Removing by index from a C++ vector using remove_i

问题:

We can use remove_if in C++ to remove elements from a vector in linear time based on a predicate that operates on the elements.

bool condition(double d) {...}

vector<double> data = ...
std::remove_if (data.begin(), data.end(), condition);

What if my condition depends not on the values, but on the indices? In other words, if I wanted to remove all the odd-indexed elements, or some arbitrary index set, etc?

bool condition(int index) {//returns whether this index should be removed}

vector<double> data = ...
std::remove_if (data.begin(), data.end(), ???);

回答1:

You can use pointer arithmetic to find the index of a specific element that std::remove_if passes to the predicate:

std::remove_if(data.begin(), data.end(),
               [](const double& d) { return (&d - &*data.begin()) % 2); });

Note that remove_if passes the result of dereferencing an iterator, and that's guaranteed to be a reference per Table 106 - Iterator requirements in the Standard.

回答2:

I actually made an account only for this. Use awesomeyi answer. Is way cleaner.

int count = 0;
auto final = std::remove_if (data.begin(), data.end(), [&count](const double d) {
    return (count++) % 2;
});

The standard does say that the predicate is applied exactly last - first times. And remove_if works with ForwardIterators.

This implies that the predicate is applied only once in the same order they originally appear in the sequence.

Unless of course, the library is trolling you, by keeping internal copies of the ForwardIterator.

回答3:

Take advantage of the fact that lambas can capture variables. A quick example:

vector<double> data = {5, 3, 6, 7, 8};

int count = 0;
auto final = std::remove_if (data.begin(), data.end(), [&](const double d) {
    bool b = false;
    if(count % 2) b = true;
    ++count;
    return b;
});

for(auto beg = data.begin(); beg != final; ++beg)
    cout << *beg << endl;

Code will print: 5 6 8