如何检查任何键值对是否在字典列表中多次出现？

Question

For ex:

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

How do I check if any key-value pair appears more than once in this list of dictionaries?

I am aware of the counter 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:
        ... 

Just not sure how to make it look and count for a specific key value pair in this list of dictionaries and check if it appears more than once.

Answer 1

You want to take all the key/value pairs in each dictionary and add them to a Counter instance whose key is (key, value) and whose value will end up being the number of times this pair exists:

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)

Prints:

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

If you want to create a new list of dictionaries where the 'name' key values are all the distinct name values in the original list of dictionaries and whose value for the 'age' key is the sum of the ages for that name, then:

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)

Prints:

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

Answer 2

You can use collections.Counter and update like the below:

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

If you want to get as output "joe", because it appears more than once for the key "name", then:

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

You can use a Counter to count the number of occurrences of each value and check if any key-value pair appears more than once.

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")

Gives:

joe appears more than once

Answer 5

I would convert each pair into a string representation and then put it on a list. After that you can use the Counter logic that is already present in this question. Granted that this is not the most effective way to treat the data, but it is still a simple solution.

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']