如何使用Underscore.js按键对数据集合进行分组？

Question

考虑到以下数据格式，如何通过密钥对数据进行分组？

数据格式为：

data=[
    {
        "billingid": 138,
        "amount": "800",
        "billamount": "200",
        "balance": "1",
        "paid": "810",
        "billdate": "Apr 2, 2014 12:00:00 AM",
        "tax": "11",
        "patientid": "TMCH17",
        "configid": 6,
        "paymenttype": "cash",
        "servicename": "Room",
        "firstname": "kannan"
    },
    {
        "billingid": 138,
        "amount": "800",
        "billamount": "500",
        "balance": "1",
        "paid": "810",
        "billdate": "Apr 2, 2014 12:00:00 AM",
        "tax": "11",
        "patientid": "TMCH17",
        "configid": 3,
        "paymenttype": "cash",
        "servicename": "Lab",
        "firstname": "kannan"
    },
    {
        "billingid": 138,
        "amount": "800",
        "billamount": "100",
        "balance": "1",
        "paid": "810",
        "billdate": "Apr 2, 2014 12:00:00 AM",
        "tax": "11",
        "patientid": "TMCH17",
        "configid": 1,
        "paymenttype": "cash",
        "servicename": "Consultation",
        "firstname": "kannan"
    }]

预期输出为：

 [{
        "billingid": 138,
        "amount": "800",
        bamount:[{
            "billamount": "100",
            "billamount": "200",
            "billamount": "300"
         }],
        "balance": "1",
        "paid": "810",
        "tax": "11",
        "patientid": "TMCH17",
        "configid": 1,
        "paymenttype": "cash",
        service[{
            "servicename": "Consultation",
            "servicename": "room",
            "servicename": "lab",
        }],
        "firstname": "kannan"
    }]

Answer 1

伪代码

以下是集合中每个对象的解决方案，

1. 验证结果集中是否存在每个密钥。
2. 如果密钥存在且结果不等于该值，则
   1. 将值转换为数组并保留所有唯一值。
   2. 否则，请将此键的值设置为值。

问题

您的问题中未定义您是否需要独特的结果。我怀疑结果集对于某些键不应该是唯一的。在这种情况下，您应该修改each中的逻辑以反映该逻辑。例如，您可能希望billamount进行所有重复，但您希望billingid是唯一的。

代码

var summary = _.reduce(data, function(result, object) {

    _.each(object, function(value, key) {
        result[key] = ( _.has(result, key) && result[key] !== value ?
                      _.uniq(_.flatten([result[key], value])) : value);
    });

    return result;
} );