I have 10,000 vector<pair<unsigned,unsigned>> and I want to merge them into a single vector such that it is lexicographically sorted and does not contain duplicates. In order to do so I wrote the following code. However, to my surprise the below code is taking a lot of time. Can someone please suggest as to how can I reduce the running time of my code?
using obj = pair<unsigned, unsigned>
vector< vector<obj> > vecOfVec; // 10,000 vector<obj>, each sorted with size()=10M
vector<obj> result;
for(auto it=vecOfVec.begin(), l=vecOfVec.end(); it!=l; ++it)
{
// append vectors
result.insert(result.end(),it->begin(),it->end());
// sort result
std::sort(result.begin(), result.end());
// remove duplicates from result
result.erase(std::unique(result.begin(), result.end()), result.end());
}
I think you should use the fact that the vector in vectOfVect are sorted.
So detecting the min value in the front on the single vectors, push_back() it in the result and remove all the values detected from the front of the vectors matching the min values (avoiding duplicates in result).
If you can delete the vecOfVec variable, something like (caution: code not tested: just to give an idea)
while ( vecOfVec.size() )
{
// detect the minimal front value
auto itc = vecOfVec.cbegin();
auto lc = vecOfVec.cend();
auto valMin = itc->front();
while ( ++itc != lc )
valMin = std::min(valMin, itc->front());
// push_back() the minimal front value in result
result.push_back(valMin);
for ( auto it = vecOfVec.begin() ; it != vecOfVec.end() ; )
{
// remove all the front values equals to valMin (this remove the
// duplicates from result)
while ( (false == it->empty()) && (valMin == it->front()) )
it->erase(it->begin());
// when a vector is empty is removed
it = ( it->empty() ? vecOfVec.erase(it) : ++it );
}
}
If you can, I suggest you to switch vecOfVec from a vector< vector<obj> > to something that permit an efficient removal from the front of single containers (stacks?) and an efficient removal of single containers (a list?).
If there are lot of duplicates, you should use set rather than vector for your result, as set is the most natural thing to store something without duplicates:
set< pair<unsigned,unsigned> > resultSet;
for (auto it=vecOfVec.begin(); it!=vecOfVec.end(); ++it)
resultSet.insert(it->begin(), it->end());
If you need to turn it into a vector, you can write
vector< pair<unsigned,unsigned> > resultVec(resultSet.begin(), resultSet.end());
Note that since your code runs over 800 billion elements, it would still take a lot of time, no matter what. At least hours, if not days.
Other ideas are:
recursively merge vectors (10000 -> 5000 -> 2500 -> ... -> 1)
to merge 10000 vectors, store 10000 iterators in a heap structure
One problem with your code is the excessive use of std::sort. Unfortunately, the quicksort algorithm (which usually is the working horse used by std::sort) is not particularly faster when encountering an already sorted array.
Moreover, you're not exploiting the fact that your initial vectors are already sorted. This can be exploited by using a heap of their next values, when you will not need to call sort again. This may be coded as follows (code tested using obj=int), but perhaps it can be made more concise.
// represents the next unused entry in one vector<obj>
template<typename obj>
struct feed
{
typename std::vector<obj>::const_iterator current, end;
feed(std::vector<obj> const&v)
: current(v.begin()), end(v.end()) {}
friend bool operator> (feed const&l, feed const&r)
{ return *(l.current) > *(r.current); }
};
// - returns the smallest element
// - set corresponding feeder to next and re-establish the heap
template<typename obj>
obj get_next(std::vector<feed<obj>>&heap)
{
auto&f = heap[0];
auto x = *(f.current++);
if(f.current == f.end) {
std::pop_heap(heap.begin(),heap.end(),std::greater<feed<obj>>{});
heap.pop_back();
} else
std::make_heap(heap.begin(),heap.end(),std::greater<feed<obj>>{});
return x;
}
template<typename obj>
std::vector<obj> merge(std::vector<std::vector<obj>>const&vecOfvec)
{
// create min heap of feed<obj> and count total number of objects
std::vector<feed<obj>> input;
input.reserve(vecOfvec.size());
size_t num_total = 0;
for(auto const&v:vecOfvec)
if(v.size()) {
num_total += v.size();
input.emplace_back(v);
}
std::make_heap(input.begin(),input.end(),std::greater<feed<obj>>{});
// append values in ascending order, avoiding duplicates
std::vector<obj> result;
result.reserve(num_total);
while(!input.empty()) {
auto x = get_next(input);
result.push_back(x);
while(!input.empty() &&
!(*(input[0].current) > x)) // remove duplicates
get_next(input);
}
return result;
}
I have two equal length vectors from which I want to remove elements based on a condition in one of the vectors. The same removal operation should be applied to both so that the indices match.
I have come up with a solution using std::erase, but it is extremely slow:
vector<myClass> a = ...;
vector<otherClass> b = ...;
assert(a.size() == b.size());
for(size_t i=0; i<a.size(); i++)
{
if( !a[i].alive() )
{
a.erase(a.begin() + i);
b.erase(b.begin() + i);
i--;
}
}
Is there a way that I can do this more efficiently and preferably using stl algorithms?
If order doesn't matter you could swap the elements to the back of the vector and pop them.
for(size_t i=0; i<a.size();)
{
if( !a[i].alive() )
{
std::swap(a[i], a.back());
a.pop_back();
std::swap(b[i], b.back());
b.pop_back();
}
else
++i;
}
If you have to maintain the order you could use std::remove_if. See this answer how to get the index of the dereferenced element in the remove predicate:
a.erase(remove_if(begin(a), end(a),
[b&](const myClass& d) { return b[&d - &*begin(a)].alive(); }),
end(a));
b.erase(remove_if(begin(b), end(b),
[](const otherClass& d) { return d.alive(); }),
end(b));
The reason it's slow is probably due to the O(n^2) complexity. Why not use list instead? As making a pair of a and b is a good idea too.
A quick win would be to run the loop backwards: i.e. start at the end of the vector. This tends to minimise the number of backward shifts due to element removal.
Another approach would be to consider std::vector<std::unique_ptr<myClass>> etc.: then you'll be essentially moving pointers rather than values.
I propose you create 2 new vectors, reserve memory and swap vectors content in the end.
vector<myClass> a = ...;
vector<otherClass> b = ...;
vector<myClass> new_a;
vector<myClass> new_b;
new_a.reserve(a.size());
new_b.reserve(b.size());
assert(a.size() == b.size());
for(size_t i=0; i<a.size(); i++)
{
if( a[i].alive() )
{
new_a.push_back(a[i]);
new_b.push_back(b[i]);
}
}
swap(a, new_a);
swap(b, new_b);
It can be memory consumed, but should work fast.
erasing from the middle of a vector is slow due to it needing to reshuffle everything after the deletion point. consider using another container instead that makes erasing quicker. It depends on your use cases, will you be iterating often? does the data need to be in order? If you aren't iterating often, consider a list. if you need to maintain order, consider a set. if you are iterating often and need to maintain order, depending on the number of elements, it may be quicker to push back all alive elements to a new vector and set a/b to point to that instead.
Also, since the data is intrinsically linked, it seems to make sense to have just one vector containing data a and b in a pair or small struct.
For performance reason need to use next.
Use
vector<pair<myClass, otherClass>>
as say #Basheba and std::sort.
Use special form of std::sort with comparision predicate. And do not enumerate from 0 to n. Use std::lower_bound instead, becouse vector will be sorted. Insertion of element do like say CashCow in this question: "how do you insert the value in a sorted vector?"
I had a similar problem where I had two :
std::<Eigen::Vector3d> points;
std::<Eigen::Vector3d> colors;
for 3D pointclouds in Open3D and after removing the floor, I wanted to delete all points and colors if the points' z coordinate is greater than 0.05. I ended up overwriting the points based on the index and resizing the vector afterward.
bool invert = true;
std::vector<bool> mask = std::vector<bool>(points.size(), invert);
size_t pos = 0;
for (auto & point : points) {
if (point(2) < CONSTANTS::FLOOR_HEIGHT) {
mask.at(pos) = false;
}
++pos;
}
size_t counter = 0;
for (size_t i = 0; i < points.size(); i++) {
if (mask[i]) {
points.at(counter) = points.at(i);
colors.at(counter) = colors.at(i);
++counter;
}
}
points.resize(counter);
colors.resize(counter);
This maintains order and at least in my case, worked almost twice as fast than the remove_if method from the accepted answer:
for 921600 points the runtimes were:
33 ms for the accepted answer
17 ms for this approach.
I want the function to return true when there is any element matching between two vectors,
Note : My vectors are not sorted
Following is my source code,
bool CheckCommon( std::vector< long > &inVectorA, std::vector< long > &inVectorB )
{
std::vector< long > *lower, *higher;
size_t sizeL = 0, sizeH = 0;
if( inVectorA.size() > inVectorB.size() )
{
lower = &inVectorA;
sizeL = inVectorA.size();
higher = &inVectorB;
sizeH = inVectorB.size();
}
else
{
lower = &inVectorB;
sizeL = inVectorB.size();
higher = &inVectorA;
sizeH = inVectorA.size();
}
size_t indexL = 0, indexH = 0;
for( ; indexH < sizeH; indexH++ )
{
bool exists = std::binary_search( lower->begin(), lower->end(), higher->at(indexH) );
if( exists == true )
return true;
else
continue;
}
return false;
}
This is working fine when the size of vector B is less than the size of vector A , but returning false even there is match when size of vector B is greater than size of vector A .
The problem with posted code is that you should not use std::binary_search when the vector is not sorted. The behaviour is defined only for sorted range.
If the input vectors are not sorted then you can use find_first_of to check for existence of first common element found.
bool CheckCommon(std::vector<long> const& inVectorA, std::vector<long> const& nVectorB)
{
return std::find_first_of (inVectorA.begin(), inVectorA.end(),
nVectorB.begin(), nVectorB.end()) != inVectorA.end();
}
Complexity of find_first_of is up to linear in inVectorA.size()*inVectorB.size(); it compares elements until a match is found.
If you want to fix your original algorithm then you can make a copy of one of vectors and std::sort it, then std::binary_search works with it.
In actual programs that do lot of such matching between containers the containers are usually kept sorted. On such case std::set_intersection can be used. Then the complexity of search is up to linear in inVectorA.size()+inVectorB.size().
std::find_first_of is more efficient than to sort both ranges and then to search for matches with std::set_intersection when both ranges are rather short or second range is shorter than binary logarithm of length of first range.
You can use a well-defined algorithm called as std::set_intersection to check if there is any common element between these vectors.
Pre-condition :- Both vectors be sorted.
You could do something like the following. Iterate over the first vector. For each element, use std::find to see if it exists in the other vector. If you find it, they have at least one common element so return true. Otherwise, move to the next element of the first vector and repeat this process. If you make it all the way through the first vector without finding a common element, there is no intersection so return false.
bool CheckCommon(std::vector<long> const& inVectorA, std::vector<long> const& nVectorB)
{
for (auto const& num : inVectorA)
{
auto it = std::find(begin(nVectorB), end(nVectorB), num);
if (it != end(nVectorB))
{
return true;
}
}
return false;
}
Usage of std::set_intersection is one option. Since the vector's elements are sorted, the code can be simplified to this:
#include <algorithm>
#include <iterator>
bool CheckCommon( const std::vector< long > &inVectorA, const std::vector< long > &inVectorB )
{
std::vector< long > temp;
std::set_intersection(inVectorA.begin(), inVectorA.end(),
inVectorB.begin(), inVectorB.end(),
std::back_inserter(temp));
return !temp.empty()
}
The drawback is that a temporary vector is being created while the set_intersection is being executed (but maybe in the future, this can be considered a "feature" if you want to know what elements are common).
Here is an implementation which uses sorted vectors, doesn't construct a new container, and which has only linear complexity (more detailed: O(container1.size()+ container2.size()):
template< class ForwardIt1, class ForwardIt2 >
bool has_common_elements( ForwardIt1 first, ForwardIt1 last, ForwardIt2 s_first, ForwardIt2 s_last )
{
auto it=first;
auto s_it=s_first;
while(it<last && s_it<s_last)
{
if(*it==*s_it)
{
return true;
}
*it<*s_it ? ++it : ++s_it; //increase the smaller of both
}
return false;
}
DEMO
Your code uses std::binary_search, whose pre-condition is that (From http://en.cppreference.com/w/cpp/algorithm/binary_search):
For std::binary_search to succeed, the range [first, last) must be at least partially ordered, i.e. it must satisfy all of the following requirements:
partitioned with respect to element < value or comp(element, value)
partitioned with respect to !(value < element) or !comp(value, element)
for all elements, if element < value or comp(element, value) is true then !(value < element) or !comp(value, element) is also true
A fully-sorted range meets these criteria, as does a range resulting from a call to std::partition.
The sample data you used for testing (as posted at http://ideone.com/XCYdM8) do not meet that requirement. Instead of using:
vectorB.push_back(11116);
vectorB.push_back(11118);
vectorB.push_back(11112);
vectorB.push_back(11120);
vectorB.push_back(11190);
vectorB.push_back(11640);
vectorB.push_back(11740);
if you use a sorted vector like below
vectorB.push_back(11112);
vectorB.push_back(11116);
vectorB.push_back(11118);
vectorB.push_back(11120);
vectorB.push_back(11190);
vectorB.push_back(11640);
vectorB.push_back(11740);
your function will work just fine.
PS The you have designed your code, if the longer std::vector is sorted, the function will work fine.
PS2 Another option is to sort the longer std::vector before calling the function.
std::sort(B.begin(), B.end());
I have two data structures with data in them.
One is a vector std::vector<int> presentStudents And other is a
char array char cAllowedStudents[256];
Now I have to compare these two such that checking every element in vector against the array such that all elements in the vector should be present in the array or else I will return false if there is an element in the vector that's not part of the array.
I want to know the most efficient and simple solution for doing this. I can convert my int vector into a char array and then compare one by one but that would be lengthy operation. Is there some better way of achieving this?
I would suggest you use a hash map (std::unordered_map). Store all the elements of the char array in the hash map.
Then simply sequentially check each element in your vector whether it is present in the map or not in O(1).
Total time complexity O(N), extra space complexity O(N).
Note that you will have to enable C++11 in your compiler.
Please refer to function set_difference() in c++ algorithm header file. You can use this function directly, and check if result diff set is empty or not. If not empty return false.
A better solution would be adapting the implementation of set_difference(), like in here: http://en.cppreference.com/w/cpp/algorithm/set_difference, to return false immediately after you get first different element.
Example adaption:
while (first1 != last1)
{
if (first2 == last2)
return false;
if (*first1 < *first2)
{
return false;
}
else
{
if (*first2 == *first1)
{
++first1;
}
++first2;
}
}
return true;
Sort cAllowedstudents using std::sort.
Iterate over the presentStudents and look for each student in the sorted cAllowedStudents using std::binary_search.
If you don't find an item of the vector, return false.
If all the elements of the vector are found, return true.
Here's a function:
bool check()
{
// Assuming hou have access to cAllowedStudents
// and presentStudents from the function.
char* cend = cAllowedStudents+256;
std::sort(cAllowedStudents, cend);
std::vector<int>::iterator iter = presentStudents.begin();
std::vector<int>::iterator end = presentStudents.end();
for ( ; iter != end; ++iter )
{
if ( !(std::binary_search(cAllowedStudents, cend, *iter)) )
{
return false;
}
}
return true;
}
Another way, using std::difference.
bool check()
{
// Assuming hou have access to cAllowedStudents
// and presentStudents from the function.
char* cend = cAllowedStudents+256;
std::sort(cAllowedStudents, cend);
std::vector<int> diff;
std::set_difference(presentStudents.begin(), presentStudents.end(),
cAllowedStudents, cend,
std::back_inserter(diff));
return (diff.size() == 0);
}
Sort both lists with std::sort and use std::find iteratively on the array.
EDIT: The trick is to use the previously found position as a start for the next search.
std::sort(begin(pS),end(pS))
std::sort(begin(aS),end(aS))
auto its=begin(aS);
auto ite=end(aS);
for (auto s:pS) {
its=std::find(its,ite,s);
if (its == ite) {
std::cout << "Student not allowed" << std::cout;
break;
}
}
Edit: As legends mentiones, it usually might be more efficient to use binary search (as in R Sahu's answer). However, for small arrays and if the vector contains a significant fraction of students from the array (I'd say at least one tenths), the additional overhead of binary search might (or might not) outweight its asymptotic complexity benefits.
Using C++11. In your case, size is 256. Note that I personally have not tested this, or even put it into a compiler. It should, however, give you a good idea of what to do yourself. I HIGHLY recommend testing the edge cases with this!
#include <algorithm>
bool check(const std::vector<int>& studs,
char* allowed,
unsigned int size){
for(auto x : studs){
if(std::find(allowed, allowed+size-1, x) == allowed+size-1 && x!= *(allowed+size))
return false;
}
return true;
}
This question already has answers here:
How do I sort a std::vector by the values of a different std::vector? [duplicate]
(13 answers)
Closed 12 months ago.
I'd like to reorder the items in a vector, using another vector to specify the order:
char A[] = { 'a', 'b', 'c' };
size_t ORDER[] = { 1, 0, 2 };
vector<char> vA(A, A + sizeof(A) / sizeof(*A));
vector<size_t> vOrder(ORDER, ORDER + sizeof(ORDER) / sizeof(*ORDER));
reorder_naive(vA, vOrder);
// A is now { 'b', 'a', 'c' }
The following is an inefficient implementation that requires copying the vector:
void reorder_naive(vector<char>& vA, const vector<size_t>& vOrder)
{
assert(vA.size() == vOrder.size());
vector vCopy = vA; // Can we avoid this?
for(int i = 0; i < vOrder.size(); ++i)
vA[i] = vCopy[ vOrder[i] ];
}
Is there a more efficient way, for example, that uses swap()?
This algorithm is based on chmike's, but the vector of reorder indices is const. This function agrees with his for all 11! permutations of [0..10]. The complexity is O(N^2), taking N as the size of the input, or more precisely, the size of the largest orbit.
See below for an optimized O(N) solution which modifies the input.
template< class T >
void reorder(vector<T> &v, vector<size_t> const &order ) {
for ( int s = 1, d; s < order.size(); ++ s ) {
for ( d = order[s]; d < s; d = order[d] ) ;
if ( d == s ) while ( d = order[d], d != s ) swap( v[s], v[d] );
}
}
Here's an STL style version which I put a bit more effort into. It's about 47% faster (that is, almost twice as fast over [0..10]!) because it does all the swaps as early as possible and then returns. The reorder vector consists of a number of orbits, and each orbit is reordered upon reaching its first member. It's faster when the last few elements do not contain an orbit.
template< typename order_iterator, typename value_iterator >
void reorder( order_iterator order_begin, order_iterator order_end, value_iterator v ) {
typedef typename std::iterator_traits< value_iterator >::value_type value_t;
typedef typename std::iterator_traits< order_iterator >::value_type index_t;
typedef typename std::iterator_traits< order_iterator >::difference_type diff_t;
diff_t remaining = order_end - 1 - order_begin;
for ( index_t s = index_t(), d; remaining > 0; ++ s ) {
for ( d = order_begin[s]; d > s; d = order_begin[d] ) ;
if ( d == s ) {
-- remaining;
value_t temp = v[s];
while ( d = order_begin[d], d != s ) {
swap( temp, v[d] );
-- remaining;
}
v[s] = temp;
}
}
}
And finally, just to answer the question once and for all, a variant which does destroy the reorder vector (filling it with -1's). For permutations of [0..10], It's about 16% faster than the preceding version. Because overwriting the input enables dynamic programming, it is O(N), asymptotically faster for some cases with longer sequences.
template< typename order_iterator, typename value_iterator >
void reorder_destructive( order_iterator order_begin, order_iterator order_end, value_iterator v ) {
typedef typename std::iterator_traits< value_iterator >::value_type value_t;
typedef typename std::iterator_traits< order_iterator >::value_type index_t;
typedef typename std::iterator_traits< order_iterator >::difference_type diff_t;
diff_t remaining = order_end - 1 - order_begin;
for ( index_t s = index_t(); remaining > 0; ++ s ) {
index_t d = order_begin[s];
if ( d == (diff_t) -1 ) continue;
-- remaining;
value_t temp = v[s];
for ( index_t d2; d != s; d = d2 ) {
swap( temp, v[d] );
swap( order_begin[d], d2 = (diff_t) -1 );
-- remaining;
}
v[s] = temp;
}
}
In-place reordering of vector
Warning: there is an ambiguity about the semantic what the ordering-indices mean. Both are answered here
move elements of vector to the position of the indices
Interactive version here.
