Count python list item occurrences

Last Updated on July 17, 2021 by Editorial Team

Author(s): Vivek Chaudhary

Programming

From a list of integers, count the occurrence of each element and add items and count as a sub-list.

Input list: inp = [7, 9, 7, 4, 3, 5, 3, 6, 9, 3]
Output list: out= [[7, 2], [9, 2], [4, 1], [3, 3], [5, 1], [6, 1]]

#Note: where 7 is list item and 2 is the count of occurrence

Using nested loops

def count_items(inp):

    for i in range(0, len(inp)):
        a = 0
        row =[]
        if i not in l:
            for j in range(0, len(inp)):
                if inp[i]== inp[j]:
                a = a + 1
            row.append(inp[i])
            row.append(a)
            l.append(row)
     for j in l:
        if j not in out:
           out.append(j)

return out

Call the function and check the output:

#call func() 
inp = [7, 9, 7, 4, 3, 5, 3, 6, 9, 3]
l = []
out = []
print(count_items(inp))

Output: [[7, 2], [9, 2], [4, 1], [3, 3], [5, 1], [6, 1]]

Using count() method

def count_items(inp):

    for i in inp:
       row =[]
       ct = 0
       ct = inp.count(i)
       row.append(i)
       row.append(ct)
       l.append(row)

   for j in l:
       if j not in out:
       out.append(j)
   return out

Call the function and check the output:

#call func()
inp = [7, 9, 7, 4, 3, 5, 3, 6, 9, 3]
l = []
out = []
print(count_items(inp))

Output: [[7, 2], [9, 2], [4, 1], [3, 3], [5, 1], [6, 1]]

Using counter() method

from collections import Counter
def count_items(inp):
     coll = Counter(inp)
     out = []
     for key,val in coll.items():
          out.append([key,val])
 return out

Call the function and check the output:

#call func() 
inp = [7, 9, 7, 4, 3, 5, 3, 6, 9, 3]
print(count_items(inp))

Output: [[7, 2], [9, 2], [4, 1], [3, 3], [5, 1], [6, 1]]

To summarize, we covered several ways to count occurrences of list items as below:

• count() method
• nested for loops
• collections library counter() method

That’s all with count occurrences of list items.

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