在列表推导中解开拆分 [英] unpacking a split inside a list comprehension

问题描述

如果我想基于文档行的元素生成元组列表，我可以这样做:

If I want to generate a list of tuples based on elements of lines of a document, i can do :

[(line.split()[0], line.split()[-1][3:8]) for line in open("doc.txt")]  

例如

(我添加了切片以显示我可能要对split的元素使用一些操作).

for example (i added the slicing to show that I might want use some operations on the elements of the split).

还是要避免两次使用split，因为这样效率不高.
所以我想使用

Still I would like to avoid using split two times, because that's unefficient.
So I wanted to use something like unpacking, with

[(linesplit0, linesplit1[3:8]) for line in open("doc.txt") for (linesplit0, linesplit1) in line.split()]  

但这不起作用，因为拆分中没有元组，因此拆分的每个元素都将缺少一个元素.

but that can't work since there are no tuples in the split, so at each element of the split we will be lacking one element.

我想要的是一种允许使用占位符名称作为拆分结果的列表的名称(如splittedlist或其他内容)，并且可以与索引(splittedlist [0])或拆包或两者一起使用，这将与理解列表语法兼容.

What I would like is something that allows to use a placeholder name for the list resulting of the split (like splittedlist or whatever), and that could be used with indexing (splittedlist[0]), or unpacking or both), and that would be compatible with the comprehension list syntax.

可行吗?

推荐答案

您可以在打开时使用map(python3)或itertools.imap(python2):

You can use map (python3) or itertools.imap (python2) over open:

[(line[0], line[-1][3:8]) for line in map(str.split, open("doc.txt"))]

或使用生成器:

[(line[0], line[-1][3:8]) for line in ( l.split() for l in open("doc.txt"))]  

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