如何对给定组大小的列表进行分区? [英] How to partition a list with a given group size?

问题描述

我正在寻找对列表(或seq)进行分区的最佳方法，以使组具有给定的大小. 对于前.假设我想将尺寸2分组(虽然可以是其他任何数字):

I'm looking for the best way to partition a list (or seq) so that groups have a given size. for ex. let's say I want to group with size 2 (this could be any other number though):

let xs = [(a,b,c); (a,b,d); (y,z,y); (w,y,z); (n,y,z)]
let grouped = partitionBySize 2 input
// => [[(a,b,c);(a,b,d)]; [(y,z,y);(w,y,z)]; [(n,y,z)]]

实现partitionBySize的明显方法是将位置添加到输入列表中的每个元组，使其变为

The obvious way to implement partitionBySize would be by adding the position to every tuple in the input list so that it becomes

[(0,a,b,c), (1,a,b,d), (2,y,z,y), (3,w,y,z), (4,n,y,z)]

，然后将GroupBy与

and then use GroupBy with

xs |> Seq.ofList |> Seq.GroupBy (function | (i,_,_,_) -> i - (i % n))

但是，这种解决方案在我看来并不十分优雅. 有没有更好的方法来实现此功能(也许带有内置功能)?

However this solution doesn't look very elegant to me. Is there a better way to implement this function (maybe with a built-in function)?

推荐答案

这是一个尾部递归函数，它遍历列表一次.

Here's a tail-recursive function that traverses the list once.

let chunksOf n items =
  let rec loop i acc items = 
    seq {
      match i, items, acc with
      //exit if chunk size is zero or input list is empty
      | _, [], [] | 0, _, [] -> ()
      //counter=0 so yield group and continue looping
      | 0, _, _::_ -> yield List.rev acc; yield! loop n [] items 
      //decrement counter, add head to group, and loop through tail
      | _, h::t, _ -> yield! loop (i-1) (h::acc) t
      //reached the end of input list, yield accumulated elements
      //handles items.Length % n <> 0
      | _, [], _ -> yield List.rev acc
    }
  loop n [] items

用法

[1; 2; 3; 4; 5]
|> chunksOf 2 
|> Seq.toList //[[1; 2]; [3; 4]; [5]]

我喜欢Tomas方法的优雅之处，但是我使用了1000万个元素的输入列表对我们的两个函数进行了基准测试.这个时间是9秒，而他的时间是22秒.当然，正如他所承认的那样，最有效的方法可能涉及数组/循环.

I like the elegance of Tomas' approach, but I benchmarked both our functions using an input list of 10 million elements. This one clocked in at 9 secs vs 22 for his. Of course, as he admitted, the most efficient method would probably involve arrays/loops.

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