合并矢量矢量到单个矢量中

Question

我有T矢量矢量:

std::vector<std::vector<T>> vector_of_vectors_of_T;

我想将它们全部合并到T的单个向量中:

std::vector<T> vector_of_T;

我目前正在使用这种方法:

size_t total_size{ 0 };
for (auto const& items: vector_of_vectors_of_T){
    total_size += items.size();
}
vector_of_T.reserve(total_size);
for (auto const& items: vector_of_vectors_of_T){
    vector_of_T.insert(end(vector_of_T), begin(items), end(items));
}

有更直接的方法吗？就像一个准备好的std函数？如果没有,是否有更有效的方法手动完成？

Answer 1

使用back_inserter和move;

size_t total_size{ 0 };
for (auto const& items: vector_of_vectors_of_T){
    total_size += items.size();
}

vector_of_T.reserve(total_size);
for (auto& items: vector_of_vectors_of_T){    
    std::move(items.begin(), items.end(), std::back_inserter(vector_of_T));
}

而不是copying,std::move给它一点性能增强.

Answer 2

尝试编写泛型是一个很好的练习join.下面的代码采用嵌套容器R1<R2<T>并返回一个连接容器R1<T>.请注意,由于标准库中的分配器参数,这有点麻烦.没有尝试检查分配器兼容性等.

幸运的是,action::joinEric Niebler即将推出的range-v3库中有这个功能,它已经非常强大,今天在Clang上工作:

#include <range/v3/all.hpp>
#include <algorithm>
#include <iostream>
#include <iterator>
#include <numeric>
#include <vector>

// quick prototype
template<template<class, class...> class R1, template<class, class...> class R2, class T, class... A1, class... A2>
auto join(R1<R2<T, A2...>, A1...> const& outer)
{
    R1<T, A2...> joined;
    joined.reserve(std::accumulate(outer.begin(), outer.end(), std::size_t{}, [](auto size, auto const& inner) {
        return size + inner.size();
    }));
    for (auto const& inner : outer)
        joined.insert(joined.end(), inner.begin(), inner.end());
    return joined;
}

int main()
{
    std::vector<std::vector<int>> v = { { 1, 2 }, { 3, 4 } };

    // quick prototype
    std::vector<int> w = join(v);
    std::copy(w.begin(), w.end(), std::ostream_iterator<int>(std::cout, ",")); std::cout << "\n";

    // Eric Niebler's range-v3
    std::vector<int> u = ranges::action::join(v);
    std::copy(u.begin(), u.end(), std::ostream_iterator<int>(std::cout, ",")); std::cout << "\n";
}

实例