Python – Elements frequency count in multiple lists
Last Updated :
27 Feb, 2023
Sometimes while working with Python lists we can have a problem in which we need to extract the frequency of elements in list. But this can be added work if we have more than 1 list we work on. Let’s discuss certain ways in which this task can be performed.
Method #1: Using dictionary comprehension + set() + count() This is one of the way in which this task can be performed. In this, we perform the task of counting using count() and set() and the extension of logic is done using dictionary comprehension.
Python3
test_list1 = [1, 3, 2, 4, 5, 1, 2]
test_list2 = [4, 1, 4, 3, 4, 2, 4]
test_list3 = [1, 4, 5, 3, 4, 5, 4]
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
print("The original list 3 is : " + str(test_list3))
res = {idx : [test_list1.count(idx), test_list2.count(idx), test_list3.count(idx)]
for idx in set(test_list1 + test_list2 + test_list3)}
print ("The frequency of each element is : " + str(res))
|
Output
The original list 1 is : [1, 3, 2, 4, 5, 1, 2]
The original list 2 is : [4, 1, 4, 3, 4, 2, 4]
The original list 3 is : [1, 4, 5, 3, 4, 5, 4]
The frequency of each element is : {1: [2, 1, 1], 2: [2, 1, 0], 3: [1, 1, 1], 4: [1, 4, 3], 5: [1, 0, 2]}
Time Complexity: O(n^2) and the space complexity is O(n) where n is the total number of elements in all the lists.
Auxiliary Space: O(n), where n is the number of elements in the input lists. The res dictionary and the set created for storing the unique elements of the input lists use O(n) space.
Method #2: Using Counter() + map() + dictionary comprehension The combination of above functionalities can be used to solve this problem. In this, the task of finding the frequency is done using Counter() and map().
Python3
from collections import Counter
test_list1 = [1, 3, 2, 4, 5, 1, 2]
test_list2 = [4, 1, 4, 3, 4, 2, 4]
test_list3 = [1, 4, 5, 3, 4, 5, 4]
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
print("The original list 3 is : " + str(test_list3))
freq = list(map(Counter, (test_list1, test_list2, test_list3)))
res = {ele: [cnt[ele] for cnt in freq] for ele in {ele for cnt in freq for ele in cnt}}
print ("The frequency of each element is : " + str(res))
|
Output
The original list 1 is : [1, 3, 2, 4, 5, 1, 2]
The original list 2 is : [4, 1, 4, 3, 4, 2, 4]
The original list 3 is : [1, 4, 5, 3, 4, 5, 4]
The frequency of each element is : {1: [2, 1, 1], 2: [2, 1, 0], 3: [1, 1, 1], 4: [1, 4, 3], 5: [1, 0, 2]}
Time complexity: O(N*M), where N is the number of elements in each list and M is the number of lists.
Space complexity: O(N), where N is the number of unique elements across all the lists.
Method #3: Using extend(),list(),set() and count() methods
Python3
test_list1 = [1, 3, 2, 4, 5, 1, 2]
test_list2 = [4, 1, 4, 3, 4, 2, 4]
test_list3 = [1, 4, 5, 3, 4, 5, 4]
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
print("The original list 3 is : " + str(test_list3))
x=[]
res=dict()
x.extend(test_list1)
x.extend(test_list2)
x.extend(test_list3)
x=list(set(x))
for i in x:
a=[]
a.append(test_list1.count(i))
a.append(test_list2.count(i))
a.append(test_list3.count(i))
res[i]=a
print ("The frequency of each element is : " + str(res))
|
Output
The original list 1 is : [1, 3, 2, 4, 5, 1, 2]
The original list 2 is : [4, 1, 4, 3, 4, 2, 4]
The original list 3 is : [1, 4, 5, 3, 4, 5, 4]
The frequency of each element is : {1: [2, 1, 1], 2: [2, 1, 0], 3: [1, 1, 1], 4: [1, 4, 3], 5: [1, 0, 2]}
Time Complexity : O(N)
Auxiliary Space : O(N)
Method #4: Using dictionary comprehension + set() + operator.countOf() This is one of the way in which this task can be performed. In this, we perform the task of counting using operator.countOf() and set() and the extension of logic is done using dictionary comprehension.
Python3
import operator as op
test_list1 = [1, 3, 2, 4, 5, 1, 2]
test_list2 = [4, 1, 4, 3, 4, 2, 4]
test_list3 = [1, 4, 5, 3, 4, 5, 4]
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
print("The original list 3 is : " + str(test_list3))
res = {idx : [op.countOf(test_list1,idx), op.countOf(test_list2,idx), op.countOf(test_list3,idx)]
for idx in set(test_list1 + test_list2 + test_list3)}
print ("The frequency of each element is : " + str(res))
|
Output
The original list 1 is : [1, 3, 2, 4, 5, 1, 2]
The original list 2 is : [4, 1, 4, 3, 4, 2, 4]
The original list 3 is : [1, 4, 5, 3, 4, 5, 4]
The frequency of each element is : {1: [2, 1, 1], 2: [2, 1, 0], 3: [1, 1, 1], 4: [1, 4, 3], 5: [1, 0, 2]}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 5: Using defaultdict
Use the defaultdict from the collections module to count the frequency of each element in the lists.
Python3
from collections import defaultdict
test_list1 = [1, 3, 2, 4, 5, 1, 2]
test_list2 = [4, 1, 4, 3, 4, 2, 4]
test_list3 = [1, 4, 5, 3, 4, 5, 4]
freq = defaultdict(lambda: [0, 0, 0])
for i in range(len(test_list1)):
freq[test_list1[i]][0] += 1
freq[test_list2[i]][1] += 1
freq[test_list3[i]][2] += 1
print(dict(freq))
|
Output
{1: [2, 1, 1], 4: [1, 4, 3], 3: [1, 1, 1], 2: [2, 1, 0], 5: [1, 0, 2]}
Time Complexity: O(n)
Auxiliary Space: O(n)
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