在Haskell的列表中查找元素的索引？ [英] Finding index of element in a list in Haskell?

问题描述

我在 Haskell 中有一个函数，它可以找到指数的最大值一个列表：

  prob99 =最大$ map（\ xs  - >（head xs）^（head（tail xs） ））数字
  

我需要找到的是这个最大值在结果列表中的位置。编辑：我找到了这样的解决方案：

  n = [[519432,525806]，[632382,518061] .... 
 prob99b [a，b] = b *（log a）
 answer = snd $最大（zip（map prob99b n）[1 ..]）
  

解决方案

如何找到最大元素的索引？如何尝试所有索引并检查它们是否是最大值？

  ghci>让maxIndex xs = head $ filter（（==最大xs）。（xs !!））[0 ..] 
  

但是这听起来像某个函数已经存在的东西。如果我使用现有的函数，我的代码将更具可读性，可维护性，甚至更高效。

所以我应该问一下如何去做， 15分钟我会得到一个答案和一些snarky评论。或者 - 我可以问hoogle并立即得到一个有用的回应（如Will建议）

  $ hoogleOrd a => [ a]  - > Int|头部
 
<无关>> 
 
 $＃hmm，所以没有函数给我指定最大值，
 $＃但是如何找到一个特定的元素，并且我给它最大值？ 
 $ hooglea  - > [a]  - > Int| head 
 Data.List elemIndex :: Eq a => a  - > [a]  - >也许Int 
 Data.List elemIndices :: Eq a => a  - > [a]  - > [Int] 
  

I have a function in Haskell which finds the maximum value of an exponentiation from a list:

prob99 = maximum $ map (\xs -> (head xs)^(head (tail xs))) numbers

What I need to find is the location of this maximum value in the resultant list. How would I go about this?

Edit: I found a solution that goes like this:

n = [[519432,525806],[632382,518061]....
prob99b [a,b] = b* (log a)
answer = snd $ maximum (zip  (map prob99b n) [1..])

解决方案

How to find the index of the maximum element? How about trying all indexes and checking whether they are the maximum?

ghci> let maxIndex xs = head $ filter ((== maximum xs) . (xs !!)) [0..]

But this sounds like something for which a function already exists. My code will be more readable, maintainable, and probably even more efficient, if I used the existing function.

So I should just ask SO how to do it, and in 15 minutes I'll get an answer and some snarky comments. Or - I could ask hoogle and get a useful response immidiately (as Will suggested)

$ hoogle "Ord a => [a] -> Int" | head

<Nothing relevant>

$ # hmm, so no function to give me the index of maximum outright,
$ # but how about finding a specific element, and I give it the maximum?
$ hoogle "a -> [a] -> Int" | head
Data.List elemIndex :: Eq a => a -> [a] -> Maybe Int
Data.List elemIndices :: Eq a => a -> [a] -> [Int]

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