#include <iostream>
#include <vector>
#include <assert.h>
using namespace std;
void REORDER(vector<double>& vA, vector<size_t>& vOrder)
{
assert(vA.size() == vOrder.size());
// for all elements to put in place
for( int i = 0; i < vA.size() - 1; ++i )
{
// while the element i is not yet in place
while( i != vOrder[i] )
{
// swap it with the element at its final place
int alt = vOrder[i];
swap( vA[i], vA[alt] );
swap( vOrder[i], vOrder[alt] );
}
}
}
int main()
{
std::vector<double> vec {7, 5, 9, 6};
std::vector<size_t> inds {1, 3, 0, 2};
REORDER(vec, inds);
for (size_t vv = 0; vv < vec.size(); ++vv)
{
std::cout << vec[vv] << std::endl;
}
return 0;
}
output
9
7
6
5
note that you can save one test because if n-1 elements are in place the last nth element is certainly in place.
On exit vA and vOrder are properly ordered.
This algorithm performs at most n-1 swapping because each swap moves the element to its final position. And we'll have to do at most 2N tests on vOrder.
draw the elements of vector from the position of the indices
Try it interactively here.
#include <iostream>
#include <vector>
#include <assert.h>
template<typename T>
void reorder(std::vector<T>& vec, std::vector<size_t> vOrder)
{
assert(vec.size() == vOrder.size());
for( size_t vv = 0; vv < vec.size() - 1; ++vv )
{
if (vOrder[vv] == vv)
{
continue;
}
size_t oo;
for(oo = vv + 1; oo < vOrder.size(); ++oo)
{
if (vOrder[oo] == vv)
{
break;
}
}
std::swap( vec[vv], vec[vOrder[vv]] );
std::swap( vOrder[vv], vOrder[oo] );
}
}
int main()
{
std::vector<double> vec {7, 5, 9, 6};
std::vector<size_t> inds {1, 3, 0, 2};
reorder(vec, inds);
for (size_t vv = 0; vv < vec.size(); ++vv)
{
std::cout << vec[vv] << std::endl;
}
return 0;
}
Output
5
6
7
9
It appears to me that vOrder contains a set of indexes in the desired order (for example the output of sorting by index). The code example here follows the "cycles" in vOrder, where following a sub-set (could be all of vOrder) of indexes will cycle through the sub-set, ending back at the first index of the sub-set.
Wiki article on "cycles"
https://en.wikipedia.org/wiki/Cyclic_permutation
In the following example, every swap places at least one element in it's proper place. This code example effectively reorders vA according to vOrder, while "unordering" or "unpermuting" vOrder back to its original state (0 :: n-1). If vA contained the values 0 through n-1 in order, then after reorder, vA would end up where vOrder started.
template <class T>
void reorder(vector<T>& vA, vector<size_t>& vOrder)
{
assert(vA.size() == vOrder.size());
// for all elements to put in place
for( size_t i = 0; i < vA.size(); ++i )
{
// while vOrder[i] is not yet in place
// every swap places at least one element in it's proper place
while( vOrder[i] != vOrder[vOrder[i]] )
{
swap( vA[vOrder[i]], vA[vOrder[vOrder[i]]] );
swap( vOrder[i], vOrder[vOrder[i]] );
}
}
}
This can also be implemented a bit more efficiently using moves instead swaps. A temp object is needed to hold an element during the moves. Example C code, reorders A[] according to indexes in I[], also sorts I[] :
void reorder(int *A, int *I, int n)
{
int i, j, k;
int tA;
/* reorder A according to I */
/* every move puts an element into place */
/* time complexity is O(n) */
for(i = 0; i < n; i++){
if(i != I[i]){
tA = A[i];
j = i;
while(i != (k = I[j])){
A[j] = A[k];
I[j] = j;
j = k;
}
A[j] = tA;
I[j] = j;
}
}
}
If it is ok to modify the ORDER array then an implementation that sorts the ORDER vector and at each sorting operation also swaps the corresponding values vector elements could do the trick, I think.
A survey of existing answers
You ask if there is "a more efficient way". But what do you mean by efficient and what are your requirements?
Potatoswatter's answer works in O(N²) time with O(1) additional space and doesn't mutate the reordering vector.
chmike and rcgldr give answers which use O(N) time with O(1) additional space, but they achieve this by mutating the reordering vector.
Your original answer allocates new space and then copies data into it while Tim MB suggests using move semantics. However, moving still requires a place to move things to and an object like an std::string has both a length variable and a pointer. In other words, a move-based solution requires O(N) allocations for any objects and O(1) allocations for the new vector itself. I explain why this is important below.
