Python - Find occurrences for each value of a particular key
Last Updated : 30 Apr, 2023
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)) OutputThe 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:
- Create an empty dictionary and set the default value to 0.
- Iterate over the list of dictionaries.
- For each dictionary, get the value of key K.
- Use this value to update the count in the dictionary created in step 1.
- 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. OutputThe 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.
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