How to loop n items at a time but not stepping only 1 each time? python

Question

I have a generator, let's say:

x = iter([1,2,3,4,5,6])

And if i want to loop 3 items at a time but step only 1 each time, i want to get:

I have tried looping two at a time but it steps 2 each time:

x = iter([1,2,3,4,5,6])
x = list(x)
for i,j in zip(x[::2],x[1::2]):
    print i,j

[out]:

1 2
3 4
5 6

I have tried looping n at a time but it also steps n:

from itertools import izip_longest

def grouper(n, iterable, fillvalue=None):
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

x = iter([1,2,3,4,5,6])

for i,j in grouper(3,x):
    print i,j
print

[out]:

Traceback (most recent call last):
  File "test.py", line 11, in <module>
    for i,j in grouper(3,x):
ValueError: too many values to unpack

I have tried just access +n each time i loop:

x = iter([1,2,3,4,5,6])
x = list(x)

for i,j in enumerate(x):
    print x[i], x[i+1], x[i+2]

[out]:

1 2 3
2 3 4
3 4 5
4 5 6
5 6
Traceback (most recent call last):
  File "test.py", line 26, in <module>
    print x[i], x[i+1], x[i+2]
IndexError: list index out of range

Question:

Can I not change the generator into a list before looping and accessing more than one value at a time but stepping only 1 each time?
Other than using the last solution i have, how to else achieve the desired iteration?

The desired output appears to be incorrect (which sounds daft, I know). Shouldn't the last line be `4 5 6`? — ebarr, Jun 03 '14 at 09:48
@jonrsharpe good dupe target... I'll make sure we add that to the sopython list :) — Jon Clements, Jun 03 '14 at 10:04

Jon Clements · Accepted Answer · 2014-06-03T09:52:54.740

2

You can use tee, islice and izip_longest as a base, and adapt the "stop" condition, eg:

from itertools import tee, islice, izip_longest

data = [1, 2, 3, 4, 5, 6]
iters = [islice(it, n, None) for n, it in enumerate(tee(data, 3))]
res = list(izip_longest(*iters))
# [(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, None), (6, None, None)]

An example stop condition might be:

from itertools import takewhile
res = list(takewhile(lambda L: None not in L, izip_longest(*iters)))
# [(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6)]

edited Jun 03 '14 at 09:52

answered Jun 03 '14 at 09:43

Jon Clements

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doesn't it destroy the generator structure and create the whole object in memory instead? – lejlot Jun 03 '14 at 09:45
@lejlot not following... – Jon Clements Jun 03 '14 at 09:45
generator's are space efficient objects, they do not actually store the whole list in the memory. Isn;t your solution breaking this feature? – lejlot Jun 03 '14 at 09:45
1

@lejlot if you're referring to the `list` part - that's purely to demonstrate the output - the `izip_longest` could be iterated over normally if needs be – Jon Clements Jun 03 '14 at 09:46
and you could switch the iters from a list to a generator to avoid generating that list. – M4rtini Jun 03 '14 at 09:48
@M4rtini I suppose one could - but for 3 items, it hardly seems worth it :) – Jon Clements Jun 03 '14 at 10:00

How to loop n items at a time but not stepping only 1 each time? python

1 Answers1