Find all matching elements in std::list - c++

I was wondering if there's any built-in or well-established way (i.e. via lambda) to go through the elements of an std::list and find all the ones that match a given value? I know I can iterate through all of them, but I thought I'd ask if there's a way to get an iterator that iterates through just the elements that match a given criteria? My sample below only gives me the iterator to the first matching element.
#include <list>
#include <algorithm>
#include <stdio.h>
int main()
{
std::list<int> List;
List.push_back(100);
List.push_back(200);
List.push_back(300);
List.push_back(100);
int findValue = 100;
auto it = std::find_if(List.begin(), List.end(), [findValue](const int value)
{
return (value == findValue);
});
if (it != List.end())
{
for (; it != List.end(); ++it)
{
printf("%d\n", * it);
}
}
return 0;
}
Thanks for any feedback.

Updated answer
With the advent of C++20 just around the corner, the standard library has now introduced the concept of ranges which comes with view adapters and are simply lazy views over collections and their transformations.
This means you can now have an "iterator" which can be used to obtain a filtered and transformed view of an underlying container/collection, without having to create several iterators or even allocate memory.
Having said that, this is a way to create a view over just the filtered elements of your list:
// List is your std::list
auto matching_100 = List | std::views::filter([](auto &v) {
return v == 100;
});
How sweet is that? All you need to use all that?
#include <ranges>
Try it out
Previous answer
Using copy_if and iterators:
#include <list>
#include <algorithm>
#include <iterator>
#include <iostream>
int main()
{
std::list<int> List;
List.push_back(100);
List.push_back(200);
List.push_back(300);
List.push_back(100);
int findValue = 100;
std::copy_if(List.begin(), List.end(), std::ostream_iterator<int>(std::cout, "\n"), [&](int v) {
return v == findValue;
});
return 0;
}
If you don't want to directly output the results and want to fill another container with the matches:
std::vector<int> matches;
std::copy_if(List.begin(), List.end(), std::back_inserter(matches), [&](int v) {
return v == findValue;
});

boost::filter_iterator allows you to work with only the elements of a iterable that satisfy a predicate. Given a predicate Pred and a container Cont,
auto begin_iter = boost::make_filter_iterator(Pred, std::begin(Cont), std::end(Cont));
auto end_iter = boost::make_filter_iterator(Pred, std::end(Cont), std::end(Cont));
You can now use begin_iter and end_iter as if they were the begin and end iterators of a container containing only those elements of Cont that satisfied Pred. Another added advantage is that you can wrap the iterators in a boost::iterator_range and use it in places which expect a iterable object, like a range-based for loop like this:
auto range = boost::make_iterator_range(begin_iter, end_iter);
for(auto x : range) do_something(x);
In particular, setting Pred to a functor(could be a lambda) that checks for equality with your fixed value will give you the iterators you need.

std::find_if is a generalisation of std::find for when you need a function to check for the elements you want, rather than a simple test for equality. If you just want to do a simple test for equality then there's no need for the generalised form, and the lambda just adds complexity and verbosity. Just use std::find(begin, end, findValue) instead:
std::vector<std::list<int>::const_iterator> matches;
auto i = list.begin(), end = list.end();
while (i != end)
{
i = std::find(i, end, findValue);
if (i != end)
matches.push_back(i++);
}
But rather than calling find in a loop I'd just write the loop manually:
std::vector<std::list<int>::const_iterator> matches;
for (auto i = list.begin(), toofar = list.end(); i != toofar; ++i)
if (*i == findValue)
matches.push_back(i);

std::partition lets you simply move all elements matching the predicate to the front of the container (first partition). The return value is an iterator pointing to the first element of the second partition (containing the non matching elements). That's pretty much all you need to "filter" a container.

