Python: Count Repeated Elements in the List

Python: count repeated elements in the list

You can do that using count:

my_dict = {i:MyList.count(i) for i in MyList}

>>> print my_dict     #or print(my_dict) in python-3.x
{'a': 3, 'c': 3, 'b': 1}

Or using collections.Counter:

from collections import Counter

a = dict(Counter(MyList))

>>> print a           #or print(a) in python-3.x
{'a': 3, 'c': 3, 'b': 1}

How do I count the occurrences of a list item?

If you only want a single item's count, use the count method:

>>> [1, 2, 3, 4, 1, 4, 1].count(1)
3

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Important: this is very slow if you are counting multiple different items

Each count call goes over the entire list of n elements. Calling count in a loop n times means n * n total checks, which can be catastrophic for performance.

If you want to count multiple items, use Counter, which only does n total checks.

How to count the frequency of the elements in an unordered list?

If the list is sorted, you can use groupby from the itertools standard library (if it isn't, you can just sort it first, although this takes O(n lg n) time):

from itertools import groupby

a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]
[len(list(group)) for key, group in groupby(sorted(a))]

Output:

[4, 4, 2, 1, 2]

How can I get a count of repeated elements in a Python list?

Solved!

I used itertools

here's the code:

list4=[]
a = [list(g) for k, g in itertools.groupby(lista)]
for i in range(len(a)):
    if 0 in a[i]:
        list4.append([0,len(a[i]),i])
    elif 1 in a[i]:
        list4.append([1,len(a[i]),i])

Count adjacent repeated elements in the list

Your code is not accounting for the last item in the list because you are avoiding an index out-of-bound error in the evaluation of lst[index] == lst[index+1] by iterating over a range of one less than the length of the list with while(index < len(lst)-1):.

An easy workaround without having to duplicate the counting code after the loop is to unconditionally append a dummy item to the list first, and optionally pop it after the counting is done, if you still need the original list intact. The example below uses an instance of object as a dummy item because it helps make sure that the last evaluation of lst[index] == lst[index+1] would never be True:

lst.append(object()) # append a dummy item first
while(index < len(lst)-1):
    count = 1
    result.append(lst[index])
    while(lst[index] == lst[index+1]):
        count += 1
        index += 1

    result.append(str(count))
    index += 1
    print("step: ", result)

lst.pop() # add this line only if you still need the original lst for later use

Trying to get a count of duplicate elements

If the Count is greater than one, check if a is not in the dup list, then add a to it.
Finally, print the length of the dup list

n=int(input("Enter the number of products to be stored in a list : "))

list1=[]

for i in range(n):
    item = int(input("value for product " + str(i+1) + " : "))
    list1.append(item)
    
dup = []

for a in list1:
    if (list1.count(a) > 1) and (a not in dup):
        dup.append(a)

print("Count of duplicate elements in the list: ", len(dup))

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