How to conditionally skip number of iteration steps in a for loop in python?

Question

We have a list item_list,

item_list = ["a", "b", "XYZ", "c", "d", "e", "f", "g"]

We iterate over its items with a for loop, if item is "XYZ", skip items "c", "d", "e" and continue with "f":

for item in item_list:
    if item == "XYZ":
       do_something()
       skip_3_items() ----> skip items "c", "d", "e"
    else:
        do_something_else()

What could be the most pythonic way to achieve this?

You want to skip those particular items or any next three items in the list ? — jramirez, Jan 16 '14 at 18:00
You could use a counter and `continue` until it's 0, but i'm pretty sure the most pythonic way would be rethinking your loop entirely — loopbackbee, Jan 16 '14 at 18:01

kindall · Accepted Answer · 2014-01-16T18:08:46.567

7

list_iter = iter(item_list)

for item in list_iter:
    if item == "XYZ":
        do_something()
        for _ in range(3):   # skip next 3 items
            next(list_iter, None)

# etc.

Basically, rather than iterating over the list directly, you create an abstraction for it called an iterator and iterate over that. You can tell the iterator to advance to the next item by calling next(...) which we do three times to skip the next three items. The next time through the loop, it picks up at the next item after that.

edited Jan 16 '14 at 18:08

answered Jan 16 '14 at 18:01

kindall

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What does next(list_iter) do? – alwbtc Jan 16 '14 at 18:01
1

@alwbtc it advances the iterator one position – loopbackbee Jan 16 '14 at 18:02
2

You should put those `next` statements in a `try ... except StopIteration` – Steinar Lima Jan 16 '14 at 18:03
@alwbtc if you're new to iterators, you should [read this](http://stackoverflow.com/a/9884259/1595865) – loopbackbee Jan 16 '14 at 18:03
3

Instead of catching a StopIteration, since you're ignoring the value anyway, you can simply use the default arg, e.g. `next(list_iter, None)` – DSM Jan 16 '14 at 18:09
What if it is a dictionary, but not a list? – alwbtc Jan 16 '14 at 18:11
@DSM has the better idea. also changed it to a loop rather than writing `next()` three times. – kindall Jan 16 '14 at 18:11

score 2 · Answer 2 · answered Jan 16 '14 at 18:09

2

Since nobody has mentioned a while loop, I will:

item_list = ["a", "b", "XYZ", "c", "d", "e", "f", "g"]
i = 0
while i < len(item_list):
    item = item_list[i]
    if item == "XYZ":
        do_something()
        i += 3
    else:
        do_something_else()
    i += 1

answered Jan 16 '14 at 18:09

inspectorG4dget

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I believe no one has mentioned because it is usually very "non-pythonic" to use while. The whole point of using iterators is to get away from counting integers and stuff. Your code works, of course – lsoranco Mar 31 '17 at 16:57

damienfrancois · Answer 3 · 2014-01-16T18:11:04.343

1

Use an iterator:

$ cat t.py 
item_list = ["a", "b", "XYZ", "c", "d", "e", "f", "g"]

it = iter(item_list)
for item in it:
    if item == "XYZ":
       print item
       for _ in range(3):
           next(it, None)
    else:
        print item

This gives:

$ python t.py 
a
b
XYZ
f
g

edited Jan 16 '14 at 18:11

answered Jan 16 '14 at 18:04

damienfrancois

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score 1 · Answer 4 · answered Jan 16 '14 at 18:12

I would split the processing it two parts for sake of readability

>>> def foo(item_list,key = "XYZ", skip = 3):
    from itertools import takewhile, islice
    def do_something():
        return "do_something()"
    def do_something_else():
        return "do_something_else()"
    it = iter(item_list)
    for items in takewhile(lambda e: e != key, it):
        print items, do_something_else()
    print do_something()
    it = islice(it,skip, None)
    for items in it:
        print items, do_something_else()


>>> foo(item_list)
a do_something_else()
b do_something_else()
do_something()
f do_something_else()
g do_something_else()

Chris Arena · Answer 5 · 2014-01-16T18:13:12.407

I'd store a counter that handles skipping items.

def skipper(item_list):
    skip_count = 0
    for item in item_list:
        if item == "XYZ":
            skip_count = 3
        else:
            if skip_count:
                skip_count -= 1
            else:
                # do_something()
                print item,

Example:

In [23]: item_list
Out[23]: ['a', 'b', 'XYZ', 'c', 'd', 'e', 'f', 'g']

In [24]: skipper(item_list)
a b f g

roippi · Answer 6 · 2014-01-16T18:17:34.540

0

I'd probably write it out like this, personally:

xyz_i = item_list.index('XYZ')

do_something('XYZ') #or do_something(item_list[xyz_i]) but.. just save yourself the list lookup

for x in item_list[xyz_i+4:]:
    do_something_else(x)

edited Jan 16 '14 at 18:17

answered Jan 16 '14 at 18:05

roippi

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How to conditionally skip number of iteration steps in a for loop in python?

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