Related

Erasing many vector elements while going through it with 'auto'

Let's say that I have vector of pairs, where each pair corresponds to indexes (row and column) of certain matrix I am working on
using namespace std;
vector<pair<int, int>> vec;
I wanted to, using auto, go through the whole vector and delete at once all the pairs that fulfill certain conditions, for example something like
for (auto& x : vec) {
if (x.first == x.second) {
vec.erase(x);
}
}
but it doesn't work, as I suppose vec.erase() should have an iterator as an argument and x is actually a pair that is an element of vector vec, not iterator. I tried to modify it in few ways, but I am not sure how going through container elements with auto exactly works and how can I fix this.
Can I easily modify the code above to make it work and to erase multiple elements of vector, while going through it with auto? Or I should modify my approach?
For now it's just a vector of pairs, but it will be much worse later on, so I would like to use auto for simplicity.
vector::erase() invalidates any outstanding iterators, including the one your range based for loop is using. Use std::remove_if():
vec.erase(
std::remove_if(
vec.begin(),
vec.end(),
[](const pair<int,int> &xx) { return xx.first == xx.second; }
), vec.end()
);
std::remove_if() swaps the elements to the end of the vector and then you can safely erase them.
I would prefer something like this:
pair<int, int> pair = nullptr;
auto iter = vec.begin();
while(iter != vec.end()){
pair = (*iter);
if(pair.first == pair.second){
iter = this->vec.erase(iter);
}else{
++iter;
}
}

Want to delete a vector from another vector in C++ [duplicate]

I want to clear a element from a vector using the erase method. But the problem here is that the element is not guaranteed to occur only once in the vector. It may be present multiple times and I need to clear all of them. My code is something like this:
void erase(std::vector<int>& myNumbers_in, int number_in)
{
std::vector<int>::iterator iter = myNumbers_in.begin();
std::vector<int>::iterator endIter = myNumbers_in.end();
for(; iter != endIter; ++iter)
{
if(*iter == number_in)
{
myNumbers_in.erase(iter);
}
}
}
int main(int argc, char* argv[])
{
std::vector<int> myNmbers;
for(int i = 0; i < 2; ++i)
{
myNmbers.push_back(i);
myNmbers.push_back(i);
}
erase(myNmbers, 1);
return 0;
}
This code obviously crashes because I am changing the end of the vector while iterating through it. What is the best way to achieve this? I.e. is there any way to do this without iterating through the vector multiple times or creating one more copy of the vector?
Use the remove/erase idiom:
std::vector<int>& vec = myNumbers; // use shorter name
vec.erase(std::remove(vec.begin(), vec.end(), number_in), vec.end());
What happens is that remove compacts the elements that differ from the value to be removed (number_in) in the beginning of the vector and returns the iterator to the first element after that range. Then erase removes these elements (whose value is unspecified).
Edit: While updating a dead link I discovered that starting in C++20 there are freestanding std::erase and std::erase_if functions that work on containers and simplify things considerably.
Calling erase will invalidate iterators, you could use:
void erase(std::vector<int>& myNumbers_in, int number_in)
{
std::vector<int>::iterator iter = myNumbers_in.begin();
while (iter != myNumbers_in.end())
{
if (*iter == number_in)
{
iter = myNumbers_in.erase(iter);
}
else
{
++iter;
}
}
}
Or you could use std::remove_if together with a functor and std::vector::erase:
struct Eraser
{
Eraser(int number_in) : number_in(number_in) {}
int number_in;
bool operator()(int i) const
{
return i == number_in;
}
};
