列表列表的总和；返回总和清单 [英] Sum of list of lists; returns sum list

问题描述

让data = [[3,7,2],[1,4,5],[9,8,7]]

比方说，我想对列表中每个列表的索引元素求和，例如在矩阵列中添加数字以获得单个列表.我假设数据中的所有列表的长度都相等.

Let's say I want to sum the elements for the indices of each list in the list, like adding numbers in a matrix column to get a single list. I am assuming that all lists in data are equal in length.

    print foo(data)

   [[3,7,2],
    [1,4,5],
    [9,8,7]]
    _______
 >>>[13,19,14]

如何遍历列表列表而又不会出现索引超出范围的错误?也许lambda?谢谢！

How can I iterate over the list of lists without getting an index out of range error? Maybe lambda? Thanks!

推荐答案

您可以尝试以下方法:

In [9]: l = [[3,7,2],[1,4,5],[9,8,7]]

In [10]: [sum(i) for i in zip(*l)]
Out[10]: [13, 19, 14]

这使用zip和*的组合来解压缩列表，然后根据项目的索引对其进行压缩.然后，您可以使用列表推导来遍历相似索引的组，对其进行求和并返回其原始"位置.

This uses a combination of zip and * to unpack the list and then zip the items according to their index. You then use a list comprehension to iterate through the groups of similar indices, summing them and returning in their 'original' position.

希望更清楚一点，这是您遍历zip(*l)时发生的事情:

To hopefully make it a bit more clear, here is what happens when you iterate through zip(*l):

In [13]: for i in zip(*l):
   ....:     print i
   ....:     
   ....:     
(3, 1, 9)
(7, 4, 8)
(2, 5, 7)

对于长度不相等的列表，可以将itertools.izip_longest与fillvalue的0一起使用-这基本上用0填充缺失的索引，从而使您可以对所有列"求和:

In the case of lists that are of unequal length, you can use itertools.izip_longest with a fillvalue of 0 - this basically fills missing indices with 0, allowing you to sum all 'columns':

In [1]: import itertools

In [2]: l = [[3,7,2],[1,4],[9,8,7,10]]

In [3]: [sum(i) for i in itertools.izip_longest(*l, fillvalue=0)]
Out[3]: [13, 19, 9, 10]

在这种情况下，这是在izip_longest上进行迭代的样子:

In this case, here is what iterating over izip_longest would look like:

In [4]: for i in itertools.izip_longest(*l, fillvalue=0):
   ...:     print i
   ...:     
(3, 1, 9)
(7, 4, 8)
(2, 0, 7)
(0, 0, 10)

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