在python中创建知道其他朋友的朋友的字典 [英] Creating dictionaries of Friends that know other Friends in python

问题描述

在任何一群人中有很多对朋友。假设分享朋友的两个人是朋友。 （是的，这是现实生活中的一个不切实际的假设，但让我们来做）。换句话说，如果A和B是朋友，B是C的朋友，那么A和C也必须是朋友。使用这个规则，只要我们知道组中的友谊，就可以将任何一群人分成友谊圈。

编写一个功能网络（）参数。第一个参数是组中的人数，第二个参数是定义朋友的元组对象列表。假设人数由0到n-1。例如，元组（0,2）表示人0是与人2的朋友。该功能应该将人的分区打印成友谊圈。以下显示了函数的几个示例运行：

 >>> networks（5，[（0,1） ，（1,2），（3,4）]）＃执行
  

社交网络0是{ 0,1,2}

社交网络1是{3,4}

我真的很失落关于如何启动这个程序，任何提示将不胜感激。

解决方案

  def networks（n，lst）：
 groups = [] 
 for i in range（n）
 groups.append（{i}）
 for lst：
 union = groups [pair [0]] | groups [pair [ 1]] 
 for p in union：
 groups [p] = union 
 sets = set（）
 for g in group：
 sets.add（tuple 
i = 0 
套数：
 print（network，i，is，set（s））
 i + = 1 
  

这是我正在寻找的，如果有人关心。

In any group of people there are many pairs of friends. Assume that two people who share a friend are friends themselves. (Yes, this is an unrealistic assumption in real life, but let's make it nevertheless). In other words, if people A and B are friends and B is friends with C, then A and C must also be friends. Using this rule we can partition any group of people into friendship circles as long as we know something about the friendships in the group.

Write a function networks() that takes two parameters. The first parameter is the number of people in the group and the second parameter is a list of tuple objects that define friends. Assume that people are identified by numbers 0 through n-1. For example, tuple (0, 2) says that person 0 is friends with person 2. The function should print the partition of people into friendship circles. The following shows several sample runs of the function:

>>>networks(5,[(0,1),(1,2),(3,4)])#execute

Social network 0 is {0,1,2}

Social Network 1 is {3,4}

I am honestly pretty lost on how to start this program, any tips would be greatly appreciated.

解决方案

def networks(n,lst):
groups= []
for i in range(n)
    groups.append({i})
for pair in lst:
    union = groups[pair[0]]|groups[pair[1]]
    for p in union:
        groups[p]=union
sets= set()
for g in groups:
    sets.add(tuple(g))
i=0
for s in sets:
    print("network",i,"is",set(s))
    i+=1

This is what I was looking for if anybody cares.

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