Haskell Aeson处理丢失的数据 [英] Haskell Aeson to deal with missing data

问题描述

我有一个(有效)json编码的数组，该数组缺少数据或格式错误.我希望Aeson将其转换为Maybe [Maybe Point]并使用Nothing，其中数组元素不是有效的Point.

I have a (valid) json encoded array that has missing or malformed data. I want Aeson to turn that into Maybe [Maybe Point] and have Nothing where the array element was not a valid Point.

import Data.Aeson
decode "[1,2,3,4,null,\"hello\"]" :: (Maybe [Maybe Int])
=> Nothing

但是我更希望它评估为

=> Just [Just 1, Just 2, Just 3, Just 4, Nothing, Nothing]

如果Aeson无法做到这一点，那么还有另一个图书馆可以做到吗?

If this can't be done with Aeson, is there another library that can do it?

请注意，实际对象比简单的整数复杂得多，因此字符串操作不是可行的解决方案.

Note that the actual object is much more complex than a simple integer so string manipulations are not a viable solution.

推荐答案

我会使用Values并从中使用它们:

I would use Values and work from them:

decode "[1,2,3,4,null,\"hello\"]" :: (Maybe [Value])
Just [Number 1.0,Number 2.0,Number 3.0,Number 4.0,Null,String "hello"]

所以

> let fromNum v = case v of Number x -> Just x ; _ -> Nothing 
> maybe [] (map fromNum) (decode  "[1,2,3,4,null,\"hello\"]" :: (Maybe [Value])
[Just 1.0,Just 2.0,Just 3.0,Just 4.0,Nothing,Nothing]

一个非常无效(但安全)的解决方案是:

A very ineffective (but safe) solution would be:

> let tmp = maybe [] (map fromNum) (decode  "[1,2,3,4,null,\"hello\"]" :: (Maybe [Value])
> let keepJust l v = case v of Just i -> i:l; Nothing -> l
> reverse (foldl keepJust [] tmp)
[1.0,2.0,3.0,4.0]

如果您想使事情更有效，则可能需要使用Vector s和foldl'.

If you want to make things more effective you might want to use Vectors and foldl'.

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