从词典中的列表创建词典

问题描述

我对Python还是很陌生，我被一个看似简单的任务所困扰.在我的程序的一部分中，我想从列表内部的值创建二级字典"，这些值是一级字典"的值.

I'm quite new to Python and I have been stumped by a seemingly simple task.In part of my program, I would like to create Secondary Dictionaries from the values inside of lists, of which they are values of a Primary Dictionary.

我还想将这些值默认设置为0

I would also like to default those values to 0

为简单起见，主要字典看起来像这样:

For the sake of simplicity, the Primary Dictionary looks something like this:

primaryDict = {'list_a':['apple', 'orange'], 'list_b':['car', 'bus']}

我希望得到的结果是这样的:

What I would like my result to be is something like:

{'list_a':[{'apple':0}, {'orange':0}], 'list_b':[{'car':0}, {'bus':0}]}

我知道该过程应该是遍历primaryDict中的每个列表，然后遍历该列表中的项目，然后将它们分配为Dictionary.

I understand the process should be to iterate through each list in the primaryDict, then iterate through the items in the list and then assign them as Dictionaries.

我尝试了许多"for"循环的变体，它们看起来都类似于:

I've tried many variations of "for" loops all looking similar to:

for listKey in primaryDict:
    for word in listKey:
        {word:0 for word in listKey}

我还尝试了一些将Dictionary和List理解相结合的方法，但是当我尝试索引和打印字典时，例如:

I've also tried some methods of combining Dictionary and List comprehension,but when I try to index and print the Dictionaries with, for example:

print(primaryDict['list_a']['apple'])

我得到"TypeError:列表索引必须是整数或切片，而不是str"，我将其解释为"apple"实际上不是字典，而是列表中的字符串.我通过将'apple'替换为0进行测试，它只返回'apple'，证明它是正确的.

I get the "TypeError: list indices must be integers or slices, not str", which I interpret that my 'apple' is not actually a Dictionary, but still a string in a list. I tested that by replacing 'apple' with 0 and it just returns 'apple', proving it true.

在以下方面，我需要帮助:

I would like help with regards to:

-是否将列表中的值分配为"0"值的字典

-Whether or not the values in my list are assigned as Dictionaries with value '0'

或

-无论是我的索引错误(在循环还是在打印功能中)，以及我的误解

-Whether the mistake is in my indexing (in the loop or the print function), and what I am mistaken with

或

-我所做的一切都无法获得理想的结果，我应该尝试一种不同的方法

-Everything I've done won't get me the desired outcome and I should attempt a different approach

谢谢

推荐答案

您可以通过以下方式获取所需的数据结构:

You can get the data structure that you desire via:

primaryDict = {'list_a':['apple', 'orange'], 'list_b':['car', 'bus']}
for k, v in primaryDict.items():
    primaryDict[k] = [{e: 0} for e in v]

# primaryDict
{'list_b': [{'car': 0}, {'bus': 0}], 'list_a': [{'apple': 0}, {'orange': 0}]}

但是正确的嵌套访问将是:

But the correct nested access would be:

print(primaryDict['list_a'][0]['apple'])  # note the 0

如果您确实希望primaryDict['list_a']['apple']工作，请改为

If you actually want primaryDict['list_a']['apple'] to work, do instead

for k, v in primaryDict.items():
    primaryDict[k] = {e: 0 for e in v}

# primaryDict
{'list_b': {'car': 0, 'bus': 0}, 'list_a': {'orange': 0, 'apple': 0}}

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