3`
But that's only tangentially related

Related

How to count the equal elements of 2 different vectors in c++?

I am trying to find a way to count how many elements are equal in 2 different vectors of the same size in c++. The vectors hold structs and i want to compare the equality by a double variable of the struct shown on the example.
And to make it clear. I do NOT want to check if the 2 vectors are equal but only to count how many of their elements are.
The following doesn't work. It gives addresses instead of values. Also If I try to access the dist variable like pointsA[j].dist I get an error.
vector<struct PointWithDistance*> pointsA, pointsB;
//the struct
struct PointWithDistance {
Point *p;
double dist;
};
for (int j = 0; j < k; j++){
if (pointsA[j] == pointsB[j])
equalCount++;
}
vector<struct PointWithDistance*> pointsA, pointsB;
Did you mean to use pointers? If so, you have to do *(points[A]) (and b) because your current comparison compares the pointers, not their content.
Also, does the struct Point have an operator == so comparison on the type can be performed??
Do you want to force same positions? say, a vector {1,2,3} and a vector {2,3,4} by your algorithm will have 0 items equal, do you want that? If not, loop the first vector and use std::find (or std::upper_bound if the vector is sorted) on each element to the second vector.
Some quick code:
template <typename T=int> struct Point
{
T x,y;
bool operator==(const T& t) { return (x == t.x && y == t.y); }
};
std::vector<Point<>> p1 = {1,2,3};
std::vector<Point<>> p2 = {2,3,4};
for(auto& p : p1)
{
if (std::find(p2.begin(),p2.end(),p) != p2.end())
{
// similar++;
}
}
// or
assert(p1.size() == p2.size());
for(size_t i1 = 0 ; i1 < p1.size() ; i1++)
{
if (p1[i1] == p2[i1])
{
// equal++;
}
}
A generic solution to count the number of duplicates in 2 containers could look like this. Using std::transform_reduce to add (std::plus<>{}) the boolean result if an element was found in the container. Note how it can accep two different types of containers as long as their contained type stays the same (e.g. std::vector<int> and std::set<int>). The length of the containers don't have to be equal. There are two SFINAE implementations to differenciate between the case where T is a pointer and where it isn't:
#include <algorithm> //std::find, std::find_if
#include <cstddef> //std::size_t
#include <functional> //std::plus,
#include <iterator> //std::cbegin, std::cend
#include <numeric> //std::transform_reduce
#include <type_traits> //std::enable_if_t, std::is_pointer_v
namespace {
//core implementation for duplicate_count
template<class C, class F>
std::size_t duplicate_count_impl(const C& container, F pred) {
return std::transform_reduce(std::cbegin(container), std::cend(container), std::size_t{}, std::plus<>{}, pred);
}
}
//returns the number of duplicates in two (different) containers.
//overload for containers where T is a pointer type.
template<typename T, template <typename...> class C1, template <typename...> class C2, std::enable_if_t<std::is_pointer_v<T>>* = nullptr>
std::size_t duplicate_count(const C1<T>& a, const C2<T> &b) {
return duplicate_count_impl(b, [&](T ptr_b) -> bool {
return std::find_if(std::cbegin(a), std::cend(a), [&](T ptr_a) -> bool {
return *ptr_a == *ptr_b;
}) != std::cend(a);
});
}
//returns the number of duplicates in two (different) containers.
//overload for containers where T is not a pointer type.
template<typename T, template <typename...> class C1, template <typename...> class C2, std::enable_if_t<!std::is_pointer_v<T>>* = nullptr>
std::size_t duplicate_count(const C1<T>& a, const C2<T> &b) {
return duplicate_count_impl(b, [&](T n) -> bool {
return std::find(std::cbegin(a), std::cend(a), n) != std::cend(a);
});
}
#include <iostream>
#include <vector>
#include <list>
//[duplicate_count implementations]
struct Point {
int a, b;
bool operator==(const Point& other) const {
return this->a == a && this->b == other.b;
}
};
int main() {
{
std::list<int> v = { 1, 2, 7, 7 };
std::list<int> u = { 0, 1, 2, 7 };
std::cout << "list<int>\t number of duplicates: " << duplicate_count(v, u) << '\n';
}
{
auto[a, b, c, d] = std::make_tuple(0, 1, 2, 3);
std::vector<int*> v = { &b, &c, &d, &d };
std::vector<int*> u = { &a, &b, &c, &d };
std::cout << "vector<int*>\t number of duplicates: " << duplicate_count(v, u) << '\n';
}
{
auto[a, b, c, d] = std::make_tuple(
Point{ 0, 0 },
Point{ 1, 1 },
Point{ 2, 2 },
Point{ 4, 4 });
std::vector<Point*> v = { &b, &c, &d, &d };
std::vector<Point*> u = { &a, &b, &c, &d };
std::cout << "vector<Point*>\t number of duplicates: " << duplicate_count(v, u) << '\n';
}
}
list<int> number of duplicates: 3
vector<int*> number of duplicates: 3
vector<Point*> number of duplicates: 3
Your shown solution is good, fast and efficient.
It has some minor problem that can easily be resolved. In your definition vector<struct PointWithDistance*> pointsA, pointsB;, variables pointsA and pointsB are vectors, containing pointer to structs.
With pointsA[n] you will get a pointer to the struct. But you want the struct by itself. So you simply need to dereference the gotten pointer. And since you want to access a member of a struct (usually done with variable.member), you can use (*(pointsA[j])).dist or pointsA[j]->dist.
If the size of your vectors are guaranteed the same, then you simply need to update your code to
vector<struct PointWithDistance*> pointsA, pointsB;
//the struct
struct PointWithDistance {
Point *p;
double dist;
};
for (int j = 0; j < k; j++){
if (pointsA[j]->dist == pointsB[j]->dist)
equalCount++;
}
That is the only thing you were missing.
You can use the std::inner_product algorithm:
#include <iostream>
#include <numeric>
#include <vector>
int main () {
const std::vector a{7, 7, 7, 7};
const std::vector b{7, 6, 7, 7};
const auto equalCount = std::inner_product(
a.begin(), a.end(), b.begin(), 0,
std::plus<>(), std::equal_to<>()
);
std::cout << equalCount << " of the elements are equal.\n";
}
outputs
3 of the elements are equal.
It is a generalization of the standard inner product,
using the functions + (plus) and == (equal_to),
instead of + and *.
Thus it computes
0 + (a[0] == b[0]) + (a[1] == b[1]) + ....
This uses the fact that false/true can be interpreted as 0/1.

