# How to check if a number is repeated in a list Python | Example code

There are several ways and logics to find the number that is repeated in a list python.

Here is the linear solution for it.

``````import numpy as np

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

res = set([x for x in numbers if numbers.count(x) > 1])

print(res)``````

Output: {2, 3}

## Python check if there is a repeated value in a list example

Simple example code.

Method 1

Using for-loop and if statement.

``````numbers = [1, 2, 3, 4, 2, 3, 5]
rep = []
for n in numbers:
if numbers.count(n) > 1:
if n not in rep:
print("Repeated number: ", n)
rep.append(n)
``````

Output:

Method 2

It will find only the first repeated value.

``````numbers = [1, 2, 3, 4, 2, 3, 5]
endLoop = False

for n1 in range(0, len(numbers)):
for n2 in range(1, len(numbers)):
if numbers[n1] == numbers[n2] and n1 != n2:
print(numbers)
print(numbers[n1])
endLoop = True
if endLoop:
break

``````

Output:

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

Method 3

Using set, it also finds first repeated values.

``````number = [1, 2, 3, 4, 2, 3, 5]

def find_repeat(numbers):
seen = set()
for num in numbers:
if num in seen:
return num

res = find_repeat(number)
print(res)``````

Method 4

Using NumPy, you have to import the NumPy module. But this is a different solution, where only counting a repetition of every element of the list.

``````import numpy as np

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

counts = np.bincount(numbers)
np.where([counts > 1])[1]

print(counts)``````

Output: [0 1 2 2 1 1]

Do comment if you have any questions or suggestions on this Python list tutorial.

Note: IDE: PyCharm 2021.3.3 (Community Edition)

Windows 10

Python 3.10.1

All Python Examples are in Python 3, so Maybe its different from python 2 or upgraded versions.