Extract Duplicates from List After Applying Sort

Question

I am working on code snippet to extract duplicates from a list. I have seen several implementations/solutions on this site. However, I am not getting this line correctly - syntax wise I think. After sorting, compare index(x) with index(x+1). If it is add to the set.

print(set([i for i in a if (a[i] == a[i+1]))

a = [1,2,3,2,1,5,6,5,5,5]
print(a)
print(set(sorted(a)))
# l1[i] == l1[i+1]
print(set([i for i in a if (a[i] == a[i+1]))
print(set([i for i in a if sum([1 for item in a if item == i]) > 1]))

Expected results: {1, 2, 5}

Possible duplicate of [How do I find the duplicates in a list and create another list with them?](https://stackoverflow.com/questions/9835762/how-do-i-find-the-duplicates-in-a-list-and-create-another-list-with-them) — akaur, Apr 17 '19 at 05:08
I am specifically trying this logic print(set([i for i in a if (a[i] == a[i+1])). I am aware of many different ways to extract the dups. — RSSregex, Apr 17 '19 at 05:09

hiro protagonist · Answer 1 · 2019-04-17T05:06:42.037

5

you could use collections.Counter:

from collections import Counter

a = [1,2,3,2,1,5,6,5,5,5]
c = Counter(a)

res = [n for n, m in c.items() if m > 1]
print(res)  # [1, 2, 5]

this way you iterate once over the list and once over the counter only.

edited Apr 17 '19 at 05:06

answered Apr 17 '19 at 05:04

hiro protagonist

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1

From a complexity perspective, this is the way to go. – iz_ Apr 17 '19 at 05:05
1

I keep forgetting `Counter` exists. This is better +1 – gmds Apr 17 '19 at 05:09

Albin Paul · Answer 2 · 2019-04-17T05:23:53.517

2

From what I could gather you are trying to implement this logic, this code runs in O(nlogn) time complexity while the code which runs with counter runs in O(n) time complexity means it is faster and cleaner.

a = [1,2,3,2,1,5,6,5,5,5]
a.sort()
print(set([a[i] for i in range(len(a)-1) if (a[i] == a[i+1])]) )

OUTPUT

set([1, 2, 5])

edited Apr 17 '19 at 05:23

answered Apr 17 '19 at 05:20

Albin Paul

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How can `O(n log n)` be faster than `O(n)`? – gmds Apr 17 '19 at 05:22

score 1 · Answer 3 · answered Apr 17 '19 at 05:03

1

How about this instead?

a = [1,2,3,2,1,5,6,5,5,5]
duplicates = set(element for element in a if a.count(element) > 1)
print(duplicates)

Output:

{1, 2, 5}

answered Apr 17 '19 at 05:03

gmds

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sameera lakshitha · Answer 4 · 2019-04-17T05:38:09.050

0

Suggest a simple solution to find duplicates from List.

>>> a = [1,2,3,2,1,5,6,5,5,5]
>>> a.sort()
>>> set([x for x in a if a.count(x) > 1])

Output: {1, 2, 5}

edited Apr 17 '19 at 05:38

answered Apr 17 '19 at 05:18

sameera lakshitha

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Extract Duplicates from List After Applying Sort

4 Answers4