Is it possible to skip a fixed number of iterations in a loop in Python?

Question

Introduction:

So I know that there is already a question (Skip multiple iterations in loop) similar to mine with a really good answer, but there are still some open questions for me:

Question 1:

Is there any way of doing it without an iterator?

I am looking for something like * 3:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']

for sing in song:
    print(sing, end=" ")
    if sing == 'look':
        continue * 3

expected output:

always look side of life

Question 2:

If I have to use an iterator object, then is it possible to do it a fixed amount of time?

The original question has a solution like this:

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(next(song_iter))

But I want it to do it let's say x = 5 times. It is not possible like this:

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) * x

        print(next(song_iter))

So how would you do it? I know it is possible to use the function itertools.islice, but is there a way without any libraries?

My Approach:

This works great:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)

skip_iterations = 3

for sing in song_iter:
    print(sing)
    if sing == "look":
        while skip_iterations > 0:
            next(song_iter, "")

            skip_iterations -= 1

Output:

always look side of life

But maybe anyone else has a better idea? :)

Links:

The Question I was mentioning - The Answer for that question

your approach seems good enough, except I would use a for loop: for _ in range(skip_iterations): next(song_iter) — Matiiss
– Matiiss, Commented Dec 19, 2021 at 16:49
Every for loop uses an iterator, whether you're aware of it or not. Using iter simply gives you access to one at the code level. Anyway, it seems like your question is: given an iterator object, how can we call next on it a specified number of times? Well - how do you normally do anything a specified number of times in Python? With... a for loop, right? So.... — Karl Knechtel
– Karl Knechtel, Commented Dec 19, 2021 at 16:49
"If I have to use an iterator object, then is it possible to do it a fixed amount of time?" - with an arbitrary iterator, no. Iterators are sequential, not random-access. For example, an iterator over a linked list cannot be advanced without following links one by one, and a generator doesn't even have well-defined elements until you run it and make it yield elements one by one. — user2357112
– user2357112, Commented Dec 19, 2021 at 16:52
If you're just iterating over a list, then you can skip in constant time easily enough, but not with a standard list iterator. You would have to write your own handling. — user2357112
– user2357112, Commented Dec 19, 2021 at 16:59

Divij Mahajan · Accepted Answer · 2021-12-19 17:04:18Z

1

Ans 1 There are many ways to do the problem one of the easiest one would be by checking the index like

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
skip_iterations=3
for i,sing in enumerate(song):
    if not song.index('look') < i < song.index('look')+skip_iterations+1:
    print(sing, end=" ")

'enumerate' returns (index, element) Ans 2 Your method is just fine you may prefer a for loop instead of while loop but thats just a personal suggestion.

answered Dec 19, 2021 at 17:04

Divij Mahajan

861 silver badge6 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

dawg · Accepted Answer · 2021-12-19 17:18:35Z

1

An alternate approach is to have a C style for loop that has more complete update logic:

def cfor(i, test_i, update_i):
    while test_i(i):
        yield i
        i=update_i(i)

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']

skip=3

for i in cfor(0, 
              lambda i: i<len(song), 
              lambda i: i+skip+1 if song[i]=='look' else i+1):
    print(song[i], end=' ')

Or, that can just be a while loop:

idx=0
while(idx<len(song)):
    print(song[idx], end=' ')
    idx+=skip+1 if song[idx]=='look' else 1

If your skip is large, either of these would be faster.

You can also use enumerate with a comprehension:

l=song.index('look')
' '.join([w for i,w in enumerate(song) if i<=l or i>(l+skip)])

edited Dec 19, 2021 at 17:18

answered Dec 19, 2021 at 16:59

dawg

105k24 gold badges142 silver badges217 bronze badges

Comments

Kelly Bundy · Accepted Answer · 2021-12-19 17:34:09Z

0

With a skip counter:

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']

skip = 0
for sing in song:
    if skip:
        skip -= 1
        continue
    print(sing, end=" ")
    if sing == 'look':
        skip = 3

answered Dec 19, 2021 at 17:34

Kelly Bundy

28k7 gold badges34 silver badges76 bronze badges

Comments

jongboi · Accepted Answer · 2021-12-19 17:34:39Z

Question 1

You could employ the use of an iteration bound to keep track of when you can resume evaluation in the loop

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']

iter_bound = -1
for i, sing in enumerate(song):
    if i <= iter_bound:
        continue
    if sing == 'look':
        iter_bound = i + 3
    print(sing, end=" ")

If you use this very often in your code, you may want to code a function around it:

from typing import Iterable, Callable

def skip_n_on_condition(iterator: Iterable, condition: Callable, n_skips: int):
    """Loop over iterable and perform n_skips whenever the condition on the element is met"""
    iter_bound = -1
    for i, element in enumerate(iterator):
        if i <= iter_bound:
            continue
        if condition(element):
            iter_bound = i + 3
        yield element
    
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']

print(*skip_n_on_condition(song, lambda x: x == 'look', 3), end=" ")

Question 2

Since you know the number of times you will be looping, it may be better to use a for loop. It will also be an implementation that does not change the value of skip_iterations which you may not have expected.

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)

skip_iterations = 3

for sing in song_iter:
    print(sing)
    if sing == "look":
        for _ in range(skip_iterations):
            next(song_iter, "")

You could make the loop a one liner using a list comprehension.

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)

skip_iterations = 3

for sing in song_iter:
    print(sing)
    if sing == "look":
        [next(song_iter, "") for _ in range(skip_iterations)]

VPfB · Accepted Answer · 2021-12-19 17:53:55Z

0

I would consider the "historic" way of iterating.

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']

i = 0
while i < len(song):
    sing = song[i]
    i += 1

    print(sing, end=" ")
    if sing == 'look':
        i += 3
        continue

    # ... other code ...

answered Dec 19, 2021 at 17:53

VPfB

18.1k8 gold badges54 silver badges97 bronze badges

Collectives™ on Stack Overflow

Is it possible to skip a fixed number of iterations in a loop in Python?

Introduction:

Question 1:

Is there any way of doing it without an iterator?

Question 2:

If I have to use an iterator object, then is it possible to do it a fixed amount of time?

My Approach:

Links:

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

Introduction:

Question 1:

Is there any way of doing it without an iterator?

Question 2:

If I have to use an iterator object, then is it possible to do it a fixed amount of time?

My Approach:

Links:

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related