Python | Program to print duplicates from a list of integers

Given a list of integers with duplicate elements in it. The task is to generate another list, which contains only the duplicate elements. In simple words, the new list should contain elements that appear as more than one.

Examples:

Input : list = [10, 20, 30, 20, 20, 30, 40, 50, -20, 60, 60, -20, -20]
Output : output_list = [20, 30, -20, 60]

Input :  list = [-1, 1, -1, 8]
Output : output_list = [-1]

Method 1: Using the Brute Force approach

Python3

# Python program to print
# duplicates from a list
# of integers
def Repeat(x):
    _size = len(x)
    repeated = []
    for i in range(_size):
        k = i + 1
        for j in range(k, _size):
            if x[i] == x[j] and x[i] not in repeated:
                repeated.append(x[i])
    return repeated

# Driver Code
list1 = [10, 20, 30, 20, 20, 30, 40,
         50, -20, 60, 60, -20, -20]
print (Repeat(list1))

# This code is contributed
# by Sandeep_anand

Output

[20, 30, -20, 60]

Method 2: Using a single for loop

Python3

# Python program to print duplicates from
# a list of integers
lis = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]

uniqueList = []
duplicateList = []

for i in lis:
    if i not in uniqueList:
        uniqueList.append(i)
    elif i not in duplicateList:
        duplicateList.append(i)

print(duplicateList)

Output

[1, 2, 5, 9]

Method 3: Using Counter() function from collection module

Python3

from collections import Counter

l1 = [1,2,1,2,3,4,5,1,1,2,5,6,7,8,9,9]
d = Counter(l1)
print(d)

new_list = list([item for item in d if d[item]>1])
print(new_list)

Output

Counter({1: 4, 2: 3, 5: 2, 9: 2, 3: 1, 4: 1, 6: 1, 7: 1, 8: 1})
[1, 2, 5, 9]

Method 4: Using count() method

Python3

# program to print duplicate numbers in a given list
# provided input
list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]

new = []  # defining output list

# condition for reviewing every
# element of given input list
for a in list:

     # checking the occurrence of elements
    n = list.count(a)

    # if the occurrence is more than
    # one we add it to the output list
    if n > 1:

        if new.count(a) == 0:  # condition to check

            new.append(a)

print(new)

# This code is contributed by Himanshu Khune

Output

[1, 2, 5, 9]

Method 5: Using list comprehension method

Python3

def duplicate(input_list):
    return list(set([x for x in input_list if input_list.count(x) > 1]))

if __name__ == '__main__':
    input_list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
    print(duplicate(input_list))

# This code is contributed by saikot

Output

[1, 2, 5, 9]

Method 6: Using list-dictionary approach (without any inbuild count function)

Python3

def duplicate(input_list):
    new_dict, new_list = {}, []

    for i in input_list:
        if not i in new_dict:
            new_dict[i] = 1
        else:
            new_dict[i] += 1

    for key, values in new_dict.items():
        if values > 1:
            new_list.append(key)

    return new_list

if __name__ == '__main__':
    input_list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
    print(duplicate(input_list))

# This code is contributed by saikot

Output

[1, 2, 5, 9]

Method 7: Using in, not in operators and count() method

Python3

lis = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
x = []
y = []
for i in lis:
    if i not in x:
        x.append(i)
for i in x:
    if lis.count(i) > 1:
        y.append(i)
print(y)

Output

[1, 2, 5, 9]

Method 8: Using enumerate function

Python3

input_list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
print(list(set([x for i,x in enumerate(input_list) if input_list.count(x) > 1])))

Output

[1, 2, 5, 9]

Time Complexity: O(n)

Auxiliary Space: O(1)

Method 9: Using operator.countOf() method

Python3

import operator as op
# program to print duplicate numbers in a given list
# provided input
list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]

new = []  # defining output list

# condition for reviewing every
# element of given input list
for a in list:

     # checking the occurrence of elements
    n = op.countOf(list, a)

    # if the occurrence is more than
    # one we add it to the output list
    if n > 1:

        if op.countOf(new, a) == 0:  # condition to check

            new.append(a)

print(new)

# This code is contributed by vikkycirus

Output

[1, 2, 5, 9]

Time Complexity: O(n)

Auxiliary Space: O(1)