Take every nth block from list

Question

Given a list:

import string
a = list(string.ascii_lowercase)

What is the Pythonic way to return every nth block of m elements? Note that this is different from just returning every nth element.

Desired result of taking every 1st of 3 blocks of 3 elements (take 3, skip 6, take 3, skip 6...):

['a', 'b', 'c', 'j', 'k', 'l', 's', 't', 'u']

I can get to this as:

import itertools
s1 = a[::9]
s2 = a[1::9]
s3 = a[2::9]    
res = list(itertools.chain.from_iterable(zip(s1,s2, s3)))

Is there a cleaner way?

score 3 · Accepted Answer · edited Sep 18 '17 at 19:54

3

For a fixed order of select and skip, you can wrap indices taking the modulo on the total length of the window (9 here) and select only those beneath the given threshold, 3:

lst = [x for i, x in enumerate(a) if i % 9 < 3]
print(lst)
# ['a', 'b', 'c', 'j', 'k', 'l', 's', 't', 'u']

You can make this into a function that makes it more intuitive to use:

def select_skip(iterable, select, skip):
    return [x for i, x in enumerate(iterable) if i % (select+skip) < select]  

print(select_skip(a, select=3, skip=6))
# ['a', 'b', 'c', 'j', 'k', 'l', 's', 't', 'u']

edited Sep 18 '17 at 19:54

Nick T

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answered Sep 18 '17 at 15:21

Moses Koledoye

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Clever idea, but it takes a couple of brain cycles to see how the 'skip' is correct (6 == 9 - 3). Might be better to wrap this logic in a helper function or generator expression. – wim Sep 18 '17 at 15:31
@wim Yeah, should be more intuitive in a function. – Moses Koledoye Sep 18 '17 at 15:40

score 2 · Answer 2 · answered Sep 18 '17 at 15:22

2

Perhaps just writing a simple generator is the most readable

def thinger(iterable, take=3, skip=6):
    it = iter(iterable)
    try:
        while True:
            for i in range(take):
                yield next(it)
            for i in range(skip):
                next(it)
    except StopIteration:
        return

This has the advantage of working even if the input is infinite, or not slicable (e.g. data coming in from a socket).

answered Sep 18 '17 at 15:22

wim

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I think it may be the most approachable solution, maybe. I don't think it is the most readable in any way. Reading function names is usually nicer than reading syntax, and it is a more useful tool for a programmer in the future. – Reut Sharabani Sep 18 '17 at 15:24

pylang · Answer 3 · 2017-09-19T08:31:54.443

more_itertools is a third-party library that implements itertools recipes and other helpful tools such as more_itertools.windowed.

>  pip install more_itertools

Code

import string

from more_itertools import windowed, flatten


m, n = 3, 6
list(flatten(windowed(string.ascii_lowercase, m, step=m+n)))
# ['a', 'b', 'c', 'j', 'k', 'l', 's', 't', 'u']

windowed naturally steps one position per iteration. Given a new step by advancing beyond the overlaps (m), the windows are appropriately determined.

Reut Sharabani · Answer 4 · 2017-09-18T15:26:30.803

0

You can do it using some generic "chunks" recipe:

windows = chunks(original_iter, n=3)

Now that you've windowed you're data as you think of it, use islice's second variant for its' 'step' capabilities:

# flattens the list as well using chain
result = chain.from_iterable(islice(windows, 0, None, 2))

edited Sep 18 '17 at 15:26

answered Sep 18 '17 at 15:20

Reut Sharabani

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You're relying on there to be a common multiple between the chunk and stride size. It might be cleaner to take chunks of 9, then slice them as needed. – Nick T Sep 18 '17 at 15:22
Maybe, I just wanted to show the general approach. It can be varied easily I think. – Reut Sharabani Sep 18 '17 at 15:25
1

This is less general, less easy to vary. That's why I commented. – Nick T Sep 18 '17 at 15:34
I disagree, you stated the way to change it for a slightly different use case yourself. The point here is using almost-standard functions because we rarely do anything special. No need to re-invent the wheel. – Reut Sharabani Sep 18 '17 at 15:38
Assume the user now wants to take 4, skip 5. Or take 3, skip 7. How do they do that? – Nick T Sep 18 '17 at 15:43
"take chunks of X, then slice them as needed." – Reut Sharabani Sep 18 '17 at 16:05

DjaouadNM · Answer 5 · 2017-09-18T15:42:19.290

You can use a list comprehension and create a function that does this for any skip, take and list values:

import string
import itertools
a = list(string.ascii_lowercase)
def everyNthBlock(a, take, skip):
  res = [a[i:i + take] for i in range(0, len(a) ,skip + take)]
  return list(itertools.chain(*res))

print(everyNthBlock(a, 3, 6))
#^^^^ => ['a', 'b', 'c', 'j', 'k', 'l', 's', 't', 'u']
print(everyNthBlock(a, 4, 7))
#^^^^ => ['a', 'b', 'c', 'd', 'l', 'm', 'n', 'o', 'w', 'x', 'y', 'z']

Jeronimo · Answer 6 · 2017-09-18T15:56:28.120

0

Using incomprehensible list comprehension :D

m, n = 3, 3
[elem for blockstart in range(0, len(a), m*n) for elem in a[blockstart:blockstart+n]]    
#> ['a', 'b', 'c', 'j', 'k', 'l', 's', 't', 'u']

edited Sep 18 '17 at 15:56

answered Sep 18 '17 at 15:33

Jeronimo

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Take every nth block from list

6 Answers6