使用ramda过滤出唯一的嵌套值

Question

我收集了如下所示的UNIX时间戳：

[
  {"start_time":1540458000000, "end_time":1540472400000}, 
  {"start_time":1540458000000, "end_time":1540486800000}, 
  {"start_time":1540458000000, "end_time":1540501200000}, 
  {"start_time":1540472400000, "end_time":1540486800000}, 
  {"start_time":1540472400000, "end_time":1540501200000}, 
  {"start_time":1540486800000, "end_time":1540501200000}
]

我想从两个挑选出所有的独特的价值观start_time和end_time，所以我留下了：

[
  {"start_time":1540458000000}, 
  {"start_time":1540472400000}, 
  {"start_time":1540486800000}
  {"end_time":1540472400000},
  {"end_time":1540486800000}, 
  {"end_time":1540501200000}, 
]

我看使用类似的东西使用groupBy，pluck，zipObj多使用的答案在这里。但不幸的是没有运气。

最好的是ramda函数，该函数无需指定特定键即可工作。

Answer 1

Ramda的另一种方法：

const {pipe, map, toPairs, unnest, groupBy, head, uniqBy, last, values, apply, objOf} = R

const uniqTimes = pipe(
  map(toPairs),          //=> [[['start', 1], ['end', 2]], [['start', 1], ['end', 3]], ...]
  unnest,                //=> [['start', 1], ['end', 2], ['start', 1], ['end', 3], ...]
  groupBy(head),         //=> {start: [['start', 1], ['start', 1], ['start', 4], ...], end: [['end', 2], ...]}
  map(uniqBy(last)),     //=> {start: [['start', 1], ['start', 4], ...], end: [['end', 2], ...]}
  values,                //=> [[['start', 1], ['start', 4], ...], [['end', 2], ...]]
  unnest,                //=> [['start', 1], ['start', 4], ..., ['end', 2], ...]
  map(apply(objOf))      //=> [{"start": 1}, {"start": 4}, ..., {"end": 2}, ...]
)

const timestamps = [{"start_time":1540458000000,"end_time":1540472400000},{"start_time":1540458000000,"end_time":1540486800000},{"start_time":1540458000000,"end_time":1540501200000},{"start_time":1540472400000,"end_time":1540486800000},{"start_time":1540472400000,"end_time":1540501200000},{"start_time":1540486800000,"end_time":1540501200000}]

console.log(uniqTimes(timestamps))

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

这就是我喜欢使用Ramda的方式：构建一个函数管道，每个函数都进行简单的转换。

更新资料

有条评论询问如何生成更类似的输出

{
   "start_time": [1540458000000, 1540458000000],
   "end_time": [1540472400000, 1540486800000]
}

对于相同的输入，应该这样做：

const uniqTimes = pipe(
  map(toPairs), 
  unnest,
  groupBy(head),
  map(map(last)),
  map(uniq)
)