将JSON解组成具有唯一元素的映射切片

huangapple
go评论110阅读模式
英文:

Unmarshal JSON into slice of maps with unique elements

问题

我正在将一些JSON文件解组成[]map[string]string{}，但是很多时候源数据中有很多重复的相同对象。

输入的样子是这样的：

[{"sa1":"8172"},{"sa3":"8175"},{"sa1":"8172"},{"sa3":"8175"},{"sa3":"8175"},{"sa3":"8175"},{"sa1":"8172"},{"sa3":"8175"},{"sa3":"8175"}]

结果是：

  1. map[sa1:8172]
  2. ([]map[string]string) (len=9 cap=9) {
  3.  (map[string]string) (len=1) {
  4.   (string) (len=3) "sa1": (string) (len=4) "8172"
  5.  },
  6.  (map[string]string) (len=1) {
  7.   (string) (len=3) "sa3": (string) (len=4) "8175"
  8.  },
  9.  (map[string]string) (len=1) {
  10.   (string) (len=3) "sa1": (string) (len=4) "8172"
  11.  },
  12.  (map[string]string) (len=1) {
  13.   (string) (len=3) "sa3": (string) (len=4) "8175"
  14.  },
  15.  (map[string]string) (len=1) {
  16.   (string) (len=3) "sa3": (string) (len=4) "8175"
  17.  },
  18.  (map[string]string) (len=1) {
  19.   (string) (len=3) "sa3": (string) (len=4) "8175"
  20.  },
  21.  (map[string]string) (len=1) {
  22.   (string) (len=3) "sa1": (string) (len=4) "8172"
  23.  },
  24.  (map[string]string) (len=1) {
  25.   (string) (len=3) "sa3": (string) (len=4) "8175"
  26.  },
  27.  (map[string]string) (len=1) {
  28.   (string) (len=3) "sa3": (string) (len=4) "8175"
  29.  }
  30. }

我该如何清理映射切片，使其只包含唯一的元素？

英文:

I'm unmarshaling some json files into []map[string]string{}
but very often the source is dirty with many repeated equal objects.

The input looks like:

[{"sa1":"8172"},{"sa3":"8175"},{"sa1":"8172"},{"sa3":"8175"},{"sa3":"8175"},{"sa3":"8175"},{"sa1":"8172"},{"sa3":"8175"},{"sa3":"8175"}]

Resulting in:

  1. map[sa1:8172]
  2. ([]map[string]string) (len=9 cap=9) {
  3.  (map[string]string) (len=1) {
  4.   (string) (len=3) "sa1": (string) (len=4) "8172"
  5.  },
  6.  (map[string]string) (len=1) {
  7.   (string) (len=3) "sa3": (string) (len=4) "8175"
  8.  },
  9.  (map[string]string) (len=1) {
  10.   (string) (len=3) "sa1": (string) (len=4) "8172"
  11.  },
  12.  (map[string]string) (len=1) {
  13.   (string) (len=3) "sa3": (string) (len=4) "8175"
  14.  },
  15.  (map[string]string) (len=1) {
  16.   (string) (len=3) "sa3": (string) (len=4) "8175"
  17.  },
  18.  (map[string]string) (len=1) {
  19.   (string) (len=3) "sa3": (string) (len=4) "8175"
  20.  },
  21.  (map[string]string) (len=1) {
  22.   (string) (len=3) "sa1": (string) (len=4) "8172"
  23.  },
  24.  (map[string]string) (len=1) {
  25.   (string) (len=3) "sa3": (string) (len=4) "8175"
  26.  },
  27.  (map[string]string) (len=1) {
  28.   (string) (len=3) "sa3": (string) (len=4) "8175"
  29.  }
  30. }

How could I clean the slice of maps to contain only unique elements?

答案1

得分: 3

一种选择是将键值对直接解组为可比较的类型，比如结构体：

  1. type Elem struct {
  2. 	k string
  3. 	v string
  4. }
  6. func (e *Elem) UnmarshalJSON(d []byte) error {
  7. 	m := map[string]string{}
  8. 	if err := json.Unmarshal(d, &m); err != nil {
  9. 		return err
  10. 	}
  12. 	for k, v := range m {
  13. 		e.k = k
  14. 		e.v = v
  15. 		return nil
  16. 	}
  17. 	return nil
  18. }

一旦你可以逐个比较元素，你还可以将其包装在一个集合中，在解组时过滤元素。在这里隐式地进行过滤，或者事后进行过滤，这是一个意见问题。将过滤作为自己的方法可能更好地分离关注点，但为了简洁起见，我将其包含在UnmarshalJSON中。

  1. type Elems []Elem
  3. func (e *Elems) UnmarshalJSON(d []byte) error {
  4. 	tmp := []Elem{}
  5. 	err := json.Unmarshal(d, &tmp)
  6. 	if err != nil {
  7. 		return err
  8. 	}
  10. 	seen := map[Elem]bool{}
  11. 	for _, elem := range tmp {
  12. 		if seen[elem] {
  13. 			continue
  14. 		}
  15. 		seen[elem] = true
  16. 		*e = append(*e, elem)
  17. 	}
  18. 	return nil
  19. }

然后你可以解组为Elems：

  1. elems := Elems{}
  3. err := json.Unmarshal(js, &elems)
  4. if err != nil {
  5. 	log.Fatal(err)
  6. }
  8. fmt.Println(elems)

这将给你两个唯一的键值对：[{sa1 8172} {sa3 8175}]

https://go.dev/play/p/U0iqBAjvz-1

英文:

One option is to unmarshal the key value pairs directly into a comparable type, like a struct:

  1. type Elem struct {
  2. 	k string
  3. 	v string
  4. }
  6. func (e *Elem) UnmarshalJSON(d []byte) error {
  7. 	m := map[string]string{}
  8. 	if err := json.Unmarshal(d, &m); err != nil {
  9. 		return err
  10. 	}
  12. 	for k, v := range m {
  13. 		e.k = k
  14. 		e.v = v
  15. 		return nil
  16. 	}
  17. 	return nil
  18. }

Once you can compare the elements individually, you could also wrap that in a collection which filters the elements while unmarshaling. Whether to do this implicitly here, or after the fact is a matter of opinion. It may be a better separation of concerns to make filtering its own method, but I included it in UnmarshalJSON for brevity.

  1. type Elems []Elem
  3. func (e *Elems) UnmarshalJSON(d []byte) error {
  4. 	tmp := []Elem{}
  5. 	err := json.Unmarshal(d, &tmp)
  6. 	if err != nil {
  7. 		return err
  8. 	}
  10. 	seen := map[Elem]bool{}
  11. 	for _, elem := range tmp {
  12. 		if seen[elem] {
  13. 			continue
  14. 		}
  15. 		seen[elem] = true
  16. 		*e = append(*e, elem)
  17. 	}
  18. 	return nil
  19. }

Then you can unmarshal into Elems:

  1. 	elems := Elems{}
  3. 	err := json.Unmarshal(js, &elems)
  4. 	if err != nil {
  5. 		log.Fatal(err)
  6. 	}
  8. 	fmt.Println(elems)

Which will give you the two unique pairs: [{sa1 8172} {sa3 8175}]

https://go.dev/play/p/U0iqBAjvz-1

huangapple
  • 本文由 发表于 2022年11月11日 22:01:36
  • 转载请务必保留本文链接：https://go.coder-hub.com/74403623.html
  • duplicates
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  • json
  • slice
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