获取值的所有索引 [英] Getting all indices of a value

问题描述

我正在尝试创建一个haskell函数，该函数将列表中某个值出现的所有索引作为列表返回，就像这样

Im trying to create a haskell function where all the indices of an occurence of a value in a list are returned as a list, so like

indices 3 [1,2,3,3,7]

给出[2，3]作为输出.我对Haskell非常陌生，无法找到有用的东西.我尝试使用过滤器，但是我所能做的只是得到[3，3]的列表，而不是实际的索引.如果您能给我一些提示，那就太好了.

gives [2, 3] as output. I am very new to Haskell and couldnt find something useful. I tried using filter, but all i got working is getting a list of [3, 3] but not the actual indices. It would be cool if you could give me a little hint.

推荐答案

这是函数式编程中非常常见的模式，有时称为decorate-process-undecorate.这样的想法是，您希望将一些额外的信息附加到列表中的每个元素上，使用您通常会做的稍稍改动的过滤器版本进行过滤，然后将这些多余的信息去除.

This is a pretty common pattern in functional programming, sometimes called decorate-process-undecorate. The idea is that you want to attach some extra information to each of the elements in your list, filter using a slightly altered version of the filter you would have done normally, then strip that extra information away.

indicies n = undecorate . filter predicate . decorate where
  decorate   = ...
  predicate  = ...
  undecodate = ...

当尝试对装饰进行编码时，建议您查看一下 zip 函数:

When trying to code decorate I suggest taking a look at the function zip:

zip :: [a] -> [b] -> [(a, b)]

还要考虑其对无限列表的影响，例如 repeat 1 或 [1,3，...] .尝试对取消装饰进行编码时，您可能需要 map :

Consider its effect on infinite lists, too, such as repeat 1 or [1,3,...]. When trying to code undecorate you'll likely want to map:

map :: (a -> b) -> [a] -> [b]

最后，不必担心此过程的效率.用严格的语言来说，decorate-filter-undecorate可能需要遍历该列表的3个遍历.在像Haskell这样的非严格语言中，编译器会自动将3个内部循环合并为一个循环.

Finally, don't worry about the efficiency of this process. In a strict language, decorate-filter-undecorate might require 3 traversals of the list. In a non-strict language like Haskell the compiler will automatically merge the 3 inner loops into a single one.

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