如何根据对的第二个元素对对的向量进行排序？

Question

If I have a vector of pairs:

std::vector<std::pair<int, int> > vec;

Is there and easy way to sort the list in increasing order based on the second element of the pair?

I know I can write a little function object that will do the work, but is there a way to use existing parts of the STL and std::less to do the work directly?

EDIT: I understand that I can write a separate function or class to pass to the third argument to sort. The question is whether or not I can build it out of standard stuff. I'd really something that looks like:

std::sort(vec.begin(), vec.end(), std::something_magic<int, int, std::less>());

Answer 1

EDIT : using c++14, the best solution is very easy to write thanks to lambdas that can now have parameters of type auto . This is my current favorite solution

std::sort(v.begin(), v.end(), [](auto &left, auto &right) {
    return left.second < right.second;
});

---

ORIGINAL ANSWER :

Just use a custom comparator (it's an optional 3rd argument to std::sort )

struct sort_pred {
    bool operator()(const std::pair<int,int> &left, const std::pair<int,int> &right) {
        return left.second < right.second;
    }
};

std::sort(v.begin(), v.end(), sort_pred());

If you're using a C++11 compiler, you can write the same using lambdas:

std::sort(v.begin(), v.end(), [](const std::pair<int,int> &left, const std::pair<int,int> &right) {
    return left.second < right.second;
});

EDIT : in response to your edits to your question, here's some thoughts ... if you really wanna be creative and be able to reuse this concept a lot, just make a template:

template <class T1, class T2, class Pred = std::less<T2> >
struct sort_pair_second {
    bool operator()(const std::pair<T1,T2>&left, const std::pair<T1,T2>&right) {
        Pred p;
        return p(left.second, right.second);
    }
};

then you can do this too:

std::sort(v.begin(), v.end(), sort_pair_second<int, int>());

or even

std::sort(v.begin(), v.end(), sort_pair_second<int, int, std::greater<int> >());

Though to be honest, this is all a bit overkill, just write the 3 line function and be done with it :-P

Answer 2

You can use boost like this:

std::sort(a.begin(), a.end(), 
          boost::bind(&std::pair<int, int>::second, _1) <
          boost::bind(&std::pair<int, int>::second, _2));

I don't know a standard way to do this equally short and concise, but you can grab boost::bind it's all consisting of headers.

Answer 3

Its pretty simple you use the sort function from algorithm and add your own compare function

vector< pair<int,int > > v;
sort(v.begin(),v.end(),myComparison);

Now you have to make the comparison based on the second selection so declare you "myComparison" as

bool myComparison(const pair<int,int> &a,const pair<int,int> &b)
{
       return a.second<b.second;
}

Answer 4

With C++0x we can use lambda functions:

using namespace std;
vector<pair<int, int>> v;
        .
        .
sort(v.begin(), v.end(),
     [](const pair<int, int>& lhs, const pair<int, int>& rhs) {
             return lhs.second < rhs.second; } );

In this example the return type bool is implicitly deduced.

Lambda return types

When a lambda-function has a single statement, and this is a return-statement, the compiler can deduce the return type. From C++11, §5.1.2/4:

...

• If the compound-statement is of the form { return expression ; } { return expression ; } the type of the returned expression after lvalue-to-rvalue conversion (4.1), array-to-pointer conversion (4.2), and function-to-pointer conversion (4.3);
• otherwise, void .

To explicitly specify the return type use the form []() -> Type { } , like in:

sort(v.begin(), v.end(),
     [](const pair<int, int>& lhs, const pair<int, int>& rhs) -> bool {
             if (lhs.second == 0)
                 return true;
             return lhs.second < rhs.second; } );

Answer 5

For something reusable:

template<template <typename> class P = std::less >
struct compare_pair_second {
    template<class T1, class T2> bool operator()(const std::pair<T1, T2>& left, const std::pair<T1, T2>& right) {
        return P<T2>()(left.second, right.second);
    }
};

You can use it as

std::sort(foo.begin(), foo.end(), compare_pair_second<>());

or

std::sort(foo.begin(), foo.end(), compare_pair_second<std::less>());

Answer 6

您必须依赖非标准的select2nd

Answer 7

尝试交换对的元素，以便您可以正常使用std::sort() 。