我想根据相似的属性对元组进行分组

2022-01-20 00:00:00 python list tuples

问题描述

我有一个元组列表.[ (1, 2), (2, 3), (4, 3), (5, 6), (6, 7), (8, 2) ]

I have a list of tuples. [ (1, 2), (2, 3), (4, 3), (5, 6), (6, 7), (8, 2) ]

我想根据连接的元组(具有相关值)将它们分组到列表中.

I want to group them into lists based on which tuples are connected (have related values).

所以最终结果是两个相关元组值的列表 =[ [1, 2, 3, 4, 8], [5, 6, 7] ]

So the end result is two lists of related tuple values = [ [1, 2, 3, 4, 8], [5, 6, 7] ]

我怎样才能写一个函数来做到这一点?这是一个求职面试问题.我试图在 Python 中做到这一点，但我很沮丧，只想看看答案背后的逻辑，所以即使是伪代码也会帮助我，所以我可以看到我做错了什么.

How can I write a function to do this? This was a job interview question. I was trying to do it in Python, but I'm frustrated and just want to see the logic behind the answer, so even psuedo code would help me, so I can see what I did wrong.

我只有几分钟的时间来现场做这件事，但这是我尝试过的:

I only had a few minutes to do this on the spot, but here is what I tried:

def find_partitions(connections):
 theBigList = []     # List of Lists
 list1 = []          # The initial list to store lists
 theBigList.append(list1)

 for list in theBigList:
 list.append(connection[1[0], 1[1]])
     for i in connections:
         if i[0] in list or i[1] in list:
             list.append(i[0], i[1])

         else:
             newList = []
             theBigList.append(newList)

本质上，这家伙想要一个相关值列表的列表.我尝试使用for循环，但意识到它不起作用，然后时间用完了.

Essentially, the guy wanted an list of lists of values that were related. I tried to use a for loop, but realized that it wouldn't work, and then time ran out.

解决方案

在我们填写组件时，在每个阶段都需要考虑三种情况(因为您必须匹配重叠的组):

As we fill in the components, at each stage there are three cases to consider (as you will have to match up overlapping groups):

  1. x 或 y 均不在已找到的任何组件中.
  2. 两者都已经在不同的集合中，x 在 set_i 中，y 在 set_j 中.
  3. 其中一个或两个都在一个组件中，x 在 set_i 中或 y 在 set_i 中.
  1. Neither x or y are in any component already found.
  2. Both are already in different sets, x in set_i and y in set_j.
  3. Either one or both are in one component, x in set_i or y in a set_i.

我们可以使用内置的set来帮助.(请参阅 @jwpat 和 @DSM 的更棘手的示例):

We can use the built-in set to help. (see @jwpat's and @DSM's trickier examples):

def connected_components(lst):
    components = [] # list of sets
    for (x,y) in lst:
        i = j = set_i = set_j = None
        for k, c in enumerate(components):
            if x in c:
                i, set_i = k, c
            if y in c:
                j, set_j = k, c

        #case1 (or already in same set)
        if i == j:
             if i == None:
                 components.append(set([x,y]))
             continue

        #case2
        if i != None and j != None:
            components = [components[k] for k in range(len(components)) if k!=i and k!=j]
            components.append(set_i | set_j)
            continue

        #case3
        if j != None:
            components[j].add(x)
        if i != None:
            components[i].add(y)

    return components               

lst = [(1, 2), (2, 3), (4, 3), (5, 6), (6, 7), (8, 2)]
connected_components(lst)
# [set([8, 1, 2, 3, 4]), set([5, 6, 7])]
map(list, connected_components(lst))
# [[8, 1, 2, 3, 4], [5, 6, 7]]

connected_components([(1, 2), (4, 3), (2, 3), (5, 6), (6, 7), (8, 2)])
# [set([8, 1, 2, 3, 4]), set([5, 6, 7])] # @jwpat's example

connected_components([[1, 3], [2, 4], [3, 4]]
# [set([1, 2, 3, 4])] # @DSM's example

这肯定不是最有效的方法，但可能与他们所期望的相似.正如 Jon Clements 指出的那样，有一个用于这些类型计算的库:networkx，其中它们会更有效率.

This certainly won't be the most efficient method, but is perhaps one similar to what they would expect. As Jon Clements points out there is a library for these type of calculations: networkx, where they will be much more efficent.

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