Underscore.js group 分两层

Question

我似乎无法正确地表达这一点，因为我需要按date和分组foo：

// Using date as numbers for clear example
// date key should be grouped into a key
// Under that key (date), I should see another key of "foo"

let list = [
  { date: "1", foo: "me", result: "meFoo"},
  { date: "1", foo: "me2", result: "meYou"},
  { date: "1", foo: "me3", result: "meHe"},
  { date: "2", foo: "me", result: "meHim"},
  { date: "2", foo: "me2", result: "meHim"}
]

let grouped = _.groupBy(list, "date") // keys ie 1, 2

// I need another key under the date key, possible?
let grouped = _.groupBy(list, "date", (val) => _.groupBy(val), "foo") // result not expected

我在这里变得有点复杂，所以我使用下划线来减轻复杂性。这可能吗？

然后我会做一些类似的事情，我可以使用_.pluck：

Object.keys(grouped).map( dateKey => Object.keys(grouped[dateKey]).map(fooKey => console.log( grouped[dateKey][fookey][0] )) )

这在最高级别上看起来确实是错误的。

预期结果类似于：

DATE // key
 |
 |___ FOO // key
       |
       |__ array foo objects

Answer 1

功能（使用reduce）：

list.reduce((acc, item) => {
  const key1 = item.date
  const group1 = acc[key1] || {}
  const key2 = item.foo
  const group2 = group1[key2] || []
  return {
    ...acc,
    [key1]: {
      ...group1,
      [key2]: group2.concat(item)
    }
  }
}, {})

程序：

let grouped = {}

list.forEach((item) => {
    // Create an empty object if this key has not already been set:
    grouped[item.date] = grouped[item.date] || {}

    // Create an empty list if this key has not already been set:
    grouped[item.date][item.foo] = grouped[item.date][item.foo] || []

    // Push the item to our group:
    grouped[item.date][item.foo].push(item)
})

在 python 中，这将是：

from collections import defaultdict

_list = [
    {"date":"1","foo":"me","result":"meFoo"},
    {"date":"1","foo":"me2","result":"meYou"},
    {"date":"1","foo":"me3","result":"meHe"},
    {"date":"2","foo":"me","result":"meHim"},
    {"date":"2","foo":"me2","result":"meHim"}
]

grouped = defaultdict(lambda: defaultdict(list))

for item in _list:
    grouped[item['date']][item['foo']].append(item)