如何根据另一个向量中的条件从向量中删除元素? [英] How to remove elements from a vector based on a condition in another vector?

问题描述

我有两个等长向量，我想根据其中一个向量的条件从中删除元素.应当对两者应用相同的删除操作，以使索引匹配.

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.

我想出了一个使用 std :: erase 的解决方案，但是它非常慢:

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--;
    }
}

有没有一种方法可以更有效地执行此操作，并且最好使用stl算法?

Is there a way that I can do this more efficiently and preferably using stl algorithms?

推荐答案

如果顺序无关紧要，您可以将元素交换到向量的后面，然后将其弹出.

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;
}

如果您必须保持顺序，则可以使用 std :: remove_if .请参见此答案如何获取删除谓词中被取消引用的元素的索引:

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));

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