std::vector<int> myNumbers;
myNumbers.erase(std::remove_if(myNumbers.begin(), myNumbers.end(), Eraser(number_in)), myNumbers.end());
Instead of writing your own functor in this case you could use std::remove:
std::vector<int> myNumbers;
myNumbers.erase(std::remove(myNumbers.begin(), myNumbers.end(), number_in), myNumbers.end());
In C++11 you could use a lambda instead of a functor:
std::vector<int> myNumbers;
myNumbers.erase(std::remove_if(myNumbers.begin(), myNumbers.end(), [number_in](int number){ return number == number_in; }), myNumbers.end());
In C++17 std::experimental::erase and std::experimental::erase_if are also available, in C++20 these are (finally) renamed to std::erase and std::erase_if (note: in Visual Studio 2019 you'll need to change your C++ language version to the latest experimental version for support):
std::vector<int> myNumbers;
std::erase_if(myNumbers, Eraser(number_in)); // or use lambda
or:
std::vector<int> myNumbers;
std::erase(myNumbers, number_in);
You can iterate using the index access,
To avoid O(n^2) complexity
you can use two indices, i - current testing index, j - index to
store next item and at the end of the cycle new size of the vector.
code:
void erase(std::vector<int>& v, int num)
{
size_t j = 0;
for (size_t i = 0; i < v.size(); ++i) {
if (v[i] != num) v[j++] = v[i];
}
// trim vector to new size
v.resize(j);
}
In such case you have no invalidating of iterators, complexity is O(n), and code is very concise and you don't need to write some helper classes, although in some case using helper classes can benefit in more flexible code.
This code does not use erase method, but solves your task.
Using pure stl you can do this in the following way (this is similar to the Motti's answer):
#include <algorithm>
void erase(std::vector<int>& v, int num) {
vector<int>::iterator it = remove(v.begin(), v.end(), num);
v.erase(it, v.end());
}
Depending on why you are doing this, using a std::set might be a better idea than std::vector.
It allows each element to occur only once. If you add it multiple times, there will only be one instance to erase anyway. This will make the erase operation trivial.
The erase operation will also have lower time complexity than on the vector, however, adding elements is slower on the set so it might not be much of an advantage.
This of course won't work if you are interested in how many times an element has been added to your vector or the order the elements were added.
There are std::erase and std::erase_if since C++20 which combines the remove-erase idiom.
std::vector<int> nums;
...
std::erase(nums, targetNumber);
or
std::vector<int> nums;
...
std::erase_if(nums, [](int x) { return x % 2 == 0; });
If you change your code as follows, you can do stable deletion.
void atest(vector<int>& container,int number_in){
for (auto it = container.begin(); it != container.end();) {
if (*it == number_in) {
it = container.erase(it);
} else {
++it;
}
}
}
However, a method such as the following can also be used.
void btest(vector<int>& container,int number_in){
container.erase(std::remove(container.begin(), container.end(), number_in),container.end());
}
If we must preserve our sequence’s order (say, if we’re keeping it sorted by some interesting property), then we can use one of the above. But if the sequence is just a bag of values whose order we don’t care about at all, then we might consider moving single elements from the end of the sequence to fill each new gap as it’s created:
void ctest(vector<int>& container,int number_in){
for (auto it = container.begin(); it != container.end(); ) {
if (*it == number_in) {
*it = std::move(container.back());
container.pop_back();
} else {
++it;
}
}
}
Below are their benchmark results:
CLang 15.0:
Gcc 12.2:

