Underscore.js - _.groupBy嵌套属性

Question

Tastypie返回一个包含嵌套资源的数组，如下所示：

data = [

{"adult_price": "123", "child_price": "123", "currency": [{"abbrev": "USD", "id": "1", "name": "US Dollars", "resource_uri": "/api/v1/currency/1/", "symbol": "$"}], "day": [{"day_of_week": "TUE", "id": "2", "resource_uri": "/api/v1/days/2/"}], "description": "Please enter the tour description here", "id": "1", "important": "ex. Please contact us to negotiate a price if you want to book the Fiat for 1 person only.", "location": [{"id": "1", "name": "Dublin", "resource_uri": "/api/v1/location/1/"}], "name": "test dublin", "resource_uri": "/api/v1/tours/1/", "start_time": "23:03:51", "subtitle": "ex. These prices include... but not...", "teenager_student_price": "123", "under_6_price": "123"}, 

{"adult_price": "22", "child_price": "22", "currency": [{"abbrev": "USD", "id": "1", "name": "US Dollars", "resource_uri": "/api/v1/currency/1/", "symbol": "$"}], "day": [{"day_of_week": "WED", "id": "3", "resource_uri": "/api/v1/days/3/"}], "description": "Please enter the tour description here", "id": "2", "important": "ex. Please contact us to negotiate a price if you want to book the Fiat for 1 person only.", "location": [{"id": "2", "name": "Venice", "resource_uri": "/api/v1/location/2/"}], "name": "test Venice", "resource_uri": "/api/v1/tours/2/", "start_time": "23:09:01", "subtitle": "ex. These prices include... but not...", "teenager_student_price": "22", "under_6_price": "22"}, 

{"adult_price": "22", "child_price": "222", "currency": [{"abbrev": "USD", "id": "1", "name": "US Dollars", "resource_uri": "/api/v1/currency/1/", "symbol": "$"}], "day": [{"day_of_week": "MON", "id": "1", "resource_uri": "/api/v1/days/1/"}, {"day_of_week": "TUE", "id": "2", "resource_uri": "/api/v1/days/2/"}, {"day_of_week": "WED", "id": "3", "resource_uri": "/api/v1/days/3/"}], "description": "Please enter the tour description here", "id": "3", "important": "ex. Please contact us to negotiate a price if you want to book the Fiat for 1 person only.", "location": [{"id": "3", "name": "Rome", "resource_uri": "/api/v1/location/3/"}], "name": "test Rome", "resource_uri": "/api/v1/tours/3/", "start_time": "23:15:09", "subtitle": "ex. These prices include... but not...", "teenager_student_price": "22", "under_6_price": "222"}, 

{"adult_price": "22", "child_price": "222", "currency": [{"abbrev": "USD", "id": "1", "name": "US Dollars", "resource_uri": "/api/v1/currency/1/", "symbol": "$"}], "day": [{"day_of_week": "MON", "id": "1", "resource_uri": "/api/v1/days/1/"}, {"day_of_week": "WED", "id": "3", "resource_uri": "/api/v1/days/3/"}], "description": "Please enter the tour description here", "id": "4", "important": "ex. Please contact us to negotiate a price if you want to book the Fiat for 1 person only.", "location": [{"id": "3", "name": "Rome", "resource_uri": "/api/v1/location/3/"}], "name": "test Rome 2", "resource_uri": "/api/v1/tours/4/", "start_time": "01:01:11", "subtitle": "ex. These prices include... but not...", "teenager_student_price": "22", "under_6_price": "22"}, {"adult_price": "123", "child_price": "123", "currency": [{"abbrev": "USD", "id": "1", "name": "US Dollars", "resource_uri": "/api/v1/currency/1/", "symbol": "$"}], "day": [{"day_of_week": "TUE", "id": "2", "resource_uri": "/api/v1/days/2/"}, {"day_of_week": "THU", "id": "4", "resource_uri": "/api/v1/days/4/"}], "description": "Please enter the tour description here", "id": "5", "important": "ex. Please contact us to negotiate a price if you want to book the Fiat for 1 person only.", "location": [{"id": "2", "name": "Venice", "resource_uri": "/api/v1/location/2/"}], "name": "test Venice 2", "resource_uri": "/api/v1/tours/5/", "start_time": "01:03:27", "subtitle": "ex. These prices include... but not...", "teenager_student_price": "123", "under_6_price": "123"}

]

我想执行.groupBy操作，通过location[0].names属性进行分组，这应该返回3个数组。这可能吗？

我该如何执行相当于:

_.groupBy(data, 'location[0].name']

- snakesNbronies

Answer 1

groupBy 的第二个参数可以是一个函数或一个字符串：

groupBy _.groupBy(list, iterator)

将集合按通过iterator运行每个值的结果分成组。如果iterator是字符串而不是函数，则按每个值上由iterator命名的属性进行分组。

由于您要分组的不是简单的顶级属性，因此应使用函数形式：

_(data).groupBy(function(o) {
    return o.location[0].name;
});

或者：

_(data).groupBy(o => o.location[0].name)

示例：http://jsfiddle.net/ambiguous/YFKXC/

- mu is too short