to solve simple python list iteration

Asked May 29 '18 at 05:58

Active May 29 '18 at 07:20

Viewed 48 times

seq1 = ['1', '2', '3']
seq2 = ['1', '2']

for elt1 in seq1:
    for elt2 in seq2:
        if elt1 == elt2:
            seq1.remove(i)
print seq1

I want to leave seq1 = ['3'] but I find the result out seq1 = ['2', '3'] when I use this code.

I know how to solve it by changing this code. I just want to know why my code does not work. what is actually happening in the loop?

My analysis is:

The problem is number 2 in seq1 does not iterate . Only '1', '3' iterate. So in the end, the result comes out ['2', '3'] (if it iterates on 2 in seq1 then the result will be ['3'])

edited May 29 '18 at 07:20

DevX

asked May 29 '18 at 05:58

fluentparrot

4

You are iterating over a list while changing the lists' size (by removing from it). This is the source of your bugs. – juanpa.arrivillaga May 29 '18 at 05:59
Iterate in reverse, with the appropriate condition, or use a list comprehension: `t = set(lst2); lst1 = [x for x in lst1 if x not in t]` – cs95 May 29 '18 at 06:01
_"I know how to solve it by changing this code. I just want to know why my code does not work."_ I just can't understand this at all... – AGN Gazer May 29 '18 at 06:10
@juanpa.arrivillaga i got it thank you ^------------------------^ – fluentparrot May 30 '18 at 08:18
@AGNGazer it means i know another answer. – fluentparrot May 30 '18 at 08:18
@coldspeed thank you for your concern – fluentparrot May 30 '18 at 08:19

to solve simple python list iteration

0 Answers0