获取 Haskell 列表中元素的索引

Question

I have a list [4539, 5646, 6547, 7546] .

How can i to get the index of every element ?

index of 4539 --> 0

index of 5646 --> 1

index of 6547 --> 2

index of 7546 --> 3

Answer 1

-- >> elemIndex 4539 [4539, 5646, 6547, 7546]
-- 0
-- >> elemIndex 5646 [4539, 5646, 6547, 7546]
-- 1
-- >> elemIndex 6547 [4539, 5646, 6547, 7546]
-- 2
elemIndex :: Eq a => a -> [a] -> Int

What should happen when the element is not found?

>> elemIndex 20 []
*** Exception: elemIndex: empty list

The usual solution is to wrap the index in Maybe . This elemIndex function exists in Data.List .

-- >> elemIndex 4539 [4539, 5646, 6547, 7546]
-- Just 0
-- >> elemIndex 5646 [4539, 5646, 6547, 7546]
-- Just 1
-- >> elemIndex 6547 [4539, 5646, 6547, 7546]
-- Just 2
-- >> elemIndex 20 []
-- Nothing
elemIndex :: Eq a => a -> [a] -> Maybe Int

Use elemIndices if you want to return a list of occurrences.

-- >> elemIndices 'o' "ooookay"
-- [0,1,2,3]
elemIndices :: Eq a => a -> [a] -> [Int]