Preserving the reordering vector
We might want that reordering vector! Sorting costs O(N log N). But, if you know you'll be sorting several vectors in the same way, such as in a Structure of Arrays (SoA) context, you can sort once and then reuse the results. This can save a lot of time.
You might also want to sort and then unsort data. Having the reordering vector allows you to do this. A use case here is for performing genomic sequencing on GPUs where maximal speed efficiency is obtained by having sequences of similar lengths processed in batches. We cannot rely on the user providing sequences in this order so we sort and then unsort.
So, what if we want the best of all worlds: O(N) processing without the costs of additional allocation but also without mutating our ordering vector (which we might, after all, want to reuse)? To find that world, we need to ask:
Why is extra space bad?
There are two reasons you might not want to allocate additional space.
The first is that you don't have much space to work with. This can occur in two situations: you're on an embedded device with limited memory. Usually this means you're working with small datasets, so the O(N²) solution is probably fine here. But it can also happen when you are working with really large datasets. In this case O(N²) is unacceptable and you have to use one of the O(N) mutating solutions.
The other reason extra space is bad is because allocation is expensive. For smaller datasets it can cost more than the actual computation. Thus, one way to achieve efficiency is to eliminate allocation.
Outline
When we mutate the ordering vector we are doing so as a way to indicate whether elements are in their permuted positions. Rather than doing this, we could use a bit-vector to indicate that same information. However, if we allocate the bit vector each time that would be expensive.
Instead, we could clear the bit vector each time by resetting it to zero. However, that incurs an additional O(N) cost per function use.
Rather, we can store a "version" value in a vector and increment this on each function use. This gives us O(1) access, O(1) clear, and an amoritzed allocation cost. This works similarly to a persistent data structure. The downside is that if we use an ordering function too often the version counter needs to be reset, though the O(N) cost of doing so is amortized.
This raises the question: what is the optimal data type for the version vector? A bit-vector maximizes cache utilization but requires a full O(N) reset after each use. A 64-bit data type probably never needs to be reset, but has poor cache utilization. Experimenting is the best way to figure this out.
Two types of permutations
We can view an ordering vector as having two senses: forward and backward. In the forward sense, the vector tell us where elements go to. In the backward sense, the vector tells us where elements are coming from. Since the ordering vector is implicitly a linked list, the backward sense requires O(N) additional space, but, again, we can amortize the allocation cost. Applying the two senses sequentially brings us back to our original ordering.
Performance
Running single-threaded on my "Intel(R) Xeon(R) E-2176M CPU # 2.70GHz", the following code takes about 0.81ms per reordering for sequences 32,767 elements long.
Code
Fully commented code for both senses with tests:
#include <algorithm>
#include <cassert>
#include <random>
#include <stack>
#include <stdexcept>
#include <vector>
///#brief Reorder a vector by moving its elements to indices indicted by another
/// vector. Takes O(N) time and O(N) space. Allocations are amoritzed.
///
///#param[in,out] values Vector to be reordered
///#param[in] ordering A permutation of the vector
///#param[in,out] visited A black-box vector to be reused between calls and
/// shared with with `backward_reorder()`
template<class ValueType, class OrderingType, class ProgressType>
void forward_reorder(
std::vector<ValueType> &values,
const std::vector<OrderingType> &ordering,
std::vector<ProgressType> &visited
){
if(ordering.size()!=values.size()){
throw std::runtime_error("ordering and values must be the same size!");
}
//Size the visited vector appropriately. Since vectors don't shrink, this will
//shortly become large enough to handle most of the inputs. The vector is 1
//larger than necessary because the first element is special.
if(visited.empty() || visited.size()-1<values.size());
visited.resize(values.size()+1);
//If the visitation indicator becomes too large, we reset everything. This is
//O(N) expensive, but unlikely to occur in most use cases if an appropriate
//data type is chosen for the visited vector. For instance, an unsigned 32-bit
//integer provides ~4B uses before it needs to be reset. We subtract one below
//to avoid having to think too much about off-by-one errors. Note that
//choosing the biggest data type possible is not necessarily a good idea!
//Smaller data types will have better cache utilization.
if(visited.at(0)==std::numeric_limits<ProgressType>::max()-1)
std::fill(visited.begin(), visited.end(), 0);
//We increment the stored visited indicator and make a note of the result. Any
//value in the visited vector less than `visited_indicator` has not been
//visited.
const auto visited_indicator = ++visited.at(0);
//For doing an early exit if we get everything in place
auto remaining = values.size();
//For all elements that need to be placed
for(size_t s=0;s<ordering.size() && remaining>0;s++){
assert(visited[s+1]<=visited_indicator);
//Ignore already-visited elements
if(visited[s+1]==visited_indicator)
continue;
//Don't rearrange if we don't have to
if(s==visited[s])
continue;
//Follow this cycle, putting elements in their places until we get back
//around. Use move semantics for speed.
auto temp = std::move(values[s]);
auto i = s;
for(;s!=(size_t)ordering[i];i=ordering[i],--remaining){
std::swap(temp, values[ordering[i]]);
visited[i+1] = visited_indicator;
}
std::swap(temp, values[s]);
visited[i+1] = visited_indicator;
}
}
///#brief Reorder a vector by moving its elements to indices indicted by another
/// vector. Takes O(2N) time and O(2N) space. Allocations are amoritzed.
