使用ramda過濾掉唯一的嵌套值

Question

我有一組UNIX時間戳，如下所示：

[
  {"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}
]

我想從start_time和end_time選出所有唯一值，所以我留下：

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

我看使用類似的東西使用groupBy ， pluck ， zipObj多使用這里的答案 。 但遺憾的是沒有運氣。

一個很好的將是一個ramda函數，無需給定特定的鍵即可工作。

Answer 1

另一個Ramda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> 

這就是我喜歡使用Ramda的方法：構建一個函數管道，每個函數都進行簡單的轉換。

---

更新

評論詢問如何生成更像的輸出

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

這應該為相同的輸入做到這一點：

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

Answer 2

不確定ramda，但在普通的js函數下面會這樣做

 const arr = [ {"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} ]; function foo(arr) { return [...arr.reduce((a, b) => { Object.entries(b).forEach(e => a.set(String(e), e)); return a; }, new Map())].map(([_,e]) => ({ [e[0]]: e[1] })) } console.log(foo(arr)); 

Answer 3

如果所需的屬性未知但出現在所有對象中，則可以將每個對象轉換為對，轉置結果數組，獲取每個數組的唯一值，將它們排除在單個數組中，然后轉換回對象：

 const { pipe, map, toPairs, transpose, uniqBy, last, unnest, objOf, apply } = R; const data = [ {"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} ]; const getUniqueProps = pipe( map(toPairs), transpose, map(uniqBy(last)), unnest, map(apply(objOf)) ); console.log(getUniqueProps(data)); 

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

如果你知道你想要的屬性，你可以groupBy的屬性，得到各組的第一個對象，然后接你從每個對象所需的屬性：

 const { groupBy, prop, concat, pipe, map, head, pick, values } = R; const data = [ {"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} ]; const getUniqueProp = (propName) => pipe( groupBy(prop(propName)), map(pipe(head, pick([propName]))), values, ); const getStartEnd = (data) => concat( getUniqueProp('start_time')(data), getUniqueProp('end_time')(data), ); console.log(getStartEnd(data)); 

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