如何解析具有相同键的多个字典值

Question

我有很多行数据（我不能手动修改），它们被表示为一个 key/value 对的字典。问题是有一个字典键可以多次出现（对于未定义的数字：可能是两次、三次、十次等），具有不同的值。

我需要提取所有这些值。

这是一个简单的记录，有两个键值 Key-Word:

{"Date": "Fri, 19 Apr 2019 00:54:46 GMT", "Vary": "Host,Accept-Encoding", "Key-Word": "00a", "Cache-Control": "private", "Key-Word": "xn"}

我写了这个 python 脚本来提取记录的值。

import ast
import re
import json


inFile = open("sample.txt","r",errors="replace") 


cP=0 # key found flag
cV=0 # hold the key's value


try:
    myDict = {"Date": "Fri, 19 Apr 2019 00:54:46 GMT", "Vary": "Host,Accept-Encoding", "Key-Word": "00a", "Cache-Control": "private", "Key-Word": "xn"}
    smallmyDict= {}

except (ValueError, SyntaxError) as E:
    cV="error"
except Exception as E:
    cV="error"

# convert the header's key to small letter
for key, value in myDict.items():
    smallmyDict[key.lower()] = value

# store all keys
smallmyDictKeys =smallmyDict.keys()



# search for a specific key
if 'key-word' in smallmyDictKeys: 
    cP=1
    cV = smallmyDict['key-word']
    print("Found!")
    print(cV) #print the key's value
else:
    print("NOT Found!")

我得到的输出是：

Found! xn

问题是它只打印最后一个键的值。

如果我要查找的键多次出现并单独打印每个值而不是用最后一个值覆盖它，我如何让我的代码迭代它？

Answer 1

字典中不能有 2 个同名的键。一个会覆盖另一个。在运行时，只有一对该密钥存在（最后一个条目）。

https://www.python-course.eu/dictionaries.php - 是阅读字典的好资源。

Answer 2

您可以解析字符串并将值作为列表存储在字典中：

import ast
from pprint import pprint

def parse_dict_multikey(s):
    p = ast.parse(s)
    exp_dict = p.body[0].value
    keys = list(map(ast.literal_eval, exp_dict.keys))
    values = list(map(ast.literal_eval, exp_dict.values))
    d = {}
    for k, v in zip(keys, values):
        d.setdefault(k, []).append(v)
    return d

s = ('{"Date": "Fri, 19 Apr 2019 00:54:46 GMT",'
     ' "Vary": "Host,Accept-Encoding",'
     ' "Key-Word": "00a",'
     ' "Cache-Control": "private",'
     ' "Key-Word": "xn"}')
pprint(parse_dict_multikey(s))
# {'Cache-Control': ['private'],
#  'Date': ['Fri, 19 Apr 2019 00:54:46 GMT'],
#  'Key-Word': ['00a', 'xn'],
#  'Vary': ['Host,Accept-Encoding']}

但是，这使得每个值都成为一个列表，而不仅仅是那些具有重复键的值。如果你使用 Counter，你可以避免这种情况，因为建议：

def parse_dict_multikey(s):
    p = ast.parse(s)
    exp_dict = p.body[0].value
    keys = list(map(ast.literal_eval, exp_dict.keys))
    values = list(map(ast.literal_eval, exp_dict.values))
    c = Counter(keys)
    d = {}
    for k, v in zip(keys, values):
        if c[k] > 1:
            d.setdefault(k, []).append(v)
        else:
            d[k] = v
    return d

哪个会给你：

{'Cache-Control': 'private',
 'Date': 'Fri, 19 Apr 2019 00:54:46 GMT',
 'Key-Word': ['00a', 'xn'],
 'Vary': 'Host,Accept-Encoding'}

您还可以查看更高级的内容，例如 multidict。

Answer 3

由于重复的键，您的数据无法直接加载到 json，试试这个：

from collections import defaultdict

string = '{"Date": "Fri, 19 Apr 2019 00:54:46 GMT", "Vary": "Host,Accept-Encoding", "Key-Word": "00a", "Cache-Control": "private", "Key-Word": "xn"}'

pieces = string.split('",')

for each_piece in pieces:
    key, value = each_piece.split(':', maxsplit=1)
    actual_key = key.strip('{"')
    actual_value = value.strip(' "')
    data[actual_key].append(actual_value)

print(data)

输出:

defaultdict(list,
            {' "Cache-Control': ['private'],
             ' "Key-Word': ['00a', 'xn"}'],
             ' "Vary': ['Host,Accept-Encoding'],
             'Date': ['Fri, 19 Apr 2019 00:54:46 GMT']})

Answer 4

您可以使用 json 来解析您的数据，并使用 json.loads 的 object_pairs_hook 参数对数据进行个性化处理。在下面的示例中，我将相同键的不同值分组在一个列表中（并且，按照您评论中的要求，将它们连接成一个字符串）：

import json
from collections import Counter, defaultdict

data = """{"Date": "Fri, 19 Apr 2019 00:54:46 GMT", "Vary": "Host,Accept-Encoding", "Key-Word": "00a", "Cache-Control": "private", "Key-Word": "xn"}

"""

def duplicate_keys(pairs):
    out = {}
    dups = defaultdict(list)
    key_count = Counter(key for key, value in pairs)

    for key, value in pairs:
        if key_count[key] == 1:
            out[key] = value
        else:
            dups[key].append(value)

    # Concatenate the lists of values in a string, enclosed in {} and separated by ';'
    # rather than in a list:       
    dups = {key: ';'.join('{' + v + '}' for v in values) for key, values in dups.items()}

    out.update(dups)
    return out

decoded = json.loads(data, object_pairs_hook=duplicate_keys)
print(decoded)

# {'Date': 'Fri, 19 Apr 2019 00:54:46 GMT', 
#  'Vary': 'Host,Accept-Encoding', 
#  'Cache-Control': 'private', 
#  'Key-Word': '{00a};{xn}'}

Answer 5

当您定义字典 myDict = {"Date": "Fri, 19 Apr 2019 00:54:46 GMT", "Vary": "Host,Accept-Encoding", "Key-Word": "00a", "Cache-Control": "private", "Key-Word": "xn"} 时，您需要有不同的值键：00a 和 xn。

你能 use/convert 到字符串 some_str = '{"Date": "Fri, 19 Apr 2019 00:54:46 GMT", "Vary": "Host,Accept-Encoding", "Key-Word": "00a", "Cache-Control": "private", "Key-Word": "xn"}'.

如何解析具有相同键的多个字典值

How to parse multiple dictionary values that have the same key

python

parsing

json

dictionary

abstract-syntax-tree