从F#中的列表中切片类似的功能 [英] Slice like functionality from a List in F#

问题描述

对于数组let foo = [|1;2;3;4|]，我可以使用以下任意项从数组中返回切片.

With an array let foo = [|1;2;3;4|] I can use any of the following to return a slice from an array.

foo.[..2] 
foo.[1..2] 
foo.[2..]

我如何对列表let foo2 = [1;2;3;4]做同样的事情?当我尝试与数组相同的语法时，得到error FS00039: The field, constructor or member 'GetSlice' is not defined.

How can I do the same thing for List let foo2 = [1;2;3;4]? When I try the same syntax as the array I get error FS00039: The field, constructor or member 'GetSlice' is not defined.

获取List子节的首选方法是什么?为什么不构建它们以支持GetSlice?

What's the preferred method of getting a subsection of a List and why aren't they built to support GetSlice?

推荐答案

获取的首选方法是什么 列表的一个小节，为什么不 是为支持GetSlice而构建的?

What's the preferred method of getting a subsection of a List and why aren't built to support GetSlice?

让我们首先讨论最后一个问题，最后一个问题:

Let's make the last question first and the first question last:

为什么列表不支持GetSlice

列表被实现为链接列表，因此我们没有对它们的有效索引访问.相对而言，foo.[|m..n|]对数组花费O(n-m)时间，等效语法在列表上花费O(n)时间.这是一个很大的问题，因为它使我们无法在大多数有用的情况下有效使用切片语法.

Lists are implemented as linked lists, so we don't have efficient indexed access to them. Comparatively speaking, foo.[|m..n|] takes O(n-m) time for arrays, an equivalent syntax takes O(n) time on lists. This is a pretty big deal, because it prevents us from using slicing syntax efficiently in the vast majority of cases where it would be useful.

例如，我们可以在线性时间内将数组切成相等大小的片段:

For example, we can cut up an array into equal sized pieces in linear time:

let foo = [|1 .. 100|]
let size = 4
let fuz = [|for a in 0 .. size .. 100 do yield foo.[a..a+size] |]

但是如果我们改用列表怎么办?每次调用foo.[a..a+size]都将花费越来越长的时间，整个操作为O(n^2)，因此非常不适合该工作.

But what if we were using a list instead? Each call to foo.[a..a+size] would take longer and longer and longer, the whole operation is O(n^2), making it pretty unsuitable for the job.

在大多数情况下，切片列表是错误的方法.通常，我们使用模式匹配来遍历和操纵列表.

Most of the time, slicing a list is the wrong approach. We normally use pattern matching to traverse and manipulate lists.

切片列表的首选方法?

请尽可能使用模式匹配.否则，您可以依靠Seq.skip和Seq.take为您剪切列表和序列:

Wherever possible, use pattern matching if you can. Otherwise, you can fall back on Seq.skip and Seq.take to cut up lists and sequences for you:

> [1 .. 10] |> Seq.skip 3 |> Seq.take 5 |> Seq.toList;;
val it : int list = [4; 5; 6; 7; 8]

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