Most frequent item in a list - Trick

Example 1:

l1 = [1, 2, 3, 1, 2, 1, 3, 2, 1, 1, 4,5]

#newList = max(l1) 

#if we use the max() method then return max value 4 only but in case of a number of occurrence items in the list then we can use the key keyword and count property

newList = max(l1,key = l1.count)  

if you want to optimize the above code then we can use the set

newList = max(set(l1),key = l1.count) 

print(newList) # 1

Example 2:

from functools import reduce 

l1 = [1, 2, 3, 1, 2, 1,1,1,1,1,1,1,1 ,3, 2, 1, 1, 4,5,5,5,5,5,5]

ar = {}

for i in l1:

    if i in ar:

        ar[i] += 1

    else:

        ar[i] = 1

print(ar)

# {1: 12, 2: 3, 3: 2, 4: 1, 5: 6}

# Use reduce to find the greatest value

gt_val = reduce(lambda x, y: x if x > y else y, ar.values())

#  list comprehension 

x = [key for key, val in ar.items() if val == gt_val]

print(x)

the above example is lengthy but here we can learn loop, list comprehension, reduce, lambda and many more 

Example 3:

l1 = [1, 2, 3, 1, 2, 1, 3, 2, 1, 1, 4]

#  let's consider 

most_common_item = 0

max_count = 0

# Iterate through the list

for item in l1:

    # Count the occurrences of the current item in the list

    count = l1.count(item)

    #print(count,'',item)

    if count > max_count:

        most_common_item = item # here we picked up the value

        max_count = count 

print("Most frequent item:", most_common_item)