Question

我有一组UNIX时间戳，看起来像这样：

[
  {"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start_time和end_time中挑选出所有唯一值，所以我留下了：

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

我考虑使用groupBy，pluck，zipObj和类似的东西，以及使用answer here的东西。但不幸的是没有运气。

一个很棒的功能是可以在无需指定特定键的情况下运行的ramda函数。

Answer 1

另一种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>

这就是我喜欢使用Ramda的方式：建立一个函数管道，每个函数都进行简单的转换。

更新

一条评论询问如何生成更像

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)
)

Answer 2

不确定ramda，但是下面的普通js函数会做到这一点

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

如果所需的属性未知，但出现在所有对象中，则可以将每个对象转换成对，转置结果数组，获取每个数组的唯一值，将它们嵌套到单个数组，然后转换回对象：

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>

如果知道所需的属性，则可以groupBy一个属性，从每个组中获取第一个对象，然后从每个对象中选择所需的属性：

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>

使用ramda过滤出唯一的嵌套值

3 个答案:

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