Python – Find occurrences for each value of a particular key

Given a List of dictionaries, for a particular key, find the number of occurrences for each value of that key.

Input : test_list = [{‘gfg’ : 3, ‘best’ : 4}, {‘gfg’ : 3, ‘best’ : 5}, 
{‘gfg’ : 4, ‘best’ : 4}, {‘gfg’ : 7, ‘best’ : 4} ], K = ‘gfg’ 
Output : [{3: 2}, {4: 1}, {7: 1}] 
Explanation : gfg has 2 occurrences of 3 as values.

Input : test_list = [{‘gfg’ : 3, ‘best’ : 4}, {‘gfg’ : 3, ‘best’ : 5}, 
{‘gfg’ : 4, ‘best’ : 4}, {‘gfg’ : 7, ‘best’ : 4} ], K = ‘best’ 
Output : [{4: 3}, {5: 1}] 
Explanation : best has 3 occurrences of 4 as values.

Method #1 : Using groupby() + dictionary comprehension

In this, we perform grouping of key’s values using groupby() and values frequency is assembled and extracted using dictionary comprehension and len().

Python3

# Python3 code to demonstrate working of
# Values Frequency grouping of K in dictionaries
# Using groupby() + dictionary comprehension
from itertools import groupby

# initializing list
test_list = [{'gfg' : 3, 'best' : 4}, {'gfg' : 3, 'best' : 5},
             {'gfg' : 4, 'best' : 4}, {'gfg' : 7, 'best' : 4} ]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 'gfg'

# groupby() used to group values and len() to compute Frequency
res = [{key: len(list(val))} for key, val in groupby(test_list, lambda sub: sub[K])]

# printing result
print("The Values Frequency : " + str(res))

Output:

The original list is : [{‘gfg’: 3, ‘best’: 4}, {‘gfg’: 3, ‘best’: 5}, {‘gfg’: 4, ‘best’: 4}, {‘gfg’: 7, ‘best’: 4}] The Values Frequency : [{3: 2}, {4: 1}, {7: 1}]

Time Complexity: O(n) where n is the number of elements in the list “test_list”.  groupby() + dictionary comprehension performs n number of operations.
Auxiliary Space: O(n), extra space is required where n is the number of elements in the list 

Method #2 : Using Counter()

In this, the task of performing frequency check is done using Counter(). Returns result in single dictionary.

Python3

# Python3 code to demonstrate working of
# Values Frequency grouping of K in dictionaries
# Using Counter()
from collections import Counter

# initializing list
test_list = [{'gfg' : 3, 'best' : 4}, {'gfg' : 3, 'best' : 5},
             {'gfg' : 4, 'best' : 4}, {'gfg' : 7, 'best' : 4} ]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 'gfg'

# groupby() used to group values and len() to compute Frequency
res = dict(Counter(sub[K] for sub in test_list))

# printing result
print("The Values Frequency : " + str(res))

Output:

The original list is : [{‘gfg’: 3, ‘best’: 4}, {‘gfg’: 3, ‘best’: 5}, {‘gfg’: 4, ‘best’: 4}, {‘gfg’: 7, ‘best’: 4}] The Values Frequency : [{3: 2}, {4: 1}, {7: 1}]

Method #3: Using keys(),list(),set() and count() methods

Python3

# Python3 code to demonstrate working of
# Values Frequency grouping of K in dictionaries

# initializing list
test_list = [{'gfg' : 3, 'best' : 4}, {'gfg' : 3, 'best' : 5},
            {'gfg' : 4, 'best' : 4}, {'gfg' : 7, 'best' : 4} ]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 'gfg'
x=[]
for i in test_list:
    if K in i.keys():
        x.append(i[K])
p=list(set(x))
nl=[]
for i in p:
    d={}
    d[i]=x.count(i)
    nl.append(d)

# printing result
print("The Values Frequency : " + str(nl))

Output

The original list is : [{'gfg': 3, 'best': 4}, {'gfg': 3, 'best': 5}, {'gfg': 4, 'best': 4}, {'gfg': 7, 'best': 4}]
The Values Frequency : [{3: 2}, {4: 1}, {7: 1}]

Method#4:Using defaultdict() and loop

Algorithm:

1. Create an empty dictionary and set the default value to 0.
2. Iterate over the list of dictionaries.
3. For each dictionary, get the value of key K.
4. Use this value to update the count in the dictionary created in step 1.
5. Return the resulting dictionary.

Python3

from collections import defaultdict

# initializing list
test_list = [{'gfg' : 3, 'best' : 4}, {'gfg' : 3, 'best' : 5},
             {'gfg' : 4, 'best' : 4}, {'gfg' : 7, 'best' : 4} ]

# initializing K
K = 'gfg'

# using loop and defaultdict to group values and compute frequency
freq_dict = defaultdict(int)
for item in test_list:
    freq_dict[item[K]] += 1

# formatting the result as a list of dictionaries
res = [{key: val} for key, val in freq_dict.items()]

# printing result
print("The Values Frequency : " + str(res))

#This code is contributed by Vinay Pinjala.

Output

The Values Frequency : [{3: 2}, {4: 1}, {7: 1}]

Time Complexity: O(n), where n is the number of dictionaries in the list. This is because we need to iterate over each dictionary once.

Space Complexity: O(k), where k is the number of unique values for key K. This is because we are storing the count for each unique value in the resulting dictionary.

Method #5: Using pandas
Step-by-step approach:

Import pandas library.
Convert the list of dictionaries to a pandas DataFrame using the pd.DataFrame() function.
Use the value_counts() method on the ‘gfg’ column of the DataFrame to get the frequency of each unique value.
Convert the resulting pandas Series object to a dictionary using the to_dict() method.
Format the dictionary as a list of dictionaries using a list comprehension.
Print the resulting list of dictionaries.

Python3

import pandas as pd

# initializing list
test_list = [{'gfg' : 3, 'best' : 4}, {'gfg' : 3, 'best' : 5},
             {'gfg' : 4, 'best' : 4}, {'gfg' : 7, 'best' : 4} ]

# initializing K
K = 'gfg'

# convert list of dictionaries to pandas DataFrame
df = pd.DataFrame(test_list)

# get frequency of each unique value in 'gfg' column
freq_dict = df[K].value_counts().to_dict()

# format the result as a list of dictionaries
res = [{key: val} for key, val in freq_dict.items()]

# print the result
print("The Values Frequency : " + str(res))

OUTPUT : The Values Frequency : [{3: 2}, {4: 1}, {7: 1}]

Time complexity: O(n log n), where n is the number of items in the input list. 

Auxiliary space: O(n), where n is the number of items in the input list.