Sometimes, while working with Python, we can have a problem in which we need to get occurrence of 1 element in another. But as a modification of this, we can have a problem in which we need to count the occurrence of all elements of 1 list in another. Lets discuss certain ways in which this task can be performed.
Method #1 : Using nested loops This is one of the way in which this task can be performed. This is brute force way in which this task can be performed. In this, we iterate one list and then target list, if element match, we increase the counter.
Python3
# Python3 code to demonstrate # Sum of each List element occurrence in another# using nested loops# Initializing liststest_list1 = [1, 3, 4, 5, 1, 4, 4, 6, 7]test_list2 = [4, 6, 1]# printing original listsprint("The original list 1 is : " + str(test_list1))print("The original list 2 is : " + str(test_list2))# Sum of each List element occurrence in another# using nested loopsres = 0for ele in test_list2: for ele1 in test_list1: if ele1 == ele: res = res + 1# printing result print ("The occurrence count : " + str(res)) |
The original list 1 is : [1, 3, 4, 5, 1, 4, 4, 6, 7] The original list 2 is : [4, 6, 1] The occurrence count : 6
Time Complexity : O(n*m), where n is length of test_list1 and m is length of test_list2
Auxiliary Space : O(n+m), where n is length of test_list1 and m is length of test_list2
Method #2 : Using sum() + count() The combination of above methods can be used to perform this particular task. This is one liner alternative to the above method. In this counting is done using count() and accumulation using sum().
Python3
# Python3 code to demonstrate # Sum of each List element occurrence in another# using sum() + count()# Initializing liststest_list1 = [1, 3, 4, 5, 1, 4, 4, 6, 7]test_list2 = [4, 6, 1]# printing original listsprint("The original list 1 is : " + str(test_list1))print("The original list 2 is : " + str(test_list2))# Sum of each List element occurrence in another# using sum() + count()res = sum(test_list1.count(idx) for idx in test_list2)# printing result print ("The occurrence count : " + str(res)) |
The original list 1 is : [1, 3, 4, 5, 1, 4, 4, 6, 7] The original list 2 is : [4, 6, 1] The occurrence count : 6
Method 3: using operator.countOf() method
Python3
# Python3 code to demonstrate# Sum of each List element occurrence in anotherimport operator as op# Initializing liststest_list1 = [1, 3, 4, 5, 1, 4, 4, 6, 7]test_list2 = [4, 6, 1]# printing original listsprint("The original list 1 is : " + str(test_list1))print("The original list 2 is : " + str(test_list2))# Sum of each List element occurrence in another# using sum() + count()res = sum(op.countOf(test_list1, idx) for idx in test_list2)# printing resultprint("The occurrence count : " + str(res)) |
The original list 1 is : [1, 3, 4, 5, 1, 4, 4, 6, 7] The original list 2 is : [4, 6, 1] The occurrence count : 6
Time Complexity: O(N)
Auxiliary Space : O(N)
Approach 4: Using recursive method.
Python3
# Python3 code to demonstrate# Sum of each List element occurrence in another#using recursive approachdef occurrence_count(start,lst1,lst2,count): if start==len(lst1): #base condition return count if lst1[start] in lst2: #checking if lst1[start] present in lst2 or not count+=1 return occurrence_count(start+1,lst1,lst2,count) #calling recursive call# initializing liststest_list1 = [1, 3, 4, 5, 1, 4, 4, 6, 7]test_list2 = [4, 6, 1]# printing original listsprint("The original list 1 is : " + str(test_list1))print("The original list 2 is : " + str(test_list2))res=occurrence_count(0,test_list1,test_list2,0)# printing resultprint("The occurrence count : " + str(res))#this code contributed by tvsk |
The original list 1 is : [1, 3, 4, 5, 1, 4, 4, 6, 7] The original list 2 is : [4, 6, 1] The occurrence count : 6
Time Complexity: O(N)
Auxiliary Space : O(N)
Approach 5:Using Collections library
Python3
from collections import Counter# Initializing liststest_list1 = [1, 3, 4, 5, 1, 4, 4, 6, 7]test_list2 = [4, 6, 1]# Count occurrences of elements in first listcounts = Counter(test_list1)# Use a list comprehension to add up occurrences of elements in second listres = sum(counts[ele] for ele in test_list2)# printing resultprint ("The occurrence count : " + str(res))#This code is contributed Vinay Pinjala. |
The occurrence count : 6
Time Complexity: O(N)
Auxiliary Space : O(N)
Approach 6: Using defaultdict and sum
Python3
from collections import defaultdict#Function to get the sum of each element in the first listdef get_occurrence_sum(list1, list2): dict_count = defaultdict(int) for i in list1: dict_count[i] += 1 return sum(dict_count[i] for i in list2)#Initializing liststest_list1 = [1, 3, 4, 5, 1, 4, 4, 6, 7]test_list2 = [4, 6, 1]#printing original listsprint("The original list 1 is : " + str(test_list1))print("The original list 2 is : " + str(test_list2))#Get the sum of each element in the first listres = get_occurrence_sum(test_list1, test_list2)#printing resultprint ("The occurrence count : " + str(res))#This code is contributed by Edula Vinay Kumar Reddy |
The original list 1 is : [1, 3, 4, 5, 1, 4, 4, 6, 7] The original list 2 is : [4, 6, 1] The occurrence count : 6
Time Complexity: O(N)
Auxiliary Space : O(N)
Explanation:
We create a defaultdict of integers and store the count of each element in the first list.
Then, we use sum function to get the sum of values corresponding to the keys in the second list.
This approach is efficient as the defaultdict stores the counts in constant time and sum function runs in linear time.
Approach 7: Using List Comprehension:
Python3
# Initializing liststest_list1 = [1, 3, 4, 5, 1, 4, 4, 6, 7]test_list2 = [4, 6, 1]# printing original listsprint("The original list 1 is : " + str(test_list1))print("The original list 2 is : " + str(test_list2))# Sum of each List element occurrence in another# using List Comprehensionres = len([ele for ele in test_list1 if ele in test_list2])# printing result print ("The occurrence count : " + str(res))#This code is contributed by Jyothi pinjala |
The original list 1 is : [1, 3, 4, 5, 1, 4, 4, 6, 7] The original list 2 is : [4, 6, 1] The occurrence count : 6
Time Complexity: O(N^2)
Auxiliary Space : O(1)
