拼合名单列表 [英] Flatten list of lists

问题描述

我在用方括号在Python中的一个问题。我写了一个code产生的输出如下：

I'm having a problem with square brackets in Python. I wrote a code that produces the following output:

[[180.0], [173.8], [164.2], [156.5], [147.2], [138.2]]

不过，我想执行一些计算与，但方括号不会让我。

But I would like to perform some calculations with that, but the the square brackets won't let me.

我怎样才能去掉括号？我看到一些例子，这样做，但我不能把它们应用到这种情况。

How can I remove the brackets? I saw some examples to do that but I could not apply them to this case.

推荐答案

平展列表中的删除括号的使用嵌套列表COM prehension。这将取消窝储存在你的名单列表中的每个列表！

Flatten the list to "remove the brackets" using a nested list comprehension. This will un-nest each list stored in your list of lists!

list_of_lists = [[180.0], [173.8], [164.2], [156.5], [147.2], [138.2]]
flattened = [val for sublist in list_of_lists for val in sublist]

嵌套列表的COM prehensions中相同的方式，它们解开评估（即添加新行和标签为每个新的循环因此，在这种情况下：

Nested list comprehensions evaluate in the same manner that they unwrap (i.e. add newline and tab for each new loop. So in this case:

flattened = [val for sublist in list_of_lists for val in sublist]

相当于：

flattened = []
for sublist in list_of_lists:
    for val in sublist:
        flattened.append(val)

最大的区别是，列表补偿评估MUCH比拆开的循环更快，消除了追加来电！

The big difference is that the list comp evaluates MUCH faster than the unraveled loop and eliminates the append calls!

如果你在一个子列表有多个项目的名单补偿甚至会压平了。即

If you have multiple items in a sublist the list comp will even flatten that. ie

>>> list_of_lists = [[180.0, 1, 2, 3], [173.8], [164.2], [156.5], [147.2], [138.2]]
>>> flattened  = [val for sublist in list_of_lists for val in sublist]
>>> flattened 
[180.0, 1, 2, 3, 173.8, 164.2, 156.5, 147.2,138.2]

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