Question

我正在尝试将以下输入转换为下面显示的预期输出JSON数组。 indexCols也是

var objects = ['{"ticker":"MSFT", "key": 2, "PX_LAST":100}',
               '{"ticker":"AAPL", "key": 3, "PX_LAST":100}',
               '{"ticker":"GOOG", "key": 4, "PX_LAST":100}']

/* Expected output
  [[id, '{"ticker":"MSFT",  "key": 2, "PX_LAST":100}', 'MSFT', 2 ]
    [id, '{"ticker":"AAPL", "key": 3, "PX_LAST":100}', 'AAPL', 3 ]
    [id, '{"ticker":"GOOG", "key": 4, "PX_LAST":100}', 'GOOG', 4 ]]
*/
var indexCols = ['ticker', 'key']

var transform = function(input, indexCols, id) {
  var fn2 = R.compose(R.flip(R.map)(indexCols), R.flip(R.prop), JSON.parse)
  var fn3 = R.map(R.compose(R.concat([id]), fn2))
  return fn3(objects)
}

transform(objects, indexCols, 100)
/* result : [[100, "MSFT", 2], [100, "AAPL", 3], [100, "GOOG", 4]] */

正如您所看到的，结果数组缺少第二个元素，即json字符串。我无法以功能性的方式编织它。

Answer 1

（我认为有一个拼写错误，而不是这个：返回fn3（对象）你可能是这个意思：return fn3（input））

  var transform = function(input, indexCols, id) {
      var fn2 = R.compose(R.flip(R.map)(indexCols), R.flip(R.prop), JSON.parse);
      var valueToArray = R.flip(R.append)([]);
      var fn3 = R.map(R.compose(R.flatten, R.concat([id]), R.compose(R.ap([R.identity, fn2]), valueToArray)))
      return fn3(input)
    }

https://codepen.io/anon/pen/Jrzmda

你必须复制你的输入对象（a-＆gt; [a，a]），然后你可以随意操作第二个元素（fn2），只保持第一个元素不变（标识），然后在最后，把它们连在一起。

改进1：valueToArray（a-＆gt; [a]）是否已经存在Ramda函数？改进2：有没有办法连接id 而不是嵌套来摆脱R.flatten？

Answer 2

您可以使用R.converge，以便将字符串提供给concat和JSON.parse：

var objects = ['{"ticker":"MSFT", "key": 2, "PX_LAST":100}',
               '{"ticker":"AAPL", "key": 3, "PX_LAST":100}',
               '{"ticker":"GOOG", "key": 4, "PX_LAST":100}']

/* Expected output
  [[id, '{"ticker":"MSFT",  "key": 2, "PX_LAST":100}', 'MSFT', 2 ]
    [id, '{"ticker":"AAPL", "key": 3, "PX_LAST":100}', 'AAPL', 3 ]
    [id, '{"ticker":"GOOG", "key": 4, "PX_LAST":100}', 'GOOG', 4 ]]
*/
var indexCols = ['ticker', 'key']

var transform = function(input, indexCols, id) {
  //Have to trick R.converge into making it think that JSON.parse only has 1 argument
  var fn2 = R.compose(R.flip(R.map)(indexCols), R.flip(R.prop), R.curryN(1, JSON.parse))
  var fn3 = R.map(R.compose(R.concat([id]), R.converge(R.concat, [Array, fn2])))
  return fn3(input) //typo shouldn't be objects
}

console.log(transform(objects, indexCols, 100))
/* result : [[100, "MSFT", 2], [100, "AAPL", 3], [100, "GOOG", 4]] */

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.min.js"></script>

另一种方法是使用：

R.ap(R.compose(R.concat, Array), fn2))
// concat <$> Array <*> fn2
// R.liftN(2, R.concat)(Array, fn2) is the same

而不是：

R.converge(R.concat, [Array, fn2])

这样你就不会遇到R.converge问题，因为它猜测JSON.parse有2个参数。

Answer 3

这些看起来不必要地复杂......

怎么样：

const objects = ['{"ticker":"MSFT", "key": 2, "PX_LAST":100}',
               '{"ticker":"AAPL", "key": 3, "PX_LAST":100}',
               '{"ticker":"GOOG", "key": 4, "PX_LAST":100}']
const indexCols = ['ticker', 'key', 'PX_LAST']

const pickIndeces = pipe(JSON.parse, props(indexCols))
const format = map(x => concat([x], pickIndeces(x)))
format(objects)

注意，pipe，props，concat和map是Ramda函数。所以你可以导入它们，如下所示：

const { concat, map, pipe, props } = R

坚持拉姆达组成

3 个答案: