从列表中制作字典 [英] Make dictionary from list

问题描述

所以我创建了一个像这样的列表:

So I created a list like this:

list = [line.strip() for line in open('file.txt','r')]

这是我的列表的一部分.

This is a snippet of my list.

[
 '1 2',
 '2 3',
 '2 3',
 '4 3 1',
 '3 4',
 '5 4 2 1',
 '4 4',
 '8 3 5 2',
 '5 7',
 '15 11 8 9 6 3 4',
]

我想创建一个字典，其中第一个数字是键，而后面的数字是值，但是我希望它以int形式出现.

I want to create a dictionary where the first number is the key and the number coming after are values but I want it in int form.

我不知道如何使用字典中涉及的类.

I don't know how to use the classes involved with dictionary.

推荐答案

这应该可以解决问题:

In [3]: arrs = [map(int, line.strip().split()) for line in open('file.txt')] 
In [4]: first2rest = dict( (arr[0], arr[1:]) for arr in arrs)

In [5]: first2rest
Out[5]: {1: [2], 2: [3], 3: [4], 4: [4], 5: [7], 8: [3, 5, 2], 15: [11, 8, 9, 6, 3, 4]}

让我们拆开它.这部分将文件中的每一行分割为空格，并将其转换为整数:

Let's take it apart. This part splits up each line in your file on spaces and converts them to ints:

map(int, line.strip().split())

然后，此部分创建该行中第一项到其余项的字典:

Then this part creates the dictionary of the first item in the row to the rest:

first2rest = dict( (arr[0], arr[1:]) for arr in arrs)

但是，正如@SethMMorton指出的那样，由于列出的文件多次包含相同的密钥，因此您将丢失数据.

However, as @SethMMorton pointed out, you will lose data as the file you listed includes the same key multiple times.

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