///
///#param[in,out] values Vector to be reordered
///#param[in] ordering A permutation of the vector
///#param[in,out] visited A black-box vector to be reused between calls and
/// shared with with `forward_reorder()`
template<class ValueType, class OrderingType, class ProgressType>
void backward_reorder(
std::vector<ValueType> &values,
const std::vector<OrderingType> &ordering,
std::vector<ProgressType> &visited
){
//The orderings form a linked list. We need O(N) memory to reverse a linked
//list. We use `thread_local` so that the function is reentrant.
thread_local std::stack<OrderingType> stack;
if(ordering.size()!=values.size()){
throw std::runtime_error("ordering and values must be the same size!");
}
//Size the visited vector appropriately. Since vectors don't shrink, this will
//shortly become large enough to handle most of the inputs. The vector is 1
//larger than necessary because the first element is special.
if(visited.empty() || visited.size()-1<values.size());
visited.resize(values.size()+1);
//If the visitation indicator becomes too large, we reset everything. This is
//O(N) expensive, but unlikely to occur in most use cases if an appropriate
//data type is chosen for the visited vector. For instance, an unsigned 32-bit
//integer provides ~4B uses before it needs to be reset. We subtract one below
//to avoid having to think too much about off-by-one errors. Note that
//choosing the biggest data type possible is not necessarily a good idea!
//Smaller data types will have better cache utilization.
if(visited.at(0)==std::numeric_limits<ProgressType>::max()-1)
std::fill(visited.begin(), visited.end(), 0);
//We increment the stored visited indicator and make a note of the result. Any
//value in the visited vector less than `visited_indicator` has not been
//visited.
const auto visited_indicator = ++visited.at(0);
//For doing an early exit if we get everything in place
auto remaining = values.size();
//For all elements that need to be placed
for(size_t s=0;s<ordering.size() && remaining>0;s++){
assert(visited[s+1]<=visited_indicator);
//Ignore already-visited elements
if(visited[s+1]==visited_indicator)
continue;
//Don't rearrange if we don't have to
if(s==visited[s])
continue;
//The orderings form a linked list. We need to follow that list to its end
//in order to reverse it.
stack.emplace(s);
for(auto i=s;s!=(size_t)ordering[i];i=ordering[i]){
stack.emplace(ordering[i]);
}
//Now we follow the linked list in reverse to its beginning, putting
//elements in their places. Use move semantics for speed.
auto temp = std::move(values[s]);
while(!stack.empty()){
std::swap(temp, values[stack.top()]);
visited[stack.top()+1] = visited_indicator;
stack.pop();
--remaining;
}
visited[s+1] = visited_indicator;
}
}
int main(){
std::mt19937 gen;
std::uniform_int_distribution<short> value_dist(0,std::numeric_limits<short>::max());
std::uniform_int_distribution<short> len_dist (0,std::numeric_limits<short>::max());
std::vector<short> data;
std::vector<short> ordering;
std::vector<short> original;
std::vector<size_t> progress;
for(int i=0;i<1000;i++){
const int len = len_dist(gen);
data.clear();
ordering.clear();
for(int i=0;i<len;i++){
data.push_back(value_dist(gen));
ordering.push_back(i);
}
original = data;
std::shuffle(ordering.begin(), ordering.end(), gen);
forward_reorder(data, ordering, progress);
assert(original!=data);
backward_reorder(data, ordering, progress);
assert(original==data);
}
}
Never prematurely optimize. Meassure and then determine where you need to optimize and what. You can end with complex code that is hard to maintain and bug-prone in many places where performance is not an issue.
With that being said, do not early pessimize. Without changing the code you can remove half of your copies:
template <typename T>
void reorder( std::vector<T> & data, std::vector<std::size_t> const & order )
{
std::vector<T> tmp; // create an empty vector
tmp.reserve( data.size() ); // ensure memory and avoid moves in the vector
for ( std::size_t i = 0; i < order.size(); ++i ) {
tmp.push_back( data[order[i]] );
}
data.swap( tmp ); // swap vector contents
}
This code creates and empty (big enough) vector in which a single copy is performed in-order. At the end, the ordered and original vectors are swapped. This will reduce the copies, but still requires extra memory.
If you want to perform the moves in-place, a simple algorithm could be:
template <typename T>
void reorder( std::vector<T> & data, std::vector<std::size_t> const & order )
{
for ( std::size_t i = 0; i < order.size(); ++i ) {
std::size_t original = order[i];
while ( i < original ) {
original = order[original];
}
std::swap( data[i], data[original] );
}
}
This code should be checked and debugged. In plain words the algorithm in each step positions the element at the i-th position. First we determine where the original element for that position is now placed in the data vector. If the original position has already been touched by the algorithm (it is before the i-th position) then the original element was swapped to order[original] position. Then again, that element can already have been moved...
This algorithm is roughly O(N^2) in the number of integer operations and thus is theoretically worse in performance time as compare to the initial O(N) algorithm. But it can compensate if the N^2 swap operations (worst case) cost less than the N copy operations or if you are really constrained by memory footprint.
