We have a list whose elements are numeric. Many elements are present multiple times. We want to create sub list so the frequency of each element along with the elements itself.
With for and append
In this approach we will compare each element in the list with every other elements after it. If there is a match then count will be incremented and both the element and the count will be made into a subsist. List will be made which should contain subsists showing every element and its frequency.
Example
def occurrences(list_in):
for i in range(0, len(listA)):
a = 0
row = []
if i not in listB:
for j in range(0, len(listA)):
# matching items from both positions
if listA[i] == listA[j]:
a = a + 1
row.append(listA[i])
row.append(a)
listB.append(row)
# Eliminate repetitive list items
for j in listB:
if j not in list_uniq:
list_uniq.append(j)
return list_uniq
# Caller code
listA = [13,65,78,13,12,13,65]
listB = []
list_uniq = []
print("Number of occurrences of each element in the list:\n")
print(occurrences(listA))Output
Running the above code gives us the following result −
Number of occurrences of each element in the list: [[13, 3], [65, 2], [78, 1], [12, 1]]
With Counter
We use counter method from the collections module. It will give the count of every element in the list. Then we declare a new empty list and add the key value pair for each item in form of element and its count into the new list.
Example
from collections import Counter
def occurrences(list_in):
c = Counter(listA)
new_list = []
for k, v in c.items():
new_list.append([k, v])
return new_list
listA = [13,65,78,13,12,13,65]
print("Number of occurrences of each element in the list:\n")
print(occurrences(listA))Output
Running the above code gives us the following result −
Number of occurrences of each element in the list: [[13, 3], [65, 2], [78, 1], [12, 1]]