我有以下对象records
:
{
"notes":[
{
"id":1,
"description":"hey",
"userId":2,
"replyToId":null,
"postId":2,
"parentId":null
},
{
"id":5,
"description":"hey test",
"userId":3,
"replyToId":null,
"postId":2,
"parentId":null
},
{
"id":2,
"description":"how are you",
"userId":null,
"replyToId":2,
"postId":2,
"parentId":null,
"user":null
}
]
}
我想将其输出为:
2
object with id 1
object with id 2 (because replyToId value is same as userId
3
object with id 5
所以基本上我想在同一组下考虑UserId和replyToId值。
我在lodash下构建了自己的mixin,将groupBy方法包装为:
mixin({
splitGroupBy: function(list, groupByIter){
if (_.isArray(groupByIter)) {
function groupBy(obj) {
return _.forEach(groupByIter, function (key){
if ( !!obj[key] ) return obj[key]
});
}
} else {
var groupBy = groupByIter;
}
debugger;
var groups = _.groupBy(list, groupBy);
return groups;
}
});
通话看起来像这样:
_.splitGroupBy(data.notes,['userId', 'replyToId']);
输出没有组。即使我尝试使用_.map
代替_.forEach
,分割也没有正确发生。
答案 0 :(得分:8)
使用下划线的解决方案:
var props = ['userId', 'replyToId'];
var notNull = _.negate(_.isNull);
var groups = _.groupBy(record.notes, function(note){
return _.find(_.pick(note, props), notNull);
});
答案 1 :(得分:1)
这可能做得更漂亮,但它应该有效:
lodash.mixin({
splitGroupBy: function(list, groupByIter) {
var _ = this, groupBy;
if (lodash.isArray(groupByIter)) {
groupBy = function(obj) {
return _(obj) .pick(groupByIter)
.values()
.without(null, undefined)
.first();
};
} else {
groupBy = groupByIter;
}
var groups = _.groupBy(list, groupBy);
return groups;
}
});
答案 2 :(得分:1)
您可以将属性列表映射到各自的值,然后选择第一个非虚假值作为组密钥:
_.mixin({
splitGroupBy: function(list, groupByIter){
if (!_.isArray(groupByIter))
return _.groupBy(list, groupByIter);
return _.groupBy(list, function(o) {
var values = _.map(groupByIter, function(k) {
return o[k];
});
return _.find(values);
});
}
});
var data = {
"notes":[
{
"id":1,
"userId":2,
"replyToId":null
},
{
"id":5,
"userId":3,
"replyToId":null
},
{
"id":2,
"userId":null,
"replyToId":2
}
]
};
_.mixin({
splitGroupBy: function(list, groupByIter){
if (!_.isArray(groupByIter))
return _.groupBy(list, groupByIter);
return _.groupBy(list, function(o) {
var values = _.map(groupByIter, function(k) {
return o[k];
});
return _.find(values);
});
}
});
snippet.log(JSON.stringify(_.splitGroupBy(data.notes,['userId', 'replyToId'])));
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>
<!-- Provides the `snippet` object, see http://meta.stackexchange.com/a/242144/134069 -->
<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>
答案 3 :(得分:1)
假设userId
和replyToId
是相互排斥的(即您有userId
或replyToId
,但从不两者都有),因为它们在示例数据中,指定自定义分组功能有效:
_.groupBy(data.notes, function(note) {
return note.userId || note.replyToId;
});
答案 4 :(得分:1)
您可以将stringify对象用作键。
_.groupBy(notes, ({ userId, replyToId }) => JSON.stringify({ userId, replyToId }));
输出:
{
"{\"userId\":2,\"replyToId\":null}": [
{
"id": 1,
"description": "hey",
"userId": 2,
"replyToId": null,
"postId": 2,
"parentId": null
}
],
"{\"userId\":3,\"replyToId\":null}": [
{
"id": 5,
"description": "hey test",
"userId": 3,
"replyToId": null,
"postId": 2,
"parentId": null
}
],
"{\"userId\":null,\"replyToId\":2}": [
{
"id": 2,
"description": "how are you",
"userId": null,
"replyToId": 2,
"postId": 2,
"parentId": null,
"user": null
}
]
}