I have 3-column integer arrays, whose last 2 elements are for sorting. For example
10 0 1
11 0 2
12 1 2
13 0 1
I want them to become:
10 0 1
13 0 1
11 0 2
12 1 2
The arrays are first sorted according to the 2nd column, and then again according to 3rd column.
I have over 3000 rows, so I need something also fast. How can you do this in c++?
Note: The array will be allocated dynamically using the following templates:
template <typename T>
T **AllocateDynamic2DArray(int nRows, int nCols){
T **dynamicArray;
dynamicArray = new T*[nRows];
for( int i = 0 ; i < nRows ; i++ ){
dynamicArray[i] = new T[nCols];
for ( int j=0; j<nCols;j++){
dynamicArray[i][j]= 0;
}
}
return dynamicArray;
}
in main,
int ** lineFilter = AllocateDynamic2DArray(2*numberOfLines,3);
you can use std::sort(); however, this is complicated by your array being 2D.
In general, std::sort() can't eat 2D arrays; you have to create a class to cast around the compiler warnings and complaints:
#include <iostream>
#include <algorithm>
int data[4][3] = {
{10,0,1},
{11,0,2},
{12,1,2},
{13,0,1}
};
struct row_t { // our type alias for sorting; we know this is compatible with the rows in data
int data[3];
bool operator<(const row_t& rhs) const {
return (data[1]<rhs.data[1]) || ((data[1]==rhs.data[1]) && (data[2]<rhs.data[2]));
}
};
int main() {
std::sort((row_t*)data,(row_t*)(data+4));
for(int i=0; i<4; i++)
std::cout << i << '=' << data[i][0] << ',' << data[i][1] << ',' << data[i][2] << ';' << std::endl;
return 0;
}
It becomes much easier if you use a std::vector to hold your items that really are of type row_t or such. Vectors are dynamically sized and sortable.
I think this should work:
template<typename T>
struct compareRows {
bool operator() (T * const & a, T * const & b) {
if (a[1] == b[1])
return a[2] < b[2];
else
return a[1] < b[1];
}
};
std::sort(dynamicArray, dynamicArray+nrows, compareRows<int>());
Use a functor to implement the comparison between the rows. The sort will take pointers to the beginning of each row and swap them according to the contents of the rows. The rows will stay in the same places in memory.
OK, the OP has a three-column integer arrays, which is not straightforward to sort, because you can't assign arrays.
One option is to have arrays of structs, where the struct contains one element for each column, write a custom compare routine and use std::sort.
Another option is to pretend we have such an array of structs and employ the evilness of reinterpret_cast, like below:
#include <algorithm>
#include <iostream>
struct elt_t
{
int e0;
int e1;
int e2;
};
int
compare (const elt_t &a, const elt_t &b)
{
if (a.e1 == b.e1)
return a.e2 < b.e2;
else
return a.e1 < b.e1;
}
int a [10][3] =
{
{ 10, 0, 1 },
{ 11, 0, 2 },
{ 12, 1, 2 },
{ 13, 0, 1 }
};
int
main ()
{
std::sort (reinterpret_cast<elt_t *>(&a[0]),
reinterpret_cast<elt_t *>(&a[4]), compare);
int i, j;
for (i = 0; i < 4; ++i)
std::cout << a [i][0] << ", " << a [i][1] << ", " << a [i][2] << std::endl;
return 0;
}
Of course, whether or not this is standards compliant is highly debatable :)
EDIT:
With the added requirement for the matrix to by dynamically allocated, you can use an array of std::vector, or a vector of std::vector:
#include <algorithm>
#include <iostream>
#include <vector>
int
compare (const std::vector<int> &a, const std::vector<int> &b)
{
if (a[1] == b[1])
return a[2] < b[2];
else
return a[1] < b[1];
}
std::vector<int> *
make_vec (unsigned int r, unsigned int c)
{
std::vector<int> *v = new std::vector<int> [r];
/* Don't care for column count for the purposes of the example. */
v [0].push_back (10); v [0].push_back (0); v [0].push_back (1);
v [1].push_back (11); v [1].push_back (0); v [1].push_back (2);
v [2].push_back (12); v [2].push_back (1); v [2].push_back (2);
v [3].push_back (13); v [3].push_back (0); v [3].push_back (1);
return v;
}
int
main ()
{
std::vector<int> *v = make_vec (4, 3);
std::sort (&v[0], &v[4], compare);
int i, j;
for (i = 0; i < 4; ++i)
std::cout << v[i][0] << ", " << v [i][1] << ", " << v [i][2] << std::endl;
