Separating different elements in list

Question

I'm trying to separate same elements from list and make individual list which contains same elements.

I guess you will understand after running this program

https://code.sololearn.com/cX0ZeMVDbz5m/?ref=app

Answer 1

Because the condition if prev!=i doesn't become true, because there is no different element after 6
The last group will always be missing in final, so just append it from ls2 after the loop

Answer 2

Your code only works, if the elements are already grouped, for [1, 2, 1] it would give [[1], [2], [1]]
I don't know if that's intentional or not.
If not and the output should be [[1, 1], [2]], here are two short solutions;

print([[val] * ls1.count(val) for val in sorted(set(ls1))])

from collections import Counter
print([[val] * freq for val, freq in Counter(ls1).items()])

The one with Counter has better performance

Answer 3

def getPairs(arr, n):

ls = []

for i in range(0, n):

for j in range(i + 1, n):

if arr[i] == arr[j]:

ls.append([arr[i],arr[j]])
return ls

# Driver function

arr = [1, 1, 9, 2, 2, 8, 3, 3, 4, 4, 5, 6, 6]

n = len(arr)

print(getPairs(arr, n))

Answer 4

Shail Murtaza great!
And remember that python "comes with batteries included", meaning that for all standard tasks it most likely offers a solution or at least functions that help you a lot. Use them to not only make your code shorter, but also more readable

Answer 5

Thank you very much

You saved my time
I was unable to figure out that simple thing.

Answer 6

iTech your code tries to solve something different, and it has a problem when an element occurs more than 2 times. Also you don't need to give n to the function