如何对列表的相同索引求和? [英] How to sum over same indices of list of lists?

问题描述

如何汇总列表中所有值都为索引的列表?

How can I sum a list of lists where all the values at the a index?

例如:

let k = [ [1,1,1], [2,2,2], [3,3,3] ]
sumFoo k
> [ 6, 6, 6]

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我知道我可以将两个列表总结为:

---

I know that I can sum two lists as:

zipWith (+) [1,2,3] [2,3,4]

但是列表列表呢?我尝试过类似的事情:

But what about a list of list? I tried something like:

foldr (\xs ys -> zipWith (+) xs ys) [] k

但这给了我一个空的名单！

but that gives me an empty list!

推荐答案

您有一个表示为列表列表的矩阵，并且想要提取列总和.为此，只需转置矩阵并计算行总和:

You have a matrix represented as a list of lists and want to extract the column sums. To do this, simply transpose the matrix and compute the row sums:

Prelude> import Data.List
Prelude Data.List> map sum . transpose $ [ [1,1,1], [2,2,2], [3,3,3] ]
[6,6,6]

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