Oneliner question (Recurring Character)

Question

I have recently made a code that gets the first recurring character in a string:
https://code.sololearn.com/c65Q3VDt07Hl/?ref=app

It raises an exception due to the method I used (using list indices), though I want it to display "None".

While I can easy fix this using a try-except block, I want to keep it a oneliner. Any suggestions?

Answer 1

HonFu How about this to reduce character count?
print((lambda x=input(): ([i for i in x if x.count(i)>1] or [None])[0])())

Answer 2

Short and simple:

x=input();print(next((i for i in x if x.count(i)>1),None))

Answer 3

Sorry, my version didn't work exactly as you wanted.

This one works:

print((lambda x=input(): ([i for i in x if x.count(i)>1] or [None, None])[0])())

Answer 4

Cbr✔[ Exams ] Thomas Williams
It's a oneliner to me. Most of the times anything that isn't a function is not added to the lines/characters count.

At least that's the way golf challenges work on SL.

Answer 5

Roneel thats actually a very good solution👍🏼 Thanks😁

Answer 6

HonFu I get the 'or [None, None]' now! If list is empty, then it is replaced by [None]... Smart.

Answer 7

HonFu Nice👍🏼 Thanks man, really appreciate it!

Cbr✔[ Exams ] Nice code, but just checked it out and it didnt do the "None" thing. Thanks for the reply, though

Thank you, guys👍🏼 Ill keep this thread for a day and then delete it

Answer 8

Why delete? It's a programming-related question!

Answer 9

Yeah, absolutely!

Other people may want to learn from it as well.

And it would be a pity if the very people who actually ask something programming-related, edit themselves out of here. 😂

Answer 10

Lol that's True HonFu

Answer 11

Ha - obviously! 🤦‍♂️

Thanks for pointing that out, Roneel !

Answer 12

print((lambda x=input(): [i for i in x if x.count(i)>1][0] if len([i for i in x if x.count(i)>1]) > 0 else None)())

Here is my answer, with ternary operator!

Answer 13

Théophile oh. I was hoping there would be a simpler way to do it. Yeah, I used that one in a beta-version. Was hoping someone would have a better solution

Thanks man👍🏼

Answer 14

I don't know if a simpler way exists too. I'll see!

Answer 15

print((lambda x: x[0] if len(x) > 0 else None)((lambda x=input(): [i for i in x if x.count(i)>1])()))

Answer 16

Cbr✔[ Exams ] he wants a one liner code. In yours, you can't replace string by input.

But nice code, btw!

Answer 17

HonFu I was going to delete it since its been answered. Jn retrospect, I should probably keep it around

Answer 18

Oh. Cool👍🏼 Diego. Not oneliner, but good code anyway