How to iterate over a list repeating each element in Python

Question

I'm using Python to infinitely iterate over a list, repeating each element in the list a number of times. For example given the list:

l = [1, 2, 3, 4]

I would like to output each element two times and then repeat the cycle:

1, 1, 2, 2, 3, 3, 4, 4, 1, 1, 2, 2 ...

I've got an idea of where to start:

def cycle(iterable):
  if not hasattr(cycle, 'state'):
    cycle.state = itertools.cycle(iterable)
  return cycle.next()

 >>> l = [1, 2, 3, 4]
 >>> cycle(l)
 1
 >>> cycle(l)
 2
 >>> cycle(l)
 3
 >>> cycle(l)
 4
 >>> cycle(l)
 1

But how would I repeat each element?

Edit

To clarify this should iterate infinitely. Also I've used repeating the element twice as the shortest example - I would really like to repeat each element n times.

Update

Will your solution lead me to what I was looking for:

>>> import itertools
>>> def ncycle(iterable, n):
...   for item in itertools.cycle(iterable):
...     for i in range(n):
...       yield item
>>> a = ncycle([1,2], 2)
>>> a.next()
1
>>> a.next()
1
>>> a.next()
2
>>> a.next()
2
>>> a.next()
1
>>> a.next()
1
>>> a.next()
2
>>> a.next()
2

Thanks for the quick answers!

Community · Accepted Answer · 2017-05-23 10:32:49Z

13

How about this:

import itertools

def bicycle(iterable, repeat=1):
    for item in itertools.cycle(iterable):
        for _ in xrange(repeat):
            yield item

c = bicycle([1,2,3,4], 2)
print [c.next() for _ in xrange(10)]

EDIT: incorporated bishanty's repeat count parameter and Adam Rosenfield's list comprehension.

edited May 23, 2017 at 10:32

CommunityBot

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answered Dec 20, 2008 at 18:36

Will Harris

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1 Comment

jb. Over a year ago

>>> import itertools >>> def ncycle(iterable, n): ... for item in itertools.cycle(iterable): ... for i in range(n): ... yield item

Adam Rosenfield · Accepted Answer · 2008-12-20 18:31:49Z

6

You could do it with a generator pretty easily:

def cycle(iterable):
    while True:
        for item in iterable:
            yield item
            yield item

x=[1,2,3]
c=cycle(x)

print [c.next() for i in range(10)]  // prints out [1,1,2,2,3,3,1,1,2,2]

answered Dec 20, 2008 at 18:31

Adam Rosenfield

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3 Comments

rob Over a year ago

Again, I am not 100% sure to have understood the question, but aren't generators a bit an overkill for a simple problem like returning an item twice (or n times)?

recursive Over a year ago

It's 5 lines. How can that be overkill? And if the alternative is building a list, that's impossible, because it's infinite length.

rob Over a year ago

Why to create a function in the first place? You could reach the same results with a simple nested loop, without the need for a generator, Reading the loop, even a novice immediately understand the meaning of the code, while with generators you have to trust the name of the function.

panda-34 · Accepted Answer · 2016-04-09 10:30:56Z

2

itertools.chain.from_iterable(itertools.repeat(item, repeat) for item in itertools.cycle(l))

answered Apr 9, 2016 at 10:30

panda-34

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Comments

rob · Accepted Answer · 2008-12-20 18:41:19Z

1

Solution should be something like

iterable = [1, 2, 3, 4]
n = 2

while (True):
  for elem in iterable:
    for dummy in range(n):
      print elem # or call function, or whatever

Edit: added 'While (True)' to iterate indefinitely.

edited Dec 20, 2008 at 18:41

answered Dec 20, 2008 at 18:30

rob

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Comments

Wietse