How to get a certain element in a list, given the position?

So I've got a list:
list<Object> myList;
myList.push_back(Object myObject);
I'm not sure but I'm confident that this would be the "0th" element in the array.
Is there any function I can use that will return "myObject"?
Object copy = myList.find_element(0);
?
If you frequently need to access the Nth element of a sequence, std::list, which is implemented as a doubly linked list, is probably not the right choice. std::vector or std::deque would likely be better.
That said, you can get an iterator to the Nth element using std::advance:
std::list<Object> l;
// add elements to list 'l'...
unsigned N = /* index of the element you want to retrieve */;
if (l.size() > N)
{
std::list<Object>::iterator it = l.begin();
std::advance(it, N);
// 'it' points to the element at index 'N'
}
For a container that doesn't provide random access, like std::list, std::advance calls operator++ on the iterator N times. Alternatively, if your Standard Library implementation provides it, you may call std::next:
if (l.size() > N)
{
std::list<Object>::iterator it = std::next(l.begin(), N);
}
std::next is effectively wraps a call to std::advance, making it easier to advance an iterator N times with fewer lines of code and fewer mutable variables. std::next was added in C++11.
std::list doesn't provide any function to get element given an index. You may try to get it by writing some code, which I wouldn't recommend, because that would be inefficient if you frequently need to do so.
What you need is : std::vector. Use it as:
std::vector<Object> objects;
objects.push_back(myObject);
Object const & x = objects[0]; //index isn't checked
Object const & y = objects.at(0); //index is checked
std::list<Object> l;
std::list<Object>::iterator ptr;
int i;
for( i = 0 , ptr = l.begin() ; i < N && ptr != l.end() ; i++ , ptr++ );
if( ptr == l.end() ) {
// list too short
} else {
// 'ptr' points to N-th element of list
}
Maybe not the most efficient way. But you could convert the list into a vector.
#include <list>
#include <vector>
list<Object> myList;
vector<Object> myVector(myList.begin(), myList.end());
Then access the vector using the [x] operator.
auto x = MyVector[0];
You could put that in a helper function:
#include <memory>
#include <vector>
#include <list>
template<class T>
shared_ptr<vector<T>>
ListToVector(list<T> List) {
shared_ptr<vector<T>> Vector {
new vector<string>(List.begin(), List.end()) }
return Vector;
}
Then use the helper funciton like this:
auto MyVector = ListToVector(Object);
auto x = MyVector[0];
Not very efficient, but if you must use a list, you can deference the iterator
*myList.begin()+N