delete [] v;
return 0;
}
use this for the second column and then for the third. Now it works for single dim arrays
int *toplace(int *start, int *end)
{
int *i = start+1, *j= end-1;
while(i<=j)
{
while(*i<=*start && i<=j) {i++;}
while(*j>=*start && i<=j) {j--;}
if (i<j) std::swap(*i++,*j--);
}
std::swap(*start,*(i-1));
return i-1;
}
void quicksort(int *start, int *end)
{
if (start >= end) return;
int *temp = start;
temp = toplace(start,end);
quicksort(start,temp);
quicksort(temp+1,end);
}
You can do this using the bubble sort algorithm (http://en.wikipedia.org/wiki/Bubble_sort)
Basically iterate through all records, comparing the current record, with the next. If the current record's 2nd column is higher then swap these records. If the current record's 2nd column is equal but the 3rd column is higher, then swap also.
Continue iterating until no more swaps are made.
To use your example:
10 0 1
11 0 2
12 1 2 (swap with next)
13 0 1
10 0 1
11 0 2(swap with next)
13 0 1
12 1 2
10 0 1
13 0 1
11 0 2
12 1 2
And done!
Related
I would like to use data stored into an Eigen (https://eigen.tuxfamily.org) vector
Eigen::Vector<double, 6> vec({1,2,3,4,5,6});
as if they were a triangular matrix
1 2 3
0 4 5
0 0 6
I know how to do it for a full matrix using Eigen's Map
Eigen::Vector<double, 9> vec({1,2,3,4,5,6,7,8,9});
std::cout << Eigen::Map<Eigen::Matrix<double, 3, 3, RowMajor>>(vec.data());
which produces
1 2 3
4 5 6
7 8 9
However I do not know how to make a Map to a triangular matrix.
Is it possible?
Thanks!
[Edited for clarity]
In my opinion this cannot be done using Map only: The implementation of Map as it is relies on stride sizes that remain constant no matter their index positions, see https://eigen.tuxfamily.org/dox/classEigen_1_1Stride.html.
To implement a triangular matrix map you would have to have a Map that changes its inner stride depending on the actual column number. The interfaces in Eigen do not allow that at the moment, see https://eigen.tuxfamily.org/dox/Map_8h_source.html.
But if you are just concerned about the extra memory you can just use Eigen's sparse matrix representation:
https://eigen.tuxfamily.org/dox/group__TutorialSparse.html
(Refer to section "Filling a sparse matrix".)
This is not a direct solution to your problem but a way how to calculate the std::vector to fill in the 0 at the correct place. I think it is also possible to calculate it as a std::array if needed. I am not sure if that helps, but I guess you could use the calculated vector to fill the Eigen::Map
#include <array>
#include <cstddef>
#include <iostream>
#include <vector>
template<typename T, size_t N>
class EigenVector
{
static constexpr int CalculateRowColSize(size_t n)
{
size_t i = 1;
size_t inc = 1;
do
{
if (inc == n)
{
return static_cast<int>(i);
}
i++;
inc += i;
} while (i < n);
return -1;
}
static constexpr bool IsValid(size_t n)
{
if(CalculateRowColSize(n) == -1)
{
return false;
}
return true;
}
static_assert(IsValid(N));
public:
EigenVector() = delete;
static std::vector<T> Calculate(std::array<T, N> values)
{
constexpr size_t mRowColSize = CalculateRowColSize(N);
std::vector<T> ret;
auto count = 0;
auto valueCounter = 0;
for (size_t i = 0; i < mRowColSize; i++)
{
for (auto j = 0; j < count; j++)
{
ret.push_back(T());
}
for (size_t j = 0; j < mRowColSize - count; j++)
{
ret.push_back(values[valueCounter]);
valueCounter++;
}
count++;
}
return ret;
}
};
int main()
{
{
const std::array<int, 6> arr{ 1,2,3,4,5,6 };
const auto values = EigenVector<int, 6>::Calculate(arr);
for (auto& val : values)
{
std::cout << val << " ";
}
}
std::cout << std::endl << std::endl;
{
const std::array<int, 10> arr{ 1,2,3,4,5,6,7,8,9,10 };
const auto values = EigenVector<int, 10>::Calculate(arr);
for (auto& val : values)
{
std::cout << val << " ";
}
}
return 0;
}
Output:
1 2 3 0 4 5 0 0 6
1 2 3 4 0 5 6 7 0 0 8 9 0 0 0 10
Note that the algorithm is written that only possible matrix sizes are valid as input
I need to sort a table by column. My tables are represents by a single vector.
