如何获取值的值作为字典的键 [英] How to get the value of a value as a key of a dictionary

问题描述

我不确定是否曾经问过这个问题，因为我也不知道该怎么写:

I'm not sure if this has been asked before since I also don't know how to word it:

例如，我有一个给定的字典，其中包含键和值:

So for example i have a given dictionary with keys and values:

d = {[ 0 : 1, 2, 3], [1 : 2, 3, 4], [2: 5, 6, 7]}

，我想证明1是一个键，也是一个值，因此得出以下结论: 0连接到值1，例如[0:[2,3,4]，2、3]或类似的东西.

and I want to show that 1 is a key and is also a value, thereby getting to the conclusion that 0 is connected to the value of 1. like [0:[2,3,4], 2, 3] or something like that.

我将对大量键(每个键具有多个值)执行此操作.

And I would be doing this for a large amount of keys, each with multiple values.

这可能吗?我该怎么编码呢?顺便说一下，我是Python的新手，所以请放轻松.

Is this possible? And how would I code that? By the way, I'm very new to Python so please take it easy on me.

推荐答案

您的意思是这样的吗?

d = { 0 : (1, 2, 3) , 1 : (2, 3, 4), 2: (5, 6, 7)}
for key in d.keys():
    for val in d[key]:
        try:
            d[key]+=d[val]
        except KeyError:
            pass

给予

>>> d
{0: (1, 2, 3, 2, 3, 4, 5, 6, 7), 1: (2, 3, 4, 5, 6, 7), 2: (5, 6, 7)}

如果要使用唯一值，请在for key in d.keys()循环的末尾添加d[key] = tuple(set(d[key])).

If you want unique values then add d[key] = tuple(set(d[key])) to the end of the for key in d.keys() loop.

给予

>>> d
{0: (1, 2, 3, 4, 5, 6, 7), 1: (2, 3, 4, 5, 6, 7), 2: (5, 6, 7)}

ps:d = {[ 0 : 1, 2, 3], [1 : 2, 3, 4], [2: 5, 6, 7]}无效的python！

ps: d = {[ 0 : 1, 2, 3], [1 : 2, 3, 4], [2: 5, 6, 7]} is not valid python!

查看评论.

d = { 0 : [1, 2, 3] , 1 : [2, 3, 4], 2: [5, 6, 7]}
for key in d.keys():
    orig_vals=d[key]
    new_vals=[]
    for val in orig_vals:
        try:
            new_vals+=d[val]
        except KeyError:
            pass
    d[key] = list(set(new_vals)-set(orig_vals))

给予

>>> d
{0: [4, 5, 6, 7], 1: [5, 6, 7], 2: []}

如果要避免清除未链接到其他键的值，例如[5,6,7]中的2，然后将最后一行更改为

If you want to avoid clearing values that were not linked to other keys, e.g. [5,6,7] in 2, then change the last line to

if new_vals:
        d[key] = list(set(new_vals)-set(orig_vals))

给出

>>> d
{0: [4, 5, 6, 7], 1: [5, 6, 7], 2: [5, 6, 7]}

查看评论.

d = { 0 : [1, 2, 3] , 1 : [2, 3, 4], 2: [5, 6, 7]}
for key in d.keys():
    orig_vals=d[key]
    new_vals=[]
    count = 0
    for val in orig_vals:
        try:
            new_vals+=d[val]
            count+=1
            if count >= yournumberhere: break
        except KeyError:
            pass
    d[key] = list(set(new_vals)-set(orig_vals))

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