coffeescript - manipulating json

coffeescript - manipulating json

本文关键字：json manipulating coffeescript      更新时间：2023-09-26

如果我有数据，这只是逻辑问题：

    [
        {
            from: 'user a',
            to: 'user b'
        },
        {
            from: 'user a',
            to: 'user c'
        },
        {
            from: 'user b',
            to: 'user d'
        },
        {
            from: 'user c',
            to: 'user d'
        },
        {
            from: 'user c',
            to: 'user d'
        }
    ]

我需要将数据处理为：

    [
        {
            from: 'user a',
            to: ['user b', 'user c']
        },
        {
            from: 'user b',
            to: ['user d']
        },
        {
            from: 'user c',
            to: ['user d']
        }
    ]

我用了这个代码：

    result = []
    objTemp = obj
    from = []       
    obj.map (o) ->
        if o.from not in from
            from.push o.from
            to = []
            objTemp.map (oo) ->
                if oo.from is o.from and oo.to not in to
                    to.push oo.to
            temp =
                from: o.from
                to: to
            result.push temp

但结果并不是我所期望的，仍然有相同的"to"在相同的"from"中：

   [
        {
            from: 'user a',
            to: ['user b', 'user c']
        },
        {
            from: 'user b',
            to: ['user d']
        },
        {
            from: 'user c',
            to: ['user d', 'user d'] <-- the problem
        }
    ]

你们是怎么用coffeescript解决的？

以下是我的操作方法：

res = []
obj.map (o) ->
  for r in res when r.from is o.from
    return (r.to.push o.to unless o.to in r.to)
  res.push from: o.from, to: [o.to]
为了方便使用groupBy和uniq方法（我想还有chain），我可能会使用下划线或短划线。如果您对它们的实现感兴趣，以便在原始coffeescript中完成，可以查看带注释的源代码。

http://underscorejs.org/docs/underscore.html

chain/value只允许我内联调用这些函数，而不是让临时变量来保存它们或使函数调用不那么明显。

fromUser = (x)-> x.from
toRelationships = (memo, items, index, list)->
  key = _(items).pluck('from')[0]
  vals = _.values(list[key])
  memo[key] = _(vals).chain().pluck('to').uniq().value()
  memo
result = _(data).chain().groupBy(fromUser).reduce(toRelationships, []).value()
# without chain
grouped = _(data).groupBy( fromUser )
result = _(grouped).reduce( toRelationships, [])

jsfiddle：http://jsfiddle.net/eaL6t1cn/2/

这是我的看法，我现在为没有像我希望的那样简洁而道歉。如果我使用jQuery（或其他第三方），这会更简单一些，但这里没有任何库。

a =     [
        {
            from: 'user a',
            to: 'user b'
        },
        {
            from: 'user a',
            to: 'user c'
        },
        {
            from: 'user b',
            to: 'user d'
        },
        {
            from: 'user c',
            to: 'user d'
        },
        {
            from: 'user c',
            to: 'user d'
        }
    ]
c = {}
d = []
(value for value in a).forEach (v) -> 
     c[v.from] = do -> if not c[v.from] then [v.to] else ("#{c[v.from]},#{v.to}").
         split(',').reduce ((p,c) -> 
             if c not in p then p.concat([c]) else [c] ), [] 
d.push {'from':v,'to':k} for v,k of c
console.log d
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