It's an interesting intellectual exercise to do the reorder with O(1) space requirement but in 99.9% of the cases the simpler answer will perform to your needs:
void permute(vector<T>& values, const vector<size_t>& indices)
{
vector<T> out;
out.reserve(indices.size());
for(size_t index: indices)
{
assert(0 <= index && index < values.size());
out.push_back(std::move(values[index]));
}
values = std::move(out);
}
Beyond memory requirements, the only way I can think of this being slower would be due to the memory of out being in a different cache page than that of values and indices.
You could do it recursively, I guess - something like this (unchecked, but it gives the idea):
// Recursive function
template<typename T>
void REORDER(int oldPosition, vector<T>& vA,
const vector<int>& vecNewOrder, vector<bool>& vecVisited)
{
// Keep a record of the value currently in that position,
// as well as the position we're moving it to.
// But don't move it yet, or we'll overwrite whatever's at the next
// position. Instead, we first move what's at the next position.
// To guard against loops, we look at vecVisited, and set it to true
// once we've visited a position.
T oldVal = vA[oldPosition];
int newPos = vecNewOrder[oldPosition];
if (vecVisited[oldPosition])
{
// We've hit a loop. Set it and return.
vA[newPosition] = oldVal;
return;
}
// Guard against loops:
vecVisited[oldPosition] = true;
// Recursively re-order the next item in the sequence.
REORDER(newPos, vA, vecNewOrder, vecVisited);
// And, after we've set this new value,
vA[newPosition] = oldVal;
}
// The "main" function
template<typename T>
void REORDER(vector<T>& vA, const vector<int>& newOrder)
{
// Initialise vecVisited with false values
vector<bool> vecVisited(vA.size(), false);
for (int x = 0; x < vA.size(); x++)
{
REORDER(x, vA, newOrder, vecVisited);
}
}
Of course, you do have the overhead of vecVisited. Thoughts on this approach, anyone?
To iterate through the vector is O(n) operation. Its sorta hard to beat that.
Your code is broken. You cannot assign to vA and you need to use template parameters.
vector<char> REORDER(const vector<char>& vA, const vector<size_t>& vOrder)
{
assert(vA.size() == vOrder.size());
vector<char> vCopy(vA.size());
for(int i = 0; i < vOrder.size(); ++i)
vCopy[i] = vA[ vOrder[i] ];
return vA;
}
The above is slightly more efficient.
It is not clear by the title and the question if the vector should be ordered with the same steps it takes to order vOrder or if vOrder already contains the indexes of the desired order.
The first interpretation has already a satisfying answer (see chmike and Potatoswatter), I add some thoughts about the latter.
If the creation and/or copy cost of object T is relevant
template <typename T>
void reorder( std::vector<T> & data, std::vector<std::size_t> & order )
{
std::size_t i,j,k;
for(i = 0; i < order.size() - 1; ++i) {
j = order[i];
if(j != i) {
for(k = i + 1; order[k] != i; ++k);
std::swap(order[i],order[k]);
std::swap(data[i],data[j]);
}
}
}
If the creation cost of your object is small and memory is not a concern (see dribeas):
template <typename T>
void reorder( std::vector<T> & data, std::vector<std::size_t> const & order )
{
std::vector<T> tmp; // create an empty vector
tmp.reserve( data.size() ); // ensure memory and avoid moves in the vector
for ( std::size_t i = 0; i < order.size(); ++i ) {
tmp.push_back( data[order[i]] );
}
data.swap( tmp ); // swap vector contents
}
Note that the two pieces of code in dribeas answer do different things.
I was trying to use #Potatoswatter's solution to sort multiple vectors by a third one and got really confused by output from using the above functions on a vector of indices output from Armadillo's sort_index. To switch from a vector output from sort_index (the arma_inds vector below) to one that can be used with #Potatoswatter's solution (new_inds below), you can do the following:
vector<int> new_inds(arma_inds.size());
for (int i = 0; i < new_inds.size(); i++) new_inds[arma_inds[i]] = i;
I came up with this solution which has the space complexity of O(max_val - min_val + 1), but it can be integrated with std::sort and benefits from std::sort's O(n log n) decent time complexity.
std::vector<int32_t> dense_vec = {1, 2, 3};
std::vector<int32_t> order = {1, 0, 2};
int32_t max_val = *std::max_element(dense_vec.begin(), dense_vec.end());
std::vector<int32_t> sparse_vec(max_val + 1);
int32_t i = 0;
for(int32_t j: dense_vec)
{
sparse_vec[j] = order[i];
i++;
}
std::sort(dense_vec.begin(), dense_vec.end(),
[&sparse_vec](int32_t i1, int32_t i2) {return sparse_vec[i1] < sparse_vec[i2];});
The following assumptions made while writing this code:
Vector values start from zero.
Vector does not contain repeated values.
We have enough memory to sacrifice in order to use std::sort
This should avoid copying the vector:
void REORDER(vector<char>& vA, const vector<size_t>& vOrder)
{
assert(vA.size() == vOrder.size());
for(int i = 0; i < vOrder.size(); ++i)
if (i < vOrder[i])
swap(vA[i], vA[vOrder[i]]);
}