计算Haskell中的列表累计和 [英] Calculating list cumulative sum in Haskell

问题描述

编写一个返回列表运行总和的函数。例如运行[1,2,3,5]是[1,3,6,11]。我写这个函数的下面，它可以返回列表中所有值的最终总和。那么我怎么能一个接一个地分开它们？

  sumlist'xx = aux xx 0 
其中aux [] a = a 
 aux（x：xs）a =辅助xs（a + x）
  a +> 
  sumlist'x 到每个步骤的结果并使用空列表作为基本情况： xx = aux xx 0 
其中aux [] a = [] 
 aux（x：xs）a =（a + x）：aux xs（a + x）
  
然而，Haskell更习惯于将这种事物表现为折叠或扫描。
 
Write a function that returns the running sum of list. e.g. running [1,2,3,5] is [1,3,6,11]. I write this function below which just can return the final sum of all the values among the list.So how can i separate them one by one?
sumlist' xx=aux xx 0
    where aux [] a=a
          aux (x:xs) a=aux xs (a+x)

解决方案
You can adjust your function to produce a list by simply prepending a+x to the result on each step and using the empty list as the base case:
sumlist' xx = aux xx 0
    where aux [] a = []
          aux (x:xs) a = (a+x) : aux xs (a+x)
However it is more idiomatic Haskell to express this kind of thing as a fold or scan.

                        
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