如何在 Python 中创建空列表的列表或元组?

问题描述

我需要逐步填充一个列表或一个列表元组.看起来像这样的东西:

I need to incrementally fill a list or a tuple of lists. Something that looks like this:

result = []
firstTime = True
for i in range(x):
    for j in someListOfElements:
        if firstTime:
            result.append([f(j)])
        else:
            result[i].append(j)

为了让它不那么冗长更优雅，我想我会预先分配一个空列表的列表

In order to make it less verbose an more elegant, I thought I will preallocate a list of empty lists

result = createListOfEmptyLists(x)
for i in range(x):
    for j in someListOfElements:
        result[i].append(j)

预分配部分对我来说并不明显.当我执行 result = [[]] * x 时，我收到了一个对同一列表的 x 引用的列表，以便输出以下

The preallocation part isn't obvious to me. When I do result = [[]] * x, I receive a list of x references to the same list, so that the output of the following

result[0].append(10)
print result

是:

[[10], [10], [10], [10], [10], [10], [10], [10], [10], [10]]

我可以使用循环(result = [[] for i in range(x)])，但我想知道是否存在无循环"解决方案.

I can use a loop (result = [[] for i in range(x)]), but I wonder whether a "loopless" solution exists.

是获得我想要的东西的唯一途径

Is the only way to get what I'm looking for

推荐答案

result = [list(someListOfElements) for _ in xrange(x)]

这将生成 x 个不同的列表，每个列表都有一个 someListOfElements 列表的副本(该列表中的每个项目都是引用，但其所在的列表是一个副本).

This will make x distinct lists, each with a copy of someListOfElements list (each item in that list is by reference, but the list its in is a copy).

如果更有意义，请考虑使用 copy.deepcopy(someListOfElements)

If it makes more sense, consider using copy.deepcopy(someListOfElements)

生成器和列表推导式和事物被认为非常pythonic.

Generators and list comprehensions and things are considered quite pythonic.

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