How to check if there is a repeated value in a list and get that value

Question

How can I find the repeated value in a list? There will always be only one repeated value

for example:

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

I need to get the value 3

This is what I was trying but it is sometimes printing the value of the same list 2 times.

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

What should happen if there is more than one repeated value? No repeated values? Does it matter if values are repeated more often than others? — Martijn Pieters, Aug 28 '14 at 19:29
There will always be only 1 repeated value. Thanks for the comment, editing the post now. — FutoRicky, Aug 28 '14 at 19:30
http://stackoverflow.com/questions/2161752/how-to-count-the-frequency-of-the-elements-in-a-list ? — Marc B, Aug 28 '14 at 19:30

Martijn Pieters · Answer 1 · 2014-08-28T19:36:46.690

7

Keep track of numbers you have seen with a set() object, the first number that is already in the set is repeated:

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

This is an efficient method to find the first repeated value, as it won't have to look at the rest of the numbers once it finds it.

edited Aug 28 '14 at 19:36

answered Aug 28 '14 at 19:30

Martijn Pieters

1,048,767
296
4,058
3,343

What exactly does set() do? and also the last line. – FutoRicky Aug 28 '14 at 19:34
@FUtoFarai creates an empty set then adds the number to the set if it meets the condition – Jose Magana Aug 28 '14 at 19:35
@FutoFarai: a set contains *unique* values, and testing for membership is very efficient. `seen.add()` adds a value to the set, `num in seen` tests if a value is already in the set. – Martijn Pieters Aug 28 '14 at 19:36
this is very inefficient, because operations related to the `set` takes a lot of the time. – ivan.mylyanyk Aug 28 '14 at 19:39
@cupidon4uk: no, they do **not** take a lot of time. Adding to a set, as well as membership tests, are O(1) constant time operations, see https://wiki.python.org/moin/TimeComplexity. Do you have evidence to the contrary? – Martijn Pieters Aug 28 '14 at 19:41
@MartijnPieters you are right, sorry. This way, it is much better approach than O(n*n) :) – ivan.mylyanyk Aug 28 '14 at 19:45
@cupidon4uk: exactly; the OP approach is a O(N^2) approach, this is a O(N) approach, both worst case. – Martijn Pieters Aug 28 '14 at 19:46
If you were doing an interview or test, this would be the only acceptable answer. – Nick Humrich Aug 28 '14 at 20:38

score 6 · Answer 2 · answered May 02 '18 at 16:24

6

there is a simpler method to do so. using .count() keyword. you can do something like this

for n in numbers:
    if numbers.count(n)>1:
        print("Repeated number: ",n)

answered May 02 '18 at 16:24

Muhammad Tuseeq

69
2
4

score 2 · Answer 3 · answered Aug 28 '14 at 19:46

2

If you don't mind using numpy:

numbers = [1,2,3,4,5,3]
counts = np.bincount( numbers )
np.where([ counts > 1 ])[1]

... Will get you array([3])

answered Aug 28 '14 at 19:46

Lukeclh

231
1
5

This does require a full scan again. It can be solved with less work because the OP only needs the *first* repeated number. – Martijn Pieters Aug 28 '14 at 19:49
You are quite right. This solution work well only if the array is reasonably short and/or speed is not a big issue. – Lukeclh Aug 28 '14 at 20:00

score 1 · Accepted Answer · answered Aug 28 '14 at 19:32

You should check that n1 != n2

because your program will check numbers[1] == numbers[1] and print the value at the position 1, even there is only one vaulue equal to numbers[1]

so the code will look like:

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

score 1 · Answer 5 · answered Aug 28 '14 at 19:37

1

Here is a one liner.

set([x for x in _list if _list.count(x) > 1])

answered Aug 28 '14 at 19:37

rocktheartsm4l

2,129
23
38

This is *very* inefficient; each `test.count()` call does a full scan of the list to count the elements. You are also repeating the count for any repeated element. – Martijn Pieters Aug 28 '14 at 19:37

score 0 · Answer 6 · answered Aug 28 '17 at 12:25

0

g=int(input("how many digits?"))
n=int(input("enter a number"))
c=0
k=0
k=n
for i in range(0,g):
    While  (n!=0):
               r=n%10
               n=n//10
               r1=n%10
               If (r==r1):
                               c=c+1
 if (c>0):
     print("reparation is there")
 else:
     print("no repeatation .....")

answered Aug 28 '17 at 12:25

Yogitha T

1

Welcome to stackoverflow. This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. – Kenzo_Gilead Aug 28 '17 at 12:44

score 0 · Answer 7 · answered Nov 11 '21 at 09:02

0

A variant to Lukeclh s' answer

numbers = [1,2,3,4,5,3]
counts = np.bincount( numbers )
np.where( counts > 1 )[0]

will give the same result

answered Nov 11 '21 at 09:02

fenaux

37
5

How to check if there is a repeated value in a list and get that value

7 Answers7