How do I delete a tuple within a vector? C++

Question

Been programming for 1 week. I have vector<tuple<string, string, T>> edgesVector. Which contains two strings and an INT. How do I delete a tuple within the vector that contains two specific strings?


remove_edge(const string& u, const string& v) {

    for (int i = 0; i<edgesVector.size(); i++ ) {
        if (get<0>(edgesVector[i]) == u && get<1>(edgesVector[i]) == v){
            edgesVector.erase(i);
        }
        if (get<1>(edgesVector[i]) == u && get<0>(edgesVector[i])== v){
            edgesVector.erase(i);
        }

    }           


}

What relation does this code have to your question? Is that your attempt? Does it not work? Why not? Please clarify. — underscore_d, Apr 26 '21 at 12:02
You should use the [erase remove idiom](https://stackoverflow.com/questions/347441/erasing-elements-from-a-vector) to remove the elements from your vector — Cory Kramer, Apr 26 '21 at 12:03
you don't. The vector manages the lifetime of its elements for you. — 463035818_is_not_an_ai, Apr 26 '21 at 12:03
Hello, welcome to SO. Please always post a [minimal and reproducible example](https://stackoverflow.com/help/minimal-reproducible-example) — Superlokkus, Apr 26 '21 at 12:04
[std::remove_if](https://en.cppreference.com/w/cpp/algorithm/remove) — Aykhan Hagverdili, Apr 26 '21 at 12:25
Should be marked as a dupe: https://stackoverflow.com/q/67185327/6119582 — sweenish, Apr 26 '21 at 12:53

Caleth · Answer 1 · 2021-04-26T12:36:32.630

3

You use std::remove_if, and pass the result to edgesVector.erase. This is called the erase-remove idiom.

void remove_edge(const string& u, const string& v) {
    auto it = std::remove_if(edgesVector.begin(), edgesVector.end(), [&](auto & tup){ return (get<0>(tup) == u) && (get<1>(tup) == v); }); // removes elements that match u, v
    it = std::remove_if(edgesVector.begin(), it, [&](auto & tup){ return (get<0>(tup) == v) && (get<1>(tup) == u); }); // removes elements that match v, u
    edgesVector.erase(it, edgesVector.end()); // resize the vector to only hold live elements
}

edited Apr 26 '21 at 12:36

answered Apr 26 '21 at 12:25

Caleth

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How about doing both tests in the same lambda? That'd make more sense – Aykhan Hagverdili Apr 26 '21 at 12:52
@AyxanHaqverdili that makes the lambda more than twice as complicated. I prefer this way for clarity. – Caleth Sep 18 '21 at 16:21
It seems less complicated to me as we don't need to keep track of an end iterator and we don't need the `remove_if` boilerplate twice: https://gcc.godbolt.org/z/v4h8jcnKh – Aykhan Hagverdili Sep 19 '21 at 05:09

acraig5075 · Answer 2 · 2021-04-26T14:37:33.857

0

The erase-remove idiom commented by others is the way to go, but the problem with your snippet is easily solved.

std::vector::erase takes an iterator as a parameter whereas you're passing it an index.

for (auto itr = edgesVector.begin(); itr != edgesVector.end();) {
    if ((get<0>(*itr) == u && get<1>(*itr) == v) ||
        (get<1>(*itr) == u && get<0>(*itr)== v))
        itr = edgesVector.erase(itr);
    else
        ++itr;
    }
}

edited Apr 26 '21 at 14:37

answered Apr 26 '21 at 12:34

acraig5075

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This only removes one element. OP may want to remove all the matching elements – Caleth Apr 26 '21 at 12:34
1

erase returns a new iterator, to the next element of the erased. So the algorithm can erase multiple elements, by some adjustments. Use `itr = edgesVector.erase(itr);`. To not skip elements, you need to increment itr only if the element is NOT erased – king_nak Apr 26 '21 at 14:30

How do I delete a tuple within a vector? C++

2 Answers2