Javascript 然后拉姆达集团将兄弟姐妹转化为子女
在ramda.js中,如何按字段对列表进行分组和排序,然后将每个组中除第一项以外的所有项移动到该项的子项中 例如,下面我按姓名分组并按日期降序排序:Javascript 然后拉姆达集团将兄弟姐妹转化为子女,javascript,functional-programming,ramda.js,Javascript,Functional Programming,Ramda.js,在ramda.js中,如何按字段对列表进行分组和排序,然后将每个组中除第一项以外的所有项移动到该项的子项中 例如,下面我按姓名分组并按日期降序排序: [{ id: 1, name: 'bob', date: '2007-03-05', count: 15}, { id: 2, name: 'bob', date: '2007-03-04', count: 32}, { id: 3, name: 'bob', date: '2007-03-01', count: 27}, { id: 4,
[{ id: 1, name: 'bob', date: '2007-03-05', count: 15},
{ id: 2, name: 'bob', date: '2007-03-04', count: 32},
{ id: 3, name: 'bob', date: '2007-03-01', count: 27},
{ id: 4, name: 'jack', date: '2007-03-04', count: 3},
{ id: 5, name: 'jack', date: '2007-02-22', count: 5}]
进入
我知道我可以用R.head抓取整个列表的顶部项,用R.tail抓取其余项,然后用R.merge将其作为子项添加,但我不知道如何只抓取列表中某个组的顶部或尾部。您还没有对它们进行分组,只是按名称排序。要将它们分组到数组中,请使用:
将其应用于数据后,使用
映射
从每组创建一个对象。以下是一种获得所需结果的方法
const groupByName = groupBy(obj => obj.name);
const sortByDate = sortBy(obj => obj.date);
const grouped = pipe(groupByName, map(sortByDate), values);
reduce( (acc, val) => {
acc.push(merge(head(val), {children: tail(val)}));
return acc;
}, [], grouped(data));
可能有更好的方法,但我认为这是一个开始:
function yourAnswer (data) {
const groupByName = groupBy((person) => person.name)
return (
Object.values(groupByName(data))
.map((g) => g.sort((a, b) => a.id - b.id)) // sort by ascending id
.map(function(g) { // for each group, sorted by id
const o = Object.assign(g[0]) // create an object from the minimal id
o['children'] = g.slice(1) // assign 'children' to be equal the other objects in the group
return o
})
)}
let data = [{ id: 1, name: 'bob', date: '2007-03-05', count: 15},
{ id: 2, name: 'bob', date: '2007-03-04', count: 32},
{ id: 3, name: 'bob', date: '2007-03-01', count: 27},
{ id: 4, name: 'jack', date: '2007-03-04', count: 3},
{ id: 5, name: 'jack', date: '2007-02-22', count: 5}]
console.log(yourAnswer(data))
请尝试另一种方法:
const fn = pipe(
groupBy(prop('name')),
values,
map(lift(assoc('children'))(tail, head))
);
如果要将排序包含在此中,可以在值之后添加此值,
:
map(sort(descend(prop('date')))),
如果这不清楚:map(lift(assoc('children'))(tail,head))
您可以将其替换为等效的:
map((group) => assoc('children', tail(group), head(group)))
你可以在上看到这一点,这太神奇了
map(sort(descend(prop('date')))),
map((group) => assoc('children', tail(group), head(group)))