What's the best way of skip N values of the iteration variable in Python?

Question

In many languages we can do something like:

for (int i = 0; i < value; i++)
{
    if (condition)
    {
        i += 10;
    }
}

How can I do the same in Python? The following (of course) does not work:

for i in xrange(value):
    if condition:
        i += 10

I could do something like this:

i = 0
while i < value:
  if condition:
    i += 10
  i += 1

but I'm wondering if there is a more elegant (pythonic?) way of doing this in Python.

Answer 1

Use continue.

for i in xrange(value):
    if condition:
        continue

If you want to force your iterable to skip forwards, you must call .next().

>>> iterable = iter(xrange(100))
>>> for i in iterable:
...     if i % 10 == 0:
...         [iterable.next() for x in range(10)]
... 
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[21, 22, 23, 24, 25, 26, 27, 28, 29, 30]
[41, 42, 43, 44, 45, 46, 47, 48, 49, 50]
[61, 62, 63, 64, 65, 66, 67, 68, 69, 70]
[81, 82, 83, 84, 85, 86, 87, 88, 89, 90]

As you can see, this is disgusting.

Answer 2

Create the iterable before the loop.

Skip one by using next on the iterator

it = iter(xrange(value))
for i in it:
    if condition:
        i = next(it)

Skip many by using itertools or recipes based on ideas from itertools.

itertools.dropwhile()

it = iter(xrange(value))
for i in it:
    if x<5:
        i = dropwhile(lambda x: x<5, it)

Take a read through the itertools page, it shows some very common uses of working with iterators.

itertools islice

it = islice(xrange(value), 10)
for i in it:
    ...do stuff with i...

Answer 3

It's a very old question, but I find the accepted answer is not totally stisfactory:

• first, after the if ... / [next()...] sequence, the value of i hasn't changed. In your first example, it has.
• second, the list comprehension is used to produce a side-effect. This should be avoided.
• third, there might be a faster way to achieve this.

Using a modified version of consume in itertools recipes, you can write:

import itertools

def consume(it, n):
    return next(itertools.islice(it, n-1, n), None)

it = iter(range(20))
for i in it:
    print(i, end='->')
    if i%4 == 0:
        i = consume(it, 5)
    print(i)

As written in the doctstring of consume, the iterator is consumed at C speed (didn't benchmark though). Output:

0->5
6->6
7->7
8->13
14->14
15->15
16->None

With a minor modification, one can get 21 instead of None, but I think this isnot a good idea because this code does work with any iterable (otherwise one would prefer the while version):

import string
it = iter(string.ascii_lowercase) # a-z
for x in it:
    print(x, end="->")
    if x in set('aeiouy'):
        x = consume(it, 2) # skip the two letters after the vowel
    print(x)

Output:

a->c
d->d
e->g
h->h
i->k
l->l
m->m
n->n
o->q
r->r
s->s
t->t
u->w
x->x
y->None

Answer 4

Itertools has a recommended way to do this: https://docs.python.org/3.7/library/itertools.html#itertools-recipes

import collections
def tail(n, iterable):
    "Return an iterator over the last n items"
    # tail(3, 'ABCDEFG') --> E F G
    return iter(collections.deque(iterable, maxlen=n))

Now you can do:

for i in tail(5, range(10)):
    print(i)

to get

5
6
7
8
9

Answer 5

I am hoping I am not answering this wrong... but this is the simplest way I have come across:

for x in range(0,10,2):
    print x

output should be something like this:

0
2
4
6
8

The 2 in the range parameter's is the jump value

Answer 6

There are a few ways to create iterators, but the custom iterator class is the most extensible:

class skip_if:   # skip_if(object) for python2
    """
    iterates through iterable, calling skipper with each value
    if skipper returns a positive integer, that many values are
    skipped
    """
    def __init__(self, iterable, skipper):
        self.it = iter(iterable)
        self.skip = skipper
    def __iter__(self):
        return self
    def __next__(self):   # def next(self): for python2
        value = next(self.it)
        for _ in range(self.skip(value)):
            next(self.it, None)
        return value

and in use:

>>> for i in skip_if(range(1,100), lambda n: 10 if not n%10 else 0):
...   print(i, end=', ')
... 
 1,  2,  3,  4,  5,  6,  7,  8,  9, 10,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30,
41, 42, 43, 44, 45, 46, 47, 48, 49, 50,
61, 62, 63, 64, 65, 66, 67, 68, 69, 70,
81, 82, 83, 84, 85, 86, 87, 88, 89, 90,

Answer 7

Does a generator function here is rebundant? Like this:

def filterRange(range, condition):
x = 0
while x < range:
    x = (x+10) if condition(x) else (x + 1)
    yield x

if __name__ == "__main__":
for i in filterRange(100, lambda x: x > 2):
    print i

Answer 8

Easiest way is to use more_itertools.consume(). See https://more-itertools.readthedocs.io/en/stable/api.html#more_itertools.consume.

import more_itertools

x = iter(range(10))

# skip the next four items
more_itertools.consume(x, 4)
y = list(x)
print(y)

will result in

[4, 5, 6, 7, 8, 9]

This becomes super helpful if you are iterating through a file and need to skip a bunch of lines. Something like this:

with open("test.txt", "r") as file:
    while line:= next(file):

        # Look for "magic_word" and then skip 10 lines.
        if "magic_word" in line:
            more_itertools.consume(file, 10)
    
        # do other stuff with line

Answer 9

I think you have to use a while loop for this...for loop loops over an iterable..and you cannot skip next item like how you want to do it here