如何在 Python 中的字典數組中查找公共鍵

Question

給定一個字典數組： [{'key1': 'valueA', 'key2': 'valueB'}, {'key1': 'valueC', 'key3': 'valueD'}, {}, ... ]

您將如何找到字典中最常用的鍵並對它們進行排名？

在此示例中，“key1”的鍵出現兩次，因此將排在第 1 位。然后，如果“key2”以下一個常見頻率出現，則將排在第 2 位，依此類推。

Answer 1

用這個：

keys = dict()
# d is your dictionary
for i in d:
    for k, v in i.items():
        if k in keys:
            keys[k] += 1
        else:
            keys[k] = 1
max = 0
max_key = ''
for k, v in keys.items():
    if v > max:
        max = v
        max_key = k
print(max, max_key)

Answer 2

您可以使用pandas來解析 dict然后描述它以獲取 stats 。

import pandas as pd

lod = [{'points': 50, 'time': '5:00', 'year': 2010}, 
{'points': 25, 'time': '6:00', 'month': "february"}, 
{'points':90, 'time': '9:00', 'month': 'january'}, 
{'points_h1':20, 'month': 'june'}]

df = pd.DataFrame(lod)

print(df.describe(include=['float', 'object']))

'''
           points  time    year    month  points_h1
count    3.000000     3     1.0        3        1.0
unique        NaN     3     NaN        3        NaN
top           NaN  9:00     NaN  january        NaN
freq          NaN     1     NaN        1        NaN
mean    55.000000   NaN  2010.0      NaN       20.0
std     32.787193   NaN     NaN      NaN        NaN
min     25.000000   NaN  2010.0      NaN       20.0
25%     37.500000   NaN  2010.0      NaN       20.0
50%     50.000000   NaN  2010.0      NaN       20.0
75%     70.000000   NaN  2010.0      NaN       20.0
max     90.000000   NaN  2010.0      NaN       20.0
'''

但是請注意，這不適用於嵌套字典！ 你需要安裝pandas和pip 。

Answer 3

那是你的答案嗎

from collections import Counter
d=[{'key1': 'valueA', 'key2': 'valueB'}, {'key1': 'valueC', 'key3': 'valueD'}, {'key1': 'valueC'}]
keys=[]
for i in d:
    for j in i.keys():
        keys.append(j)
        
Counter(keys)

output

Counter({'key1': 3, 'key2': 1, 'key3': 1})

Answer 4

您可以使用Counter來計算：

import collections

def count(array):
    counter = collections.Counter()
    for dict in array:
        counter.update(dict.keys())
    return counter

print(count([{'a': 1}, {'a': 2}])) # Counter({'a': 2})
print(count([{'f': 2}, {'f': '2'}, {'b': 10}])) # Counter({'f': 2, 'b': 1})

Answer 5

使用pandas的另一個不錯的解決方案是轉換數據幀中的字典，合並它們，然后使用value_counts 。 這是一個小例子

import pandas as pd

d1 = {'key1': '1', 'key2': '2'}
d2 = {'key1': '1', 'key3': '3'}

df1 = pd.DataFrame.from_dict(d1.items())
df2 = pd.DataFrame.from_dict(d2.items())

frames = [df1, df2]
result = pd.concat(frames)
print(result[0].value_counts())

哪個打印

key1    2
key2    1
key3    1
Name: 0, dtype: int64

Answer 6

# Extract the keys in a flat list
keys = [k for k, v in i.items for i in data]

# Sort and group
from itertools import sorted, groupby
s = sorted(keys)
g = groupby(s)

Answer 7

使用collections.Counter的另一種方法：

from collections import Counter
d=[{'key1': 'valueA', 'key2': 'valueB'}, {'key1': 'valueC', 'key3': 'valueD'}, {} ]
c = Counter(key for t in d for key in t.keys())

輸出：

Counter({'key1': 2, 'key2': 1, 'key3': 1}) 

Answer 8

有很多方法可以做到這一點，我認為最干凈的方法之一是使用計數器 class，但是我會這樣做：

my_dicts = [{"hello": 2},{"hello1": 3, "hello": 5},{"hello2": 5, "hello": 7, "hello1": 4}]


merged = {}
for d in my_dicts:
  for k in d.keys():
    if k in merged:
      merged[k] += 1
    else:
      merged[k] = 1
results = {k: v for k, v in sorted(merged.items(), key=lambda item: item[1], reverse = True)}

變量results最終包含按降序排列的已排序項目的字典，如下所示：

Output：

{'hello': 3, 'hello1': 2, 'hello2': 1}

Answer 9

嘗試這個;

代碼語法

dicts =[{'key1': 'valueA', 'key2': 'valueB'}, {'key1': 'valueC', 'key3': 'valueD'}]

lis = [list(dicts[i].keys()) for i in range(len(dicts))]
x = sorted([j for i in lis for j in i])
print(x)
for i, each in enumerate(x):
    if x[i-1] == x[i]:
        continue
    else:
        indey = x.count(x[x.index(each)])
        print(f"{each} = {indey}")

Output

['key1', 'key1', 'key2', 'key3']
key1 = 2
key2 = 1
key3 = 1

[Program finished]