example :
col_name A B C
vector : 1 2 3 6 5 4 7 8 9
that give me the table :
A B C
1 6 7
2 5 8
3 4 9
After a sort on column B , I need to obtain :
A B C
3 4 9
2 5 8
1 6 7
my code :
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
int main()
{
std::vector<std::string> vec = {"1","8","1","2","3","2","3",
"5","5","2","5","6","5","6",
"9","3","3","4","8","3","9"};
std::vector<std::string> rec = {"1","1","8","2","2","3","3",
"2","5","5","5","5","6","6",
"3","9","3","4","3","8","9"};
int size = 7;
int col_idx = 1;
for(int i = 0; i<3;++i)
{
if(i==col_idx)
continue;
std::sort(vec.begin() + i*size, vec.begin() + (i+1)*size,
[col_idx, size, i](std::string& s1,std::string& s2)
{
std::cout << s1 << " "
<< s2 << " "
<< *(&s1 +(col_idx - i)*size) << " "
<< *(&s2 +(col_idx - i)*size) << " "
<< (*(&s1 +(col_idx - i)*size) < *(&s2 +(col_idx - i)*size)) << std::endl;
return *(&s1 +(col_idx - i)*size) < *(&s2 +(col_idx - i)*size);
});
}
std::sort(vec.begin() + col_idx*size, vec.begin() + (col_idx+1)*size);
}
assert(vec==res);
I have a segmentation fault error : only the first line appear from the cout.
Honestly, your approach looks rather complicated to me. Most of its complexity is due to the fact that you have rows in your code but they are present only implicitly. Making stuff explicit in code not only helps to increase readability but also makes code easier to write.
Lets say you use std::array<std::string,3> for rows, then your code could be as leightweight as this:
#include <vector>
#include <array>
#include <algorithm>
#include <iostream>
int main() {
using row_t = std::array<std::string,3>;
std::vector<row_t> vec = { {"1","8","1"},{"2","3","2"},{"3","5","5"},{"2","5","6"}};
std::sort(vec.begin(),vec.end(),[](const row_t& a, const row_t& b) { return a[2] < b[2]; });
for (const auto& row : vec) {
for (const auto& e : row) std::cout << e << " ";
std::cout << '\n';
}
}
Output:
1 8 1
2 3 2
3 5 5
2 5 6
Ok, that's probably a good approch for this problem, and maybe it's what should I do, but I can't pass 2 months to change all the code ...
You could have made the requirements more clear in the question. I think if you have 10k lines of code that depend on this particular issue using a flat vector, when a different data structure would be more appropriate, then you have a bigger problem than how to sort rows. Anyhow...
Using a flat std::vector is usually not a bad idea. What I miss from your code is something along the line of
template <int stride>
std::string& access_at(std::vector<std::string>& vec,size_t row,size_t col) {
return vec[ row * stride + col ];
}
template <int stride>
const std::string& access_at(const std::vector<std::string>& vec,size_t row,size_t col) {
return vec[ row * stride + col ];
}
That lets you iterate the table like this:
for (size_t i=0;i < vec.size()/3;++i) {
for (size_t j=0;j<3;++j) {
std::cout << access_at<3>(vec,i,j) << " ";
}
std::cout << '\n';
}
Next I am going to shamelessly steal and modify code from this answer. The basic idea is to sort a vector of indices instead of sorting the vector directly:
using index_t = std::vector<size_t>;
template <int stride>
index_t make_sorted_index(const std::vector<std::string>& values,size_t col) {
index_t index(values.size() / stride);
std::iota(index.begin(), index.end(), 0);
std::sort(index.begin(),
index.end(),
[&values,&col](size_t a, size_t b) {
return access_at<stride>(values,a,col) < access_at<stride>(values,b,col);
}
);
return index;
}
Once you have that, the loop to print the sorted table need only a minor modification:
for (size_t i=0;i < vec.size()/3;++i) {
for (size_t j=0;j<3;++j) {
std::cout << access_at<3>(vec,index[i],j) << " ";
}
std::cout << '\n';
}
Putting everything together:
#include <vector>
#include <numeric>
#include <algorithm>
#include <iostream>
template <int stride>
std::string& access_at(std::vector<std::string>& vec,size_t row,size_t col) { return vec[ row * stride + col ]; }
template <int stride>
const std::string& access_at(const std::vector<std::string>& vec,size_t row,size_t col) { return vec[ row * stride + col ]; }
using index_t = std::vector<size_t>;
template <int stride>
index_t make_sorted_index(const std::vector<std::string>& values,size_t col) {
index_t index(values.size() / stride);
std::iota(index.begin(), index.end(), 0);
std::sort(index.begin(),
index.end(),
[&values,&col](size_t a, size_t b) { return access_at<stride>(values,a,col) < access_at<stride>(values,b,col); }
);
return index;
}
int main() {
std::vector<std::string> vec = { "1","8","1","2","3","2","3","5","5","2","5","6"};
for (size_t i=0;i < vec.size()/3;++i) {
for (size_t j=0;j<3;++j) {
std::cout << access_at<3>(vec,i,j) << " ";
}
std::cout << '\n';
}
std::cout << '\n';
auto index = make_sorted_index<3>(vec,1);
for (size_t i=0;i < vec.size()/3;++i) {
for (size_t j=0;j<3;++j) {
std::cout << access_at<3>(vec,index[i],j) << " ";
}
std::cout << '\n';
}
}
With output:
1 8 1
2 3 2
3 5 5
2 5 6
2 3 2
3 5 5
2 5 6
1 8 1
I'll leave it to you to actually copy the vector to get the sorted one, if you really need that.
PS: In the first version above I sorted with respect to column C, the last part sorts with respect to B as requested.
PPS: I still dont understand your code. I don't understand why you have std::cout inside the predicate and to be honest I have no clue how your call to sort is supposed to achieve what you want.
I have a c++ program which calculates maximum of a array provided no two consecutive elements of array can be taken.
For eg:
7 3 4 6 will result in a answer of 13 .Here we chose 7 and 6 for optimal maximum.
Here is my recursive program for it.
#include <iostream>
using namespace std;
int n;
int findMax(int x,int ar[])
{
if(x < n)
return max( ar[x]+findMax(x+2,ar), findMax(x+1,ar));
return 0;
}
int main(){
int ar[]={1,7,4,4,9,5,12};
n = sizeof(ar)/sizeof(ar[0]);
cout<<findMax(0,ar);
return 0;
}
However I am more interested in the indices of array which were chosen for this purpose by my program .How can I do that efficiently.
In the above program answer should be 1,4,6 as we chose 1st , 4th and 6th element of the array for the maximum.
Note: I am using 0 based indexing.
Thanks.
A recurrence relation R(k) for the maximum sum of the first k elements of the array (with no adjacent terms) is:
R(0) = 0, R(1) = max(0, a[0])
R(k) = max(a[k] + R(k-2), R(k-1))
This is almost the same recurrence you're using in your code, but in your code your function returns the maximum sum of elements k and later.