remove_if equivalent for std::map

I was trying to erase a range of elements from map based on particular condition. How do I do it using STL algorithms?
Initially I thought of using remove_if but it is not possible as remove_if does not work for associative container.
Is there any "remove_if" equivalent algorithm which works for map ?
As a simple option, I thought of looping through the map and erase. But is looping through the map and erasing a safe option?(as iterators get invalid after erase)
I used following example:
bool predicate(const std::pair<int,std::string>& x)
{
return x.first > 2;
}
int main(void)
{
std::map<int, std::string> aMap;
aMap[2] = "two";
aMap[3] = "three";
aMap[4] = "four";
aMap[5] = "five";
aMap[6] = "six";
// does not work, an error
// std::remove_if(aMap.begin(), aMap.end(), predicate);
std::map<int, std::string>::iterator iter = aMap.begin();
std::map<int, std::string>::iterator endIter = aMap.end();
for(; iter != endIter; ++iter)
{
if(Some Condition)
{
// is it safe ?
aMap.erase(iter++);
}
}
return 0;
}
Almost.
for(; iter != endIter; ) {
if (Some Condition) {
iter = aMap.erase(iter);
} else {
++iter;
}
}
What you had originally would increment the iterator twice if you did erase an element from it; you could potentially skip over elements that needed to be erased.
This is a common algorithm I've seen used and documented in many places.
[EDIT] You are correct that iterators are invalidated after an erase, but only iterators referencing the element that is erased, other iterators are still valid. Hence using iter++ in the erase() call.
erase_if for std::map (and other containers)
I use the following template for this very thing.
namespace stuff {
template< typename ContainerT, typename PredicateT >
void erase_if( ContainerT& items, const PredicateT& predicate ) {
for( auto it = items.begin(); it != items.end(); ) {
if( predicate(*it) ) it = items.erase(it);
else ++it;
}
}
}
This won't return anything, but it will remove the items from the std::map.
Usage example:
// 'container' could be a std::map
// 'item_type' is what you might store in your container
using stuff::erase_if;
erase_if(container, []( item_type& item ) {
return /* insert appropriate test */;
});
Second example (allows you to pass in a test value):
// 'test_value' is value that you might inject into your predicate.
// 'property' is just used to provide a stand-in test
using stuff::erase_if;
int test_value = 4; // or use whatever appropriate type and value
erase_if(container, [&test_value]( item_type& item ) {
return item.property < test_value; // or whatever appropriate test
});
Now, std::experimental::erase_if is available in header <experimental/map>.
See: http://en.cppreference.com/w/cpp/experimental/map/erase_if
Here is some elegant solution.
for (auto it = map.begin(); it != map.end();)
{
(SomeCondition) ? map.erase(it++) : (++it);
}
For those on C++20 there are built-in std::erase_if functions for map and unordered_map:
std::unordered_map<int, char> data {{1, 'a'},{2, 'b'},{3, 'c'},{4, 'd'},
{5, 'e'},{4, 'f'},{5, 'g'},{5, 'g'}};
const auto count = std::erase_if(data, [](const auto& item) {
auto const& [key, value] = item;
return (key & 1) == 1;
});
I got this documentation from the excellent SGI STL reference:
Map has the important property that
inserting a new element into a map
does not invalidate iterators that
point to existing elements. Erasing an
element from a map also does not
invalidate any iterators, except, of
course, for iterators that actually
point to the element that is being
erased.
So, the iterator you have which is pointing at the element to be erased will of course be invalidated. Do something like this:
if (some condition)
{
iterator here=iter++;
aMap.erase(here)
}
The original code has only one issue:
for(; iter != endIter; ++iter)
{
if(Some Condition)
{
// is it safe ?
aMap.erase(iter++);
}
}
Here the iter is incremented once in the for loop and another time in erase, which will probably end up in some infinite loop.
From the bottom notes of:
http://www.sgi.com/tech/stl/PairAssociativeContainer.html
a Pair Associative Container cannot provide mutable iterators (as defined in the Trivial Iterator requirements), because the value type of a mutable iterator must be Assignable, and pair is not Assignable. However, a Pair Associative Container can provide iterators that are not completely constant: iterators such that the expression (*i).second = d is valid.
First
Map has the important property that inserting a new element into a map does not invalidate iterators that point to existing elements. Erasing an element from a map also does not invalidate any iterators, except, of course, for iterators that actually point to the element that is being erased.
Second, the following code is good
for(; iter != endIter; )
{
if(Some Condition)
{
aMap.erase(iter++);
}
else
{
++iter;
}
}
When calling a function, the parameters are evaluated before the call to that function.
So when iter++ is evaluated before the call to erase, the ++ operator of the iterator will return the current item and will point to the next item after the call.
IMHO there is no remove_if() equivalent.
You can't reorder a map.
So remove_if() can not put your pairs of interest at the end on which you can call erase().
Based on Iron Savior's answer For those that would like to provide a range more along the lines of std functional taking iterators.
template< typename ContainerT, class FwdIt, class Pr >
void erase_if(ContainerT& items, FwdIt it, FwdIt Last, Pr Pred) {
for (; it != Last; ) {
if (Pred(*it)) it = items.erase(it);
else ++it;
}
}
Curious if there is some way to lose the ContainerT items and get that from the iterator.
Steve Folly's answer I feel the more efficient.
Here is another easy-but-less efficient solution:
The solution uses remove_copy_if to copy the values we want into a new container, then swaps the contents of the original container with those of the new one:
std::map<int, std::string> aMap;
...
//Temporary map to hold the unremoved elements
std::map<int, std::string> aTempMap;
//copy unremoved values from aMap to aTempMap
std::remove_copy_if(aMap.begin(), aMap.end(),
inserter(aTempMap, aTempMap.end()),
predicate);
//Swap the contents of aMap and aTempMap
aMap.swap(aTempMap);
If you want to erase all elements with key greater than 2, then the best way is
map.erase(map.upper_bound(2), map.end());
Works only for ranges though, not for any predicate.
I use like this
std::map<int, std::string> users;
for(auto it = users.begin(); it <= users.end()) {
if(<condition>){
it = users.erase(it);
} else {
++it;
}
}

