【Py】列表字典转为字典

假设我们有如下列表

[
  {
    "一级": "数码产品",
    "二级": "手机",
    "三级": "苹果",
    "四级": "iPhone11",
    "编码": "001"
  },
  {
    "一级": "数码产品",
    "二级": "电脑",
    "三级": "苹果",
    "四级": "MacBook",
    "编码": "002"
  },
  {
    "一级": "数码产品",
    "二级": "手机",
    "三级": "苹果",
    "四级": "iPhone12",
  	"编码": "003"
  },
  {
    "一级": "数码产品",
    "二级": "耳机",
    "三级": "苹果",
    "四级": "AirPods",
    "编码": "004"
  },
  {
    "一级": "数码产品",
    "二级": "手机",
    "三级": "三星",
    "四级": "Galaxy",
    "编码": "005"
  }
]

我们希望将其变为如下树形结构的字典

{
  "数码产品": {
    "手机": {
      "苹果": {
        "iPhone11": {
          "编码": "001"
        },
        "iPhone12": {
          "编码": "003"
        }
      },
      "三星": {
        "Galaxy": {
          "编码": "005"
        }
      }
    },
    "电脑": {
      "苹果": {
        "MacBook": {
          "编码": "002"
        }
      }
    },
    "耳机": {
      "苹果": {
        "AirPods": {
          "编码": "004"
        }
      }
    }
  }
}

test = [
    {"一级": "数码产品", "二级": "手机", "三级": "苹果", "四级": "iPhone", "编码": "001"},
    {"一级": "数码产品", "二级": "电脑", "三级": "苹果", "四级": "MacBook", "编码": "002"},
    {"一级": "数码产品", "二级": "耳机", "三级": "苹果", "四级": "AirPods", "编码": "003"},
    {"一级": "数码产品", "二级": "手机", "三级": "三星", "四级": "Galaxy", "编码": "004"}
]

for name_0 in set(d['一级'] for d in test):
    dic[name_0] = {}
    for name_1 in set(d['二级'] for d in test if d['一级'] == name_0):
        dic[name_0][name_1] = {}
        for name_2 in set(d['三级'] for d in test if d['一级'] == name_0 and d['二级'] == name_1):
            dic[name_0][name_1][name_2] = {}
            for name_3 in set(d['四级'] for d in test if d['一级'] == name_0 and d['二级'] == name_1 and d['三级'] == name_2):
                dic[name_0][name_1][name_2][name_3] = {}

for item in test:
   dic[item['一级']][item['二级']][item['三级']][item['四级']]['编码'] = item['编码']

但是这种方式单从代码上看就非常繁琐，而且不灵活，应该还会有更好的办法

我们发现以下代码仅是为了创建分支节点

for name_0 in set(d['一级'] for d in test):
    dic[name_0] = {}
    for name_1 in set(d['二级'] for d in test if d['一级'] == name_0):
        dic[name_0][name_1] = {}
        for name_2 in set(d['三级'] for d in test if d['一级'] == name_0 and d['二级'] == name_1):
            dic[name_0][name_1][name_2] = {}
            for name_3 in set(d['四级'] for d in test if d['一级'] == name_0 and d['二级'] == name_1 and d['三级'] == name_2):
                dic[name_0][name_1][name_2][name_3] = {}

我们可以利用defaultdict，在键不存在的时候自动创建，从而将上面的代码省略掉

import json
from collections import defaultdict

test = [
    {"一级": "数码产品", "二级": "手机", "三级": "苹果", "四级": "iPhone", "编码": "001"},
    {"一级": "数码产品", "二级": "电脑", "三级": "苹果", "四级": "MacBook", "编码": "002"},
    {"一级": "数码产品", "二级": "耳机", "三级": "苹果", "四级": "AirPods", "编码": "003"},
    {"一级": "数码产品", "二级": "手机", "三级": "三星", "四级": "Galaxy", "编码": "004"}
]
dic = defaultdict(lambda: defaultdict(lambda: defaultdict(lambda: defaultdict(dict))))
for item in test:
    dic[item['一级']][item['二级']][item['三级']][item['四级']]['编码'] = item['编码']

# 转为字典
json.loads(json.dumps(dic))

注意:

1. 转字典时，dict()方式仅能转换根节点，子节点则依然为defaultdict，故需要使用json；
2. 上面定义的dic最大能满足层数为5的树，如果层数大于5则需要在定义dic的时候再多嵌套若干层defaultdict(lambda: defaultdict(dict))

参考：
https://howchoo.com/python/nested-defaultdict-python
https://stackoverflow.com/a/32303615/7151777