如何将列表的每个项目包装到其适当的列表中 [英] How to wrap each item of a list into its proper list

问题描述

我该如何打开一个像这样的列表:

How can I turn a list like:

a = [1, 2, 3, 4, 5]

类似于:

b = [[1], [2], [3], [4], [5]]

我当时正在考虑使用 map 函数，但是那我将a精确映射到哪个函数呢?

I was thinking of using the map function, but then what function would I exactly map a to?

推荐答案

列表理解对于这种事情很有用.我只是将每个单独的项目放在理解内的列表中.为此，使用map会比较困难.

List comprehension are great for this kinda of thing. I simply put each individual item in a list inside the comprehension. Using map for this would be harder to accomplish.

a = [1, 2, 3, 4, 5]
b = [[i] for i in a]

列表推导是在循环内编写小段代码的简写.在循环某事并创建新列表"中，范例是如此普遍，以至于python决定创建特定的语法来处理这种用法.请注意，还有字典和对类似用法的理解.

List comprehensions are a shorthand for writing small pieces of code inside a loop. The "loop over something and create a new list" paradigm is so common that python decided to create specific syntax to handle this usage. Note that there's also dictionary and set comprehensions for similar usage.

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