sort indexes from a list with duplicates [duplicate]

This question already has answers here:
Why will std::sort crash if the comparison function is not as operator <?
(3 answers)
Closed 4 years ago.
I'm trying to obtain the sorted position of a string in a list
that may contain duplicates.
I don't care about an undefined order for duplicates but I would like a global sort.
Here is my best attempt so far:
#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
void display(const std::vector<int> &array)
{
for (const auto & value : array)
std::cout << value << " ";
std::cout << std::endl;
}
std::vector<int> sortIndexes(const std::vector<std::string> &values)
{
std::vector<int> indexes(values.size());
std::iota(indexes.begin(), indexes.end(), 0);
std::stable_sort(indexes.begin(), indexes.end(), [&values](const size_t first, const size_t second)
{
return values.at(first) <= values.at(second);
});
return indexes;
}
int main (void)
{
display(sortIndexes({"b", "a", "c"})); // Expecting: 1 0 2 Result: 1 0 2
display(sortIndexes({"c", "c", "a"})); // Expecting: 1 2 0 or 2 1 0 Result: 2 1 0
display(sortIndexes({"c", "a", "c"})); // Expecting: 1 0 2 or 2 0 1 Result: 1 2 0
return 0;
}
Is there another way to get the expected output ?
EDIT:
I was missing the strict comparison + 'inverseIndexes' part to solve my problem. Here is the updated code:
#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
void display(const std::vector<int> & array)
{
for (const auto & value : array)
std::cout << value << " ";
std::cout << std::endl;
}
std::vector<int> inverseIndexes(const std::vector<int> & indexes)
{
std::vector<int> inverse(indexes.size());
for (size_t i = 0; i < indexes.size(); ++i)
inverse[indexes[i]] = i;
return inverse;
}
std::vector<int> sortIndexes(const std::vector<std::string> & values)
{
std::vector<int> indexes(values.size());
std::iota(indexes.begin(), indexes.end(), 0);
std::stable_sort(indexes.begin(), indexes.end(), [&values](const size_t first, const size_t second)
{
return values.at(first) < values.at(second);
});
return indexes;
}
int main (void)
{
display(inverseIndexes(sortIndexes({"b", "a", "c"})));
display(inverseIndexes(sortIndexes({"c", "c", "a"})));
display(inverseIndexes(sortIndexes({"c", "a", "c"})));
return 0;
}
The comparison function for std::stable_sort should be a strict "less than", not "less or equal than". So, you only need to fix this line:
return values.at(first) < values.at(second);

Cartesian product from a vector in C++11? [duplicate]

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 can I skip the first iteration of range-based for loops?

I would like to do something like this:
for (int p : colourPos[i+1])
How do I skip the first iteration of my colourPos vector?
Can I use .begin() and .end()?
Since C++20 you can use the range adaptor std::views::drop from the Ranges library together with a range-based for loop for skipping the first element, as follows:
std::vector<int> colourPos { 1, 2, 3 };
for (int p : colourPos | std::views::drop(1)) {
std::cout << "Pos = " << p << std::endl;
}
Output:
Pos = 2
Pos = 3
Code on Wandbox
Note: I would not recommend using a solution that contains begin() and/or end(), because the idea of a range-based for loop is to get rid of iterators. If you need iterators, then I would stick with an interator-based for loop.
Live demo link.
#include <iostream>
#include <vector>
#include <iterator>
#include <cstddef>
template <typename T>
struct skip
{
T& t;
std::size_t n;
skip(T& v, std::size_t s) : t(v), n(s) {}
auto begin() -> decltype(std::begin(t))
{
return std::next(std::begin(t), n);
}
auto end() -> decltype(std::end(t))
{
return std::end(t);
}
};
int main()
{
std::vector<int> v{ 1, 2, 3, 4 };
for (auto p : skip<decltype(v)>(v, 1))
{
std::cout << p << " ";
}
}
Output:
2 3 4
Or simpler:
Yet another live demo link.
#include <iostream>
#include <vector>
template <typename T>
struct range_t
{
T b, e;
range_t(T x, T y) : b(x), e(y) {}
T begin()
{
return b;
}
T end()
{
return e;
}
};
template <typename T>
range_t<T> range(T b, T e)
{
return range_t<T>(b, e);
}
int main()
{
std::vector<int> v{ 1, 2, 3, 4 };
for (auto p : range(v.begin()+1, v.end()))
{
std::cout << p << " ";
}
}
Output:
2 3 4
Do this:
bool first = true;
for (int p : colourPos)
{
if (first)
{ first = false; continue; }
// ...
}

remove arbitrary list of items from std::vector<std::vector<T> >

I have a vector of vectors, representing an array. I would like to remove rows efficiently, ie with minimal complexity and allocations
I have thought about building a new vector of vectors, copying only non-deleted rows, using move semantics, like this:
//std::vector<std::vector<T> > values is the array to remove rows from
//std::vector<bool> toBeDeleted contains "marked for deletion" flags for each row
//Count the new number of remaining rows
unsigned int newNumRows = 0;
for(unsigned int i=0;i<numRows();i++)
{
if(!toBeDeleted[i])
{
newNumRows++;
}
}
//Create a new array already sized in rows
std::vector<std::vector<T> > newValues(newNumRows);
//Move rows
for(unsigned int i=0;i<numRows();i++)
{
if(!toBeDeleted[i])
{
newValues[i] = std::move(values[i]);
}
}
//Set the new array and clear the old one efficiently
values = std::move(newValues);
Is this the most effective way?
Edit : I just figured that I could avoid allocating a new array by moving rows down iteratively, this could be slightly more efficient and code is much more simple:
unsigned int newIndex = 0;
for(unsigned int oldIndex=0;oldIndex<values.size();oldIndex++)
{
if(!toBeDeleted[oldIndex])
{
if(oldIndex!=newIndex)
{
values[newIndex] = std::move(values[oldIndex]);
}
newIndex++;
}
}
values.resize(newIndex);
Thanks!
This can be solved using a variation on the usual erase-remove idiom, with a lambda inside the std::remove_if that looks up the index of the current row inside an iterator range of to be removed indices:
#include <algorithm> // find, remove_if
#include <iostream>
#include <vector>
template<class T>
using M = std::vector<std::vector<T>>; // matrix
template<class T>
std::ostream& operator<<(std::ostream& os, M<T> const& m)
{
for (auto const& row : m) {
for (auto const& elem : row)
os << elem << " ";
os << "\n";
}
return os;
}
template<class T, class IdxIt>
void erase_rows(M<T>& m, IdxIt first, IdxIt last)
{
m.erase(
std::remove_if(
begin(m), end(m), [&](auto& row) {
auto const row_idx = &row - &m[0];
return std::find(first, last, row_idx) != last;
}),
end(m)
);
}
int main()
{
auto m = M<int> { { 0, 1, 2, 3 }, { 3, 4, 5, 6 }, { 6, 7, 8, 9 }, { 1, 0, 1, 0 } };
std::cout << m << "\n";
auto drop = { 1, 3 };
erase_rows(m, begin(drop), end(drop));
std::cout << m << "\n";
}
Live Example.
Note: because from C++11 onwards, std::vector has move semantics, shuffling rows around in your std::vector<std::vector<T>> is done using simple pointer manipulations, regardless of your type T (it would be quite different if you want column-deletion, though!).