通过将两个长度不均匀的列表压缩在一起来创建字典 [英] Create a dictionary by zipping together two lists of uneven length

问题描述

我有两个不同长度的列表，L1 和 L2.L1 比 L2 长.我想得到一个字典，其中 L1 的成员作为键，L2 的成员作为值.

I have two lists different lengths, L1 and L2. L1 is longer than L2. I would like to get a dictionary with members of L1 as keys and members of L2 as values.

一旦L2的所有成员都用完.我想从头开始，从 L2[0] 开始.

As soon as all the members of L2 are used up. I would like to start over and begin again with L2[0].

L1 = ['A', 'B', 'C', 'D', 'E']    
L2 = ['1', '2', '3']    
D = dict(zip(L1, L2))    
print(D)

正如预期的那样，输出是这样的:

As expected, the output is this:

{'A': '1', 'B': '2', 'C': '3'}

我想实现的目标如下:

{'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

推荐答案

使用 itertools.cycle 循环到L2的开头:

from itertools import cycle
dict(zip(L1, cycle(L2)))
# {'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

<小时>

在您的情况下，将 L2 与其自身连接也有效.

---

In your case, concatenating L2 with itself also works.

# dict(zip(L1, L2 * 2))
dict(zip(L1, L2 + L2))
# {'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

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