在一个lin中创建n个空字符串

Question

可能是重复的（对不起）。我环顾四周，找不到答案。

我想在一行中生成一个空字符串列表。

我试过：

>>> list(str('') * 16)
# ['']
>>> list(str(' ') * 16)
# [' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ']
# anything with a char in it is working

下面的方法行得通，但是有更好的方法吗？为什么list(str('') * 16)不起作用？

>>> [str() for c in 'c' * 16]
['', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '']

Answer 1

你可以把这样的单子乘出来。因为''是不可变的，所以不必担心它们都是对同一字符串的引用。

[''] * 16

不能对可变对象（如列表或听写）使用相同的技巧。你需要用上一个版本

[mutable_thing() for c in range(16)]

或者

[[] for c in range(16)]

或者

[{} for c in range(16)]

Answer 2

请参见Python standard types page：

>>> [''] * 16
['', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '']

s * n, n * s

n shallow copies of s concatenated

其中，s是一个序列，n是一个整数。

此操作文档的完整脚注：

Values of n less than 0 are treated as 0 (which yields an empty sequence of the same type as s). Note also that the copies are shallow; nested structures are not copied. This often haunts new Python programmers; consider:

>>> lists = [[]] * 3
>>> lists
[[], [], []]
>>> lists[0].append(3)
>>> lists
[[3], [3], [3]]

What has happened is that [[]] is a one-element list containing an empty list, so all three elements of [[]] * 3 are (pointers to) this single empty list. Modifying any of the elements of lists modifies this single list. You can create a list of different lists this way:

>>> lists = [[] for i in range(3)]
>>> lists[0].append(3)
>>> lists[1].append(5)
>>> lists[2].append(7)
>>> lists
[[3], [5], [7]]