python 实现复合字段排序，如 ：身高一样比体重，如何实现呢？

Posted 2022-04-07 shiter

文章大纲

• 纯算法思路
• 面向对象思路
• 参考文献

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假设有一个题目：

现在学校排队
按照身高升序排列，身高相同按照体重降序（升序）排列，身高体重都相同，按照原有先后顺序排列

样例输入：

shengao = [150,150,160,160,170]
tizhong = [100,100,100,105,100]

输出：(体重升序)

index = [1，2，3，4，5]

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纯算法思路

if __name__ == '__main__':
    index = range(0,len(shengao))
    liebiao = zip(index,shengao,tizhong)
    bmi = list(liebiao)
    bmi.sort(key=lambda x: (x[1], x[2]))

    print(bmi)

输出

[(0, 150, 100), (1, 150, 100), (2, 160, 100), (3, 160, 105), (4, 170, 100)]

这样的思路有一个问题，无法实现，复核字段排序的负责逻辑

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面向对象思路

java ，c++ 都有Compare接口， 主要就是实现比较的接口

比如c++ 类似代码：

//对人的年龄进行升序排列，年龄相同对身高进行降序排列,年龄身高相同对体重升序排列
#include <iostream>
#include <list>
#include <string>

using namespace std;


class Person

public:
	Person( string Name, int Age,int Heigh,int weight)
	
		this->m_Name = Name;
		this->m_Age = Age;
		this->m_Heigh = Heigh;
		this->m_weight = weight;
	
	string m_Name;
	int m_Age;
	int m_Heigh;
	int m_weight;
;

bool Compare(Person& p1, Person& p2)

	if (p1.m_Age == p2.m_Age)
	
		if (p1.m_Heigh == p2.m_Heigh)
		
			return p1.m_weight < p2.m_weight;
		
		else
		
			return p1.m_Heigh > p2.m_Heigh;
		
	
	else
	
		return p1.m_Age < p2.m_Age;
	

void test01()

	list<Person> l;
	Person p1 =  "张飞", 23, 170 ,74;
	Person p2 =  "关羽", 23, 170 ,80;
	Person p3 =  "赵云", 23, 180 ,78;
	Person p4 =  "刘备", 25, 199 ,90;
	
	l.push_back(p1);
	l.push_back(p2);
	l.push_back(p3);
	l.push_back(p4);

	cout << "-----------------------" << endl;
	for (list<Person>::iterator i = l.begin(); i != l.end(); i++)
	
		cout << "姓名：" << i->m_Name << " " << "年龄：" << i->m_Age << " " << "身高：" << i->m_Heigh 
			<< "体重：" << i->m_weight <<endl;
	
	cout << "-----------------------" << endl;
	l.sort(Compare);
	for (list<Person>::iterator i = l.begin(); i != l.end(); i++)
	
		cout << "姓名：" << i->m_Name << " " << "年龄：" << i->m_Age << " " << "身高：" << i->m_Heigh 
		<< "体重：" << i->m_weight << endl;
	

int main()

	test01();
	return 0;

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参考文献

https://www.runoob.com/python3/python3-att-list-sort.html