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Python | Check if k occurs atleast n times in a list

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  • Last Updated : 26 Jan, 2023
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There are many common problems that a developer or common programmer comes across in day-day coding. One such problem is finding if an element occurs more than a certain number of times in the list. Let’s discuss certain ways to deal with this problem. 

Method #1 : Using sum() + list comprehension The list comprehension can be clubbed with the sum function for achieving this particular task. The sum function does the summation part and the logical case of returning true is dealt in list comprehension part. 

Python3




# Python3 code to demonstrate
# check for minimum N occurrences of K
# using sum() + list comprehension
 
# initializing list
test_list = [1, 3, 5, 5, 4, 5]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# initializing k
k = 5
 
# initializing N
N = 3
 
# using sum() + list comprehension
# checking occurrences of K atleast N times
res = 0
res = sum([1 for i in test_list if i == k]) >= N
if res == 1 :
    res = True
else :
    res = False
     
# printing result
print ("Does %d occur atleast %d times ? :" %(k, N) + str(res))

  Method #2 : Using next() + islice() These both functions can be used together to perform this particular task in more efficient manner than above. The islice function would handle the summation part and next function helps with iteration logic. 

Python3




# Python3 code to demonstrate
# check for minimum N occurrences of K
# using next() + islice()
from itertools import islice
 
# initializing list
test_list = [1, 3, 5, 5, 4, 5]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# initializing k
k = 5
 
# initializing N
N = 3
 
# using next() + islice()
# checking occurrences of K atleast N times
func = (True for i in test_list if i == k)
res = next(islice(func, N-1, None), False)
     
# printing result
print ("Does %d occur atleast %d times ? :" %(k, N) + str(res))

Output:

The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : True

Method #3: Using count()

Python3




# Python3 code to demonstrate
# check for minimum N occurrences of K
 
# initializing list
test_list = [1, 3, 5, 5, 4, 5]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# initializing k
k = 5
 
# initializing N
N = 3
# checking occurrences of K atleast N times
res = False
if(test_list.count(k)>=N):
    res=True
# printing result
print ("Does %d occur atleast %d times ? : " %(k, N) + str(res))

Output

The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : True

Method #4 : Using for loop

Python3




# Python3 code to demonstrate
# check for minimum N occurrences of K
 
# initializing list
test_list = [1, 3, 5, 5, 4, 5]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# initializing k
k = 5
 
# initializing N
N = 3
# checking occurrences of K atleast N times
res = False
c=0
for i in test_list:
    if(i==k):
        c+=1
if(c>=N):
    res=True
# printing result
print ("Does %d occur atleast %d times ? : " %(k, N) + str(res))

Output

The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : True

Method #5 : Using operator.countOf() method

Python3




# Python3 code to demonstrate
# check for minimum N occurrences of K
import operator as op
# initializing list
test_list = [1, 3, 5, 5, 4, 5]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing k
k = 5
 
# initializing N
N = 3
# checking occurrences of K atleast N times
res = False
if(op.countOf(test_list, k) >= N):
    res = True
# printing result
print("Does %d occur atleast %d times ? : " % (k, N) + str(res))

Output

The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : True

Time Complexity: O(n)

Auxiliary Space: O(1)


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