Erasing items from an STL list

I want to make a function which moves items from one STL list to another if they match a certain condition.
This code is not the way to do it. The iterator will most likely be invalidated by the erase() function and cause a problem:
for(std::list<MyClass>::iterator it = myList.begin(); it != myList.end(); it++)
{
if(myCondition(*it))
{
myOtherList.push_back(*it);
myList.erase(it);
}
}
So can anyone suggest a better way to do this ?
Erase returns an iterator pointing to the element after the erased one:
std::list<MyClass>::iterator it = myList.begin();
while (it != myList.end())
{
if(myCondition(*it))
{
myOtherList.push_back(*it);
it = myList.erase(it);
}
else
{
++it;
}
}
STL lists have an interesting feature: the splice() method lets you destructively move elements from one list to another.
splice() operates in constant time, and doesn't copy the elements or perform any free store allocations/deallocations. Note that both lists must be of the same type, and they must be separate list instances (not two references to the same list).
Here's an example of how you could use splice():
for(std::list<MyClass>::iterator it = myList.begin(); it != myList.end(); ) {
if(myCondition(*it)) {
std::list<MyClass>::iterator oldIt = it++;
myOtherList.splice(myOtherList.end(), myList, oldIt);
} else {
++it;
}
}
Solution 1
template<typename Fwd, typename Out, typename Operation>
Fwd move_if(Fwd first, Fwd last, Out result, Operation op)
{
Fwd swap_pos = first;
for( ; first != last; ++first ) {
if( !op(*first) ) *swap_pos++ = *first;
else *result++ = *first;
}
return swap_pos;
}
The idea is simple. What you want to do is remove elements from one container and place them in another if a predicate is true. So take the code of the std::remove() algorithm, which already does the remove part, and adapt it to your extra needs. In the code above I added the else line to copy the element when the predicate is true.
Notice that because we use the std::remove() code, the algorithm doesn't actually shrink the input container. It does return the updated end iterator of the input container though, so you can just use that and disregard the extra elements. Use the erase-remove idiom if you really want to shrink the input container.
Solution 2
template<typename Bidi, typename Out, typename Operation>
Bidi move_if(Bidi first, Bidi last, Out result, Operation op)
{
Bidi new_end = partition(first, last, not1(op));
copy(new_end, last, result);
return new_end;
}
The second approach uses the STL to implement the algorithm. I personally find it more readable than the first solution, but it has two drawbacks: First, it requires the more-powerful bidirectional iterators for the input container, rather than the forward iterators we used in the first solution. Second, and this is may or may not be an issue for you, the containers are not guaranteed to have the same ordering as before the call to std::partition(). If you wish to maintain the ordering, replace that call with a call to std::stable_partition(). std::stable_partition() might be slightly slower, but it has the same runtime complexity as std::partition().
Either Way: Calling the Function
list<int>::iterator p = move_if(l1.begin(), l1.end(),
back_inserter(l2),
bind2nd(less<int>(), 3));
Final Remarks
While writing the code I encountered a dilemma: what should the move_if() algorithm return? On the one hand the algorithm should return an iterator pointing to the new end position of the input container, so the caller can use the erase-remove idiom to shrink the container. But on the other hand the algorithm should return the position of the end of the result container, because otherwise it could be expensive for the caller to find it. In the first solution the result iterator points to this position when the algorithm ends, while in the second solution it is the iterator returned by std::copy() that points to this position. I could return a pair of iterators, but for the sake of making things simple I just return one of the iterators.
std::list<MyClass>::iterator endMatching =
partition(myList.begin(), myList.end(), myCondition);
myOtherList.splice(myOtherList.begin(), myList, endMatching, myList.end());
Note that partition() gives you enough to discriminate matching objects from non matching ones.
(list::splice() is cheap however)
See the following code on a concrete case inspired from
Now to remove elements that match a predicate?
#include <iostream>
#include <iterator>
#include <list>
#include <string>
#include <algorithm>
#include <functional>
using namespace std;
class CPred : public unary_function<string, bool>
{
public:
CPred(const string& arString)
:mString(arString)
{
}
bool operator()(const string& arString) const
{
return (arString.find(mString) == std::string::npos);
}
private:
string mString;
};
int main()
{
list<string> Strings;
Strings.push_back("213");
Strings.push_back("145");
Strings.push_back("ABC");
Strings.push_back("167");
Strings.push_back("DEF");
cout << "Original list" << endl;
copy(Strings.begin(), Strings.end(),ostream_iterator<string>(cout,"\n"));
CPred Pred("1");
// Linear. Exactly last - first applications of pred, and at most (last - first)/2 swaps.
list<string>::iterator end1 =
partition(Strings.begin(), Strings.end(), Pred);
list<string> NotMatching;
// This function is constant time.
NotMatching.splice(NotMatching.begin(),Strings, Strings.begin(), end1);
cout << "Elements matching with 1" << endl;
copy(Strings.begin(), Strings.end(), ostream_iterator<string>(cout,"\n"));
cout << "Elements not matching with 1" << endl;
copy(NotMatching.begin(), NotMatching.end(), ostream_iterator<string>(cout,"\n"));
return 0;
}
Another attempt:
for(std::list<MyClass>::iterator it = myList.begin(); it != myList.end; ) {
std::list<MyClass>::iterator eraseiter = it;
++it;
if(myCondition(*eraseiter)) {
myOtherList.push_back(*eraseiter);
myList.erase(eraseiter);
}
}
template <typename ForwardIterator, typename OutputIterator, typename Predicate>
void splice_if(ForwardIterator begin, ForwardIterator end, OutputIterator out, Predicate pred)
{
ForwardIterator it = begin;
while( it != end )
{
if( pred(*it) )
{
*begin++ = *out++ = *it;
}
++it;
}
return begin;
}
myList.erase(
splice_if( myList.begin(), myList.end(), back_inserter(myOutputList),
myCondition
),
myList.end()
)