Anyway, you can build a table of these values in linear time using dynamic programming. In pseudocode:
R = new array of length n+1
R[0] = 0
R[1] = max(0, a[0])
for i = 2 .. n
R[i] = max(a[i-1] + R[i-2], R[i-1])
If you just want the maximum sum, you can return R[n]. But you can also reconstruct the indices easily. In pseudo-code:
indices(a, R):
result = new empty vector
i = n
while i > 0
if (i == 1 and a[0] > 0) or R[i] == a[i-1] + R[i-2]
result.push_back(i-1)
i -= 2
else
i -= 1
You'll have to reverse result to get the indices in increasing order.
This is definitely not the most effective solution but probably the one with least implementation effort:
#include <iostream>
#include <vector>
using namespace std;
int n;
pair<int, vector<int> > findMax(int x, int ar[])
{
if (x < n) {
pair<int, vector<int> > max1 = findMax(x + 2, ar);
const pair<int, vector<int> > max2 = findMax(x + 1, ar);
max1.first += ar[x];
max1.second.insert(max1.second.begin(), x);
return max1.first >= max2.first ? max1 : max2;
}
return make_pair(0, vector<int>());
}
ostream& operator<<(ostream &out, const vector<int> &vec)
{
const char *sep = "";
for (int value : vec) {
out << sep << value; sep = ", ";
}
return out;
}
int main()
{
int ar[]={1,7,4,4,9,5,12};
n = sizeof ar / sizeof *ar;
const pair<int, vector<int> > maxAr = findMax(0, ar);
cout << maxAr.first << '\n'
<< maxAr.second << '\n';
return 0;
}
Output:
28
1, 4, 6
Life demo on coliru
Thereby, the return value is extended with a std::vector<int> which holds the used indices beside of the current sum.
std::max() could be used if I would provide a suitable (overloadeded) operator<() for std::pair<int, std::vector<int> >. To not make things over-complicated, I just replaced std::max() by the resp. condition.
I think below code will satisfy your need.
#include<bits/stdc++.h>
using namespace std;
int n;
void findMax(int arr[], int in, pair< int, vector<int> > tempStore,
pair< int, vector<int> > &resStore) {
if(in >=n) {
if(resStore.first < tempStore.first) {
resStore.first = tempStore.first;
resStore.second = tempStore.second;
}
return;
}
findMax(arr, in+1, tempStore, resStore);
tempStore.first += arr[in];
tempStore.second.push_back(in);
findMax(arr, in+2, tempStore, resStore);
}
int main() {
int ar[]={1,7,4,4,9,5,12};
n = sizeof(ar)/sizeof(ar[0]);
pair< int, vector<int> > resStore, tempStore;
findMax(ar, 0,tempStore,resStore);
cout<<"Result Value: "<<resStore.first;
cout<<"\nResult Index:\n";
for(int i=0; i<resStore.second.size(); i++) {
cout<<resStore.second[i]<<" ";
}
return 0;
}
I need a different version of permutations for my code. I could able to achieve what I want but it is not generic enough. my algorithm keeps going bigger along with my requirements. But that should not be.
This is not a home work for any one, I need it for one my critical projects, wondering if any pre-defined algorithms available from boost or any other.
Below is the standard version of next_permutation using c++.
// next_permutation example
#include <iostream> // std::cout
#include <algorithm> // std::next_permutation
int main ()
{
int myints[] = {1,2,3};
do
{
std::cout << myints[0] << ' ' << myints[1] << ' ' << myints[2] << '\n';
} while ( std::next_permutation(myints,myints+3) );
return 0;
}
That gives below output :
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
But my requirement is :- Let's say I have 1 to 9 numbers :
1,2,3,4,5,6,7,8,9
And I need a variable length of permutations and in only ASCENDING order and with out DUPLICATES.
Let's say i need 3 digit length of permutations then i need output as below.
123
124
125
.
.
.
128
129
134 // After 129 next one should be exactly 134
135 // ascending order mandatory
136
.
.
.
148
149
156 // exactly 156 after 149, ascending order mandatory
.
.
.
489 // exactly 567 after 489, because after 3rd digit 9, 2nd digit
567 // will be increased to 49? , so there is no possibility for
. // 3rd digit, so first digit gets incremented to 5 then 6 then
. // 7, in ascending order.
.
.
.
789 // and this should be the last set I need.
My list may contain upto couple of hundred's of numbers and variable length can be 1 to up to Size of the list.
My own algorithm is working for specific variable length, and a specific size, when they both changes, i need to write huge code. so, looking for a generic one.
I am not even sure if this is called as Permutations or there is a different name available for this kind of math/logic.
Thanks in advance.
musk's
Formally, you want to generate all m-combinations of the set [0;n-1].
#include <iostream>
#include <vector>
bool first_combination (std::vector<int> &v, int m, int n)
{
if ((m < 0) || (m > n)) {
return false;
}
v.clear ();
v.resize (m);
for (int i = 0; i < m; i++) {
v[i] = i;
}
return true;
}
bool next_combination (std::vector<int> &v, int m, int n)
{
for (int i = m - 1; i >= 0; i--) {
if (v[i] + m - i < n) {
v[i]++;
for (int j = i + 1; j < m; j++) {
v[j] = v[j - 1] + 1;
}
return true;
}
}
return false;
}
void print_combination (const std::vector<int> &v)
{
for (size_t i = 0; i < v.size(); i++) {
std::cout << v[i] << ' ';
}
std::cout << '\n';
}
int main ()
{
const int m = 3;
const int n = 5;
std::vector<int> v;
if (first_combination (v, m, n)) {
do {
print_combination (v);
} while (next_combination (v, m, n));
}
}
You can use this code and the linked article as inspiration.
This task can be done with a simple iterative algorithm. Just increment the first element that can be incremented and rescale the elements before it until there is no element to be incremented.
int a[] = {0,1,2,3,4,5,6,7,8,9}; // elements: must be ascending in this case
int n = sizeof(a)/sizeof(int);
int digits = 7; // number of elements you want to choose
vector<int> indexes; // creating the first combination
for ( int i=digits-1;i>=0;--i ){
indexes.push_back(i);
}
while (1){
/// printing the current combination
for ( int i=indexes.size()-1;i>=0;--i ){
cout << a[indexes[i]] ;
} cout << endl;
///
int i = 0;
while ( i < indexes.size() && indexes[i] == n-1-i ) // finding the first element
++i; // that can be incremented
if ( i==indexes.size() ) // if no element can be incremented, we are done
break;
indexes[i]++; // increment the first element
for ( int j=0;j<i;++j ){ // rescale elements before it to first combination
indexes[j] = indexes[i]+(i-j);
}
}
Output:
0123456
0123457
0123458
0123459
0123467
0123468
0123469
0123478
0123479
0123489
0123567
0123568
0123569
0123578
0123579
0123589
0123678
0123679
0123689
0123789
0124567
0124568
0124569
0124578
0124579
0124589
0124678
0124679
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I have a std::vector. I want to create iterators representing a slice of that vector. How do I do it? In pseudo C++:
class InterestingType;
void doSomething(slice& s) {
for (slice::iterator i = s.begin(); i != s.end(); ++i) {
std::cout << *i << endl;
}
}
int main() {
std::vector v();
for (int i= 0; i < 10; ++i) { v.push_back(i); }
slice slice1 = slice(v, 1, 5);
slice slice2 = slice(v, 2, 4);
doSomething(slice1);
doSomething(slice2);
return 0;
}
I would prefer not to have to copy the elements to a new datastructure.
You'd just use a pair of iterators:
typedef std::vector<int>::iterator vec_iter;
void doSomething(vec_iter first, vec_iter last) {
for (vec_iter cur = first; cur != last; ++cur) {
std::cout << *cur << endl;
}
}
int main() {
std::vector v();
for (int i= 0; i < 10; ++i) { v.push_back(i); }
doSomething(v.begin() + 1, v.begin() + 5);
doSomething(v.begin() + 2, v.begin() + 4);
return 0;
}
Alternatively, the Boost.Range library should allow you to represent iterator pairs as a single object, but the above is the canonical way to do it.
I learnt Python before I learnt C++. I wondered if C++ offered slicing of vectors like slicing in Python lists. Took a couple of minutes to write this function that allows you to slice a vector analogous to the way its done in Python.
vector<int> slice(const vector<int>& v, int start=0, int end=-1) {
int oldlen = v.size();
int newlen;
if (end == -1 or end >= oldlen){
newlen = oldlen-start;
} else {
newlen = end-start;
}
vector<int> nv(newlen);
for (int i=0; i<newlen; i++) {
nv[i] = v[start+i];
}
return nv;
}
Usage:
vector<int> newvector = slice(vector_variable, start_index, end_index);
The start_index element will be included in the slice, whereas the end_index will not be included.
Example:
For a vector v1 like {1,3,5,7,9}
slice(v1,2,4) returns {5,7}
Taken from here:
std::vector<myvector::value_type>(myvector.begin()+start, myvector.begin()+end).swap(myvector);
Usage example:
#include <iostream>
#include <vector>
int main ()
{
std::vector<int> indexes{3, 6, 9};
for( auto index : indexes )
{
int slice = 3;
std::vector<int> bar{1, 2, 3, 4, 5, 6, 7, 8, 9};
std::vector<int>( bar.begin() + index - slice, bar.begin() + index ).swap(bar);
std::cout << "bar index " << index << " contains:";
for (unsigned i=0; i<bar.size(); i++)
std::cout << ' ' << bar[i];
std::cout << '\n';
}
return 0;
}
Outputs:
bar index 3 contains: 1 2 3
bar index 6 contains: 4 5 6
bar index 9 contains: 7 8 9
As others have said, you can represent the "slice" as pair of iterators. If you are willing to use Boost, you can use the range concept. Then you will have even begin()/end() member functions available and the whole thing looks a lot like a container.
use boost range adapters. they are lazy:
operator|() is used to add new behaviour lazily and never modifies its
left argument.
boost::for_each(v|sliced(1,5)|transformed(doSomething));
doSomething needs to take range as input. a simple (may be lambda) wrapper would fix that.
You can represent those "slices" with a pair of iterators.
You don't need a pair of iterators to slice a vector. Three indexes will do because it allows you to create slices with steps:
static const int arr[] = {16,2,77,29,42};
vector<int> v (arr, arr + sizeof(arr) / sizeof(arr[0]) );
vector<int>::iterator i;
const int step = 2;
const int first = 0;
const int last = v.size()-1;
int counter=first;
for (i = v.begin()+first; counter<last; i+=step, counter+=step) {
// Do something with *i
cout << *i << endl;
}
Prints:
16
77
In this code, a counter is needed to track the position because not all iterators can do this.
It is possible to use slices with std::valarray. Which is an STL analogue of numpy.array in python. It support different vectorized operations like min, max, +,-, *, /, etc.
More info here.
std::slice(start, length, stride) allows to select and modify slices of an array without copying (documentation here).
The slicing would look like this:
std::valarray<int> foo (9);
for (int i=0; i<9; ++i) foo[i]=i; // 0 1 2 3 4 5 6 7 8
// | | | | |
std::slice myslice=std::slice(1,5,1); // v v v v v
foo[myslice] *= std::valarray<int>(10,3); // 0 10 20 30 40 50 6 7 8
Or with stride=2:
std::valarray<int> foo (9);
for (int i=0; i<9; ++i) foo[i]=i; // 0 1 2 3 4 5 6 7 8
// | | |
std::slice myslice=std::slice(1,3,2); // v v v
foo[myslice] *= std::valarray<int>(10,3); // 0 10 2 30 4 50 6 7 8
// | | |
foo[std::slice (0,3,3)] = 99; // v v v
// 99 10 2 99 4 50 99 7 8
std::cout << "foo:";
for (std::size_t n=0; n<foo.size(); n++)
std::cout << ' ' << foo[n];
std::cout << '\n';