如何检查任何键值对是否在字典列表中多次出现？

Question

例如：

people_list = [{"name":"joe", "age":20}, {"name":"tom", "age":35}, {"name":"joe", "age":46}]

如何检查任何键值对是否在此字典列表中出现多次？

我知道柜台 function。

from collections import Counter

for i in range(len(people_list):
    Counter(people_list[i]["name"])
for key,value in Counter:
    if Counter(key:value) > 1:
        ... 

只是不确定如何让它在这个字典列表中查找和计算特定的键值对，并检查它是否出现不止一次。

Answer 1

您想要获取每个字典中的所有键/值对，并将它们添加到一个Counter实例中，该实例的键为 (key, value)，其值最终将成为该对存在的次数：

from itertools import chain
from collections import Counter

l = [
    {"name":"joe", "age":20},
    {"name":"tom", "age":35},
    {"name":"joe", "age":46}
]

c = Counter(chain.from_iterable(d.items() for d in l))
# Now build the new list of tuples (key-name, key-value, count-of-occurrences)
# but include only those key/value pairs that occur more than once:
l = [(k[0], k[1], v) for k, v in c.items() if v > 1]
print(l)

印刷：

[('name', 'joe', 2)]

如果要创建一个新的字典列表，其中“name”键值是原始字典列表中所有不同的名称值，并且其“age”键的值是该名称的年龄总和，则：

from collections import defaultdict

l = [
    {"name":"joe", "age":20},
    {"name":"tom", "age":35},
    {"name":"joe", "age":46}
]

d = defaultdict(int)
for the_dict in l:
    d[the_dict["name"]] += the_dict["age"]
new_l = [{"name": k, "age": v} for k, v in d.items()]
print(new_l)

印刷：

[{'name': 'joe', 'age': 66}, {'name': 'tom', 'age': 35}]

Answer 2

您可以使用collections.Counter并像下面这样update ：

from collections import Counter

people_list = [{"name":"joe", "age":20}, {"name":"tom", "age":35}, {"name":"joe", "age":46}]
cnt = Counter()
for dct in people_list:
    cnt.update(dct.items())
print(cnt)

# Get the items > 1
for k,v in cnt.items():
    if v>1:
        print(k)

Output：

Counter({('name', 'joe'): 2, ('age', 20): 1, ('name', 'tom'): 1, ('age', 35): 1, ('age', 46): 1})

('name', 'joe')

Answer 3

如果你想得到 output “joe”，因为它针对键“name”出现了不止一次，那么：

from collections import Counter

people_list = [{"name":"joe", "age":20}, {"name":"tom", "age":35}, {"name":"joe", "age":46}]
counter = Counter([person["name"] for person in people_list])
print([name for name, count in counter.items() if count > 1]) # ['joe']

Answer 4

您可以使用Counter来统计每个值出现的次数，并检查是否有任何键值对出现超过一次。

from collections import Counter

people_list = [{"name":"joe", "age":20}, {"name":"tom", "age":35}, {"name":"joe", "age":46}]

# Count the number of occurrences of each value
counts = Counter(d["name"] for d in people_list)

# Check if any key-value pair appears more than once
for key, value in counts.items():
    if value > 1:
        print(f"{key} appears more than once")

给出：

joe appears more than once

Answer 5

我会将每一对转换为字符串表示形式，然后将其放在列表中。 之后，您可以使用该问题中已经存在的Counter逻辑。 虽然这不是处理数据的最有效方法，但它仍然是一个简单的解决方案。

from functools import reduce

from collections import Counter

people_list = [{"name":"joe", "age":20}, {"name":"tom", "age":35}, {"name":"joe", "age":46}]

str_lists = [[f'{k}_{v}' for k,v in value.items()] for value in people_list] 
# [['name_joe', 'age_20'], ['name_tom', 'age_35'], ['name_joe', 'age_46']]


normalized_list = list(reduce(lambda a,b: a+b,str_lists))
# ['name_joe', 'age_20', 'name_tom', 'age_35', 'name_joe', 'age_46']


dict_ = dict(Counter(normalized_list))
# {'name_joe': 2, 'age_20': 1, 'name_tom': 1, 'age_35': 1, 'age_46': 1}

print([k for k,v in  dict_.items() if v>1])
# ['name_joe']