如何根据另一个向量的排序对向量进行排序? [英] How can I sort a vector based on sort of another vector?

问题描述

我有四个向量，我想对它们进行相互排序.

I have a four vectors that I want to sort in relation to each other.

vector<string> color;
vector<string> shape;
vector<int> size;

每个向量的大小相同，每个向量元素相互关联，形成一行

Each vector is the same size each vector element is tied to each other such that it forms a row

{color[0], shape[0], size[0]}
{color[1], shape[1], size[1]}
{color[2], shape[2], size[2]}, etc

所以我想要做的是按颜色对颜色向量进行排序，并根据重新排列的颜色向量对其他两个向量进行排序.然后在每组颜色(即红色)中，我想按形状排序并根据该排序重新排列大小向量.最后我想对每组颜色和形状中的大小向量进行排序.我想我知道该怎么做，但是感觉非常非常混乱并且难以概念化/阅读(我还是 C++ 的新手).有没有简单的方法来完成这样的事情?

So what I am trying to do is sort the color vector by color and sort the other two vectors based on the rearranged color vector. Then within every group of colors (i.e red) I want to sort by shape and rearrange the size vector based on that sort. And then finally I want to sort the size vector within each group of color and shape. I think I know how to do it but it feels very very messy and difficult to conceptualize/read (I'm still new to C++). Is there an easy way to accomplish something like this?

例如我想做这样的事情:

For example I want to do something like this:

Blue    Circle      1   ->    Red   Triangle    1           
Red     Triangle    1   ->    Red   Triangle    2
Blue    Circle      3   ->    Red   Triangle    3
Red     Triangle    3   ->    Red   Circle      1
Red     Circle      2   ->    Red   Circle      2
Blue    Triangle    1   ->    Red   Circle      3
Red     Circle      1   ->    Blue  Triangle    1
Blue    Triangle    3   ->    Blue  Triangle    2
Red     Circle      3   ->    Blue  Triangle    3
Blue    Circle      2   ->    Blue  Circle      1
Blue    Triangle    2   ->    Blue  Circle      2
Red     Triangle    2   ->    Blue  Circle      3

推荐答案

你应该制作一个 Shape 类或结构体，然后制作一个 std::vector，然后您可以对颜色进行排序，主要是其他参数.您定义了一个重载的 operator< 以便 std::sort 函数能够找到它.

You should make a single Shape class or struct, and then make a std::vector<Shape>, which you can then sort on color, primarily, followed by your other parameters. You define an overloaded operator< so that the std::sort function will find it.

它看起来像这样:

#include <algorithm>
#include <iostream>
#include <string>
#include <tuple>
#include <vector>

struct Shape
{
    std::string color_;
    std::string shape_;
    int size_;

    Shape(const std::string& color, const std::string& shape, int size)
        : color_(color)
        , shape_(shape)
        , size_(size)
    {}

    // returns true if this shape is less than the other shape
    // "less than" is up to us: here we give priority to color, then shape, then size
    bool operator<(const Shape& other) const
    {
        // std::tie makes lexicographical compare of complex structures easy!
        return (std::tie(color_, shape_, size_) <
                std::tie(other.color_, other.shape_, other.size_));
    }

    friend std::ostream& operator<<(std::ostream& os, const std::vector<Shape>& shapes)
    {
        for (auto& shape : shapes)
        {
            os << shape.color_ << " " << shape.shape_ << " " << shape.size_ << "\n";
        }

        return os;
    }
};

int main(int argc, char** argv)
{
    std::vector<Shape> shapes;

    shapes.emplace_back("Blue", "Circle", 1);
    shapes.emplace_back("Red", "Triangle", 1);
    shapes.emplace_back("Blue", "Circle", 3);
    shapes.emplace_back("Red", "Triangle", 3);
    shapes.emplace_back("Red", "Circle", 2);
    shapes.emplace_back("Blue", "Triangle", 1);
    shapes.emplace_back("Red", "Circle", 1);
    shapes.emplace_back("Blue", "Triangle", 3);
    shapes.emplace_back("Red", "Circle", 3);
    shapes.emplace_back("Blue", "Circle", 2);
    shapes.emplace_back("Blue", "Triangle", 2);
    shapes.emplace_back("Red", "Triangle", 2);

    std::cout << "Pre sorted vector:\n";
    std::cout << shapes;

    // std::sort by default will use the operator< for the types
    // being sorted, if it's available
    std::sort(shapes.begin(), shapes.end());

    std::cout << "\nPost sorted vector:\n";
    std::cout << shapes;
}

这给出了输出:

Pre sorted vector:
Blue Circle 1
Red Triangle 1
Blue Circle 3
Red Triangle 3
Red Circle 2
Blue Triangle 1
Red Circle 1
Blue Triangle 3
Red Circle 3
Blue Circle 2
Blue Triangle 2
Red Triangle 2

Post sorted vector:
Blue Circle 1
Blue Circle 2
Blue Circle 3
Blue Triangle 1
Blue Triangle 2
Blue Triangle 3
Red Circle 1
Red Circle 2
Red Circle 3
Red Triangle 1
Red Triangle 2
Red Triangle 3

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