Question

我试图摆脱有时出现在我输入中的重复项以下是数据样本：

{
  "some": "very",
  "random": 0.228,
  "stuff": 31337,
  "people": [
    {
      "name": "carl",
      "height": "180cm",
      "wifes": 1,
      "sons": 0
    },
    {
      "name": "charlie",
      "height": "166cm",
      "wifes": 0,
      "sons": 2
    },
    {
      "name": "carl",
      "height": "180cm",
      "wifes": 1,
      "sons": 1
    },
    {
      "name": "carl",
      "height": "195cm",
      "wifes": 1.95,
      "sons": 12
    }
  ]
}

有很多卡尔，但其中只有2个是重复的 - 名字：carl＆amp;高度：180厘米
例如，我需要平均他的妻子的数量，并总结他的儿子这是预期的结果：

[
  {
    "name": "carl",
    "height": "180cm",
    "wifes": 1,
    "sons": 3
  },
  {
    "name": "charlie",
    "height": "166cm",
    "wifes": 0,
    "sons": 2
  },
  {
    "name": "carl",
    "height": "195cm",
    "wifes": 1.95,
    "sons": 12
  }
]

我尝试使用＆＃39;添加＆＃39;并且＆＃39;减少＆＃39;但我在jq＆gt;＆lt;

中非常新手

Answer 1

分组可以通过group_by/1使用[.name，。height]作为分组标准来实现：

def sum(f): map(f) | add;

def average(f): sum(f) / length;

def consolidate:
  .[0]
  + {wifes: average(.wifes)}
  + {sons:  sum(.sons) } ;

.people
| group_by([.name,.height])
| map(consolidate)

输出

输出符合给定的描述性要求：

[
  {
    "name": "carl",
    "height": "180cm",
    "wifes": 1,
    "sons": 1
  },
  {
    "name": "carl",
    "height": "195cm",
    "wifes": 1.95,
    "sons": 12
  },
  {
    "name": "charlie",
    "height": "166cm",
    "wifes": 0,
    "sons": 2
  }
]

如何通过jq合并（平均/总和）.json中的重复对象？

1 个答案:

输出