簡體   English   中英

從字典、嵌套字典、列表中獲取鍵並檢查它出現的次數

[英]Get keys from dictionary, nested dictionary, lists and check how many times it appears

所以基本上,我正在嘗試編寫一個從 json 文件中獲取字典的代碼。 該字典包含其他字典、列表等...我正在嘗試從 json(results,x1,x2,x3,y1,y2,a1,b1) 獲取所有鍵,並顯示每個鍵在檢查時出現的次數嵌套字典和列表。

代碼:

# open data
import json

with open('list.json') as f:
    my_dict = json.load(f)

k=int()

def count_k_v(k):
        keys=0
        if type(k)==dict:
            for value in k.keys():
                if isinstance(k[value], (dict,list)):
                    keys+=1
                    k=count_k_v(k[value])
                    keys+=k
                else:
                    keys+=1
        elif type(k)==list:
            for value in k:
                if isinstance(value,(list,dict)):
                    k=count_k_v(value)
                    keys+=k
                else:
                    keys+=1
        return keys


#['results','x1','x2','y1','x3','y2','a1','b1']     
for key, value in my_dict:
    print("{}".format(key,value))   

#FOUND 8 KEYS.
print("Found {} keys.",count_k_v(my_dict))

#RESULTS FOUND 1 TIME
#X1 KEY FOUND 1 TIME
#X2 KEY FOUND 1 TIME
#Y1 KEY FOUND 1 TIME
#X3 KEY FOUND 1 TIME
#Y2 KEY FOUND 1 TIME
#A1 KEY FOUND 1 TIME
#B1 KEY FOUND 1 TIME
for key in my_dict:

JSON:

{
   "results":[
      {
         "x1":5
      },
      {
         "x2":5,
         "y1":[
            1,
            2,
            3
         ]
      },
      {
         "x3":5,
         "y2":{
            "a1":2,
            "b1":67
         }
      }
   ]
}

嘗試打印:

['results','x1','x2','y1','x3','y2','a1','b1']
FOUND 8 KEYS.
RESULTS FOUND 1 TIME
X1 KEY FOUND 1 TIME
X2 KEY FOUND 1 TIME
Y1 KEY FOUND 1 TIME
X3 KEY FOUND 1 TIME
Y2 KEY FOUND 1 TIME
A1 KEY FOUND 1 TIME
B1 KEY FOUND 1 TIME

您可以使用遞歸方法,使用collections.Counter

from collections import Counter
def get_result(d):
    keys = []

    def recurse(d):
        if isinstance(d, dict):
            for k, v in d.items():
                keys.append(k)
                if isinstance(v, (dict, list)):
                    recurse(v)

        elif isinstance(d, list):
            for item in d:
                if isinstance(item, (dict, list)):
                    recurse(item)
        else:
            return
    recurse(d)
    print(keys)
    print(f'found {len(keys)} keys.'.upper())
    for k, v in Counter(keys).items():
        print(f'{k} key found {v} time'.upper())

get_result(my_dict)

Output:

['results', 'x1', 'x2', 'y1', 'x3', 'y2', 'a1', 'b1']
FOUND 8 KEYS.
RESULTS KEY FOUND 1 TIME
X1 KEY FOUND 1 TIME
X2 KEY FOUND 1 TIME
Y1 KEY FOUND 1 TIME
X3 KEY FOUND 1 TIME
Y2 KEY FOUND 1 TIME
A1 KEY FOUND 1 TIME
B1 KEY FOUND 1 TIME

或者,您可以使用re [不完全證明,推薦用於簡單 json ] 解析json string

import re

def get_result(json_str):
    keys = re.findall(r'"([^"]+?)"\s*:', json_str)
    print(keys)
    print(f'found {len(keys)} keys'.upper())
    for k, v in Counter(keys).items():
        print(f'{k} key found {v} time'.upper())

print(get_result(json.dumps(my_dict)))

Output:

['results', 'x1', 'x2', 'y1', 'x3', 'y2', 'a1', 'b1']
FOUND 8 KEYS.
RESULTS KEY FOUND 1 TIME
X1 KEY FOUND 1 TIME
X2 KEY FOUND 1 TIME
Y1 KEY FOUND 1 TIME
X3 KEY FOUND 1 TIME
Y2 KEY FOUND 1 TIME
A1 KEY FOUND 1 TIME
B1 KEY FOUND 1 TIME

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM