How to skip a single loop iteration in python?

Question

I have the following code:

for i in list1:
    if i == 5:
        #skip the NEXT iteration (not the end of this one)
    else:
        #do something

How do I skip the iteration that comes after the iteration that throws the skip. For example, if list1=[1, 2, 3, 4, 5, 6, 7], the loop will skip 6 and go straight to 7 because 5 triggered the skip I've seen this question and several others, but they all deal with skipping the current iteration, while I want to skip the next iteration. The answers to those questions suggest continue, which as far as I can tell will stop the remainder of the current iteration and move on to the next one, which is not what I want. How can I skip a single iteration in a loop?

Edit: Using next() has been suggested, but this does not work for me. When I run the following code:

a = [1, 2, 3, 4, 5, 6, 7, 8]
ai = iter(a)
for i in a:
    print i
    if i == 5:
        _ = next(ai)

I get

1
2
3
4
5
6 #this should not be here
7
8

Using next() is also the suggestion in this question: Skip multiple iterations in loop python

There's your bug: `for i in a:`. Should be `for i in ai:` – Moses Koledoye Jun 16 '16 at 16:06 — Moses Koledoye, Jun 16 '16 at 16:06

Moses Koledoye · Accepted Answer · 2018-10-14T23:19:18.817

16

You can create an iterator from the list. With this, you can mutate the loop items while looping:

it = iter(list1)
for i in it:
    if i == 5:
        next(it) # Does nothing, skips next item
    else:
        #do something

In case you're planning to use the value at i==5, you should do something before the evaluating the condition:

it = iter(list1)
for i in it:
    #do something
    if i == 5:
        next(it) # Does nothing, skips next item

If you're doing this in a terminal, you should assign the next item to a variable as the terminal may force an autoprint on a dangling reference:

>>> list1 = [1, 2, 3, 4, 5, 6, 7]
>>> it = iter(list1)
>>> for i in it:
...    print(i)
...    if i == 5:
...       j = next(it) # here
...
1
2
3
4
5
7

edited Oct 14 '18 at 23:19

answered Jun 16 '16 at 01:23

Moses Koledoye

77,341
8
133
139

1

This doesn't seem to actually skip the next iteration. When I add `print i` to the first line of the loop, it prints 1 though 7, when it should print 1, 2, 3, 4, 5, 7, skipping 6 – cat40 Jun 16 '16 at 01:46
I'm not using the terminal. I wrote and saved (and ran) a .py file – cat40 Jun 16 '16 at 01:55
And it still does not work after you passed the output to a variable? – Moses Koledoye Jun 16 '16 at 01:56
Yes, it still does not work – cat40 Jun 16 '16 at 15:43
Please update your question with the changes you have made. or you could post the code in the comments if you choose not to, so we can have a look. – Moses Koledoye Jun 16 '16 at 15:45

score 4 · Answer 2 · answered Jun 16 '16 at 01:19

4

Just set a flag:

>>> i
[0, 1, 2, 3, 4, 5, 6, 7]
>>> skip = False
>>> for q in i:
...    if skip:
...      skip = False
...      continue
...    if q == 5:
...       skip = True
...    print(q)
...
0
1
2
3
4
5
7

answered Jun 16 '16 at 01:19

Burhan Khalid

169,990
18
245
284

score 0 · Answer 3 · edited Jun 16 '16 at 01:21

0

runNext = True
for i in list1:
    if i == 5:
        runNext = False
    else:
        if runNext:
            #do something
        runNext = False

edited Jun 16 '16 at 01:21

Moon Cheesez

2,489
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24
38

answered Jun 16 '16 at 01:20

Robert Columbia

6,313
15
32
40

That's what I'm already doing. Is there a better method? – cat40 Jun 16 '16 at 01:47

How to skip a single loop iteration in python?

3 Answers3