Skip multiple iterations in loop

Mehdi Nellen picture Mehdi Nellen · Mar 10, 2014 · Viewed 35.6k times · Source

I have a list in a loop and I want to skip 3 elements after look has been reached. In this answer a couple of suggestions were made but I fail to make good use of them:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
for sing in song:
    if sing == 'look':
        print sing
        continue
        continue
        continue
        continue
        print 'a' + sing
    print sing

Four times continue is nonsense of course and using four times next() doesn't work.

The output should look like:

always
look
aside
of
life

Answer

Martijn Pieters picture Martijn Pieters · Mar 10, 2014

for uses iter(song) to loop; you can do this in your own code and then advance the iterator inside the loop; calling iter() on the iterable again will only return the same iterable object so you can advance the iterable inside the loop with for following right along in the next iteration.

Advance the iterator with the next() function; it works correctly in both Python 2 and 3 without having to adjust syntax:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        next(song_iter)
        next(song_iter)
        next(song_iter)
        print 'a' + next(song_iter)

By moving the print sing line up we can avoid repeating ourselves too.

Using next() this way can raise a StopIteration exception, if the iterable is out of values.

You could catch that exception, but it'd be easier to give next() a second argument, a default value to ignore the exception and return the default instead:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        next(song_iter, None)
        next(song_iter, None)
        next(song_iter, None)
        print 'a' + next(song_iter, '')

I'd use itertools.islice() to skip 3 elements instead; saves repeated next() calls:

from itertools import islice

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        print 'a' + next(islice(song_iter, 3, 4), '')

The islice(song_iter, 3, 4) iterable will skip 3 elements, then return the 4th, then be done. Calling next() on that object thus retrieves the 4th element from song_iter().

Demo:

>>> from itertools import islice
>>> song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
>>> song_iter = iter(song)
>>> for sing in song_iter:
...     print sing
...     if sing == 'look':
...         print 'a' + next(islice(song_iter, 3, 4), '')
... 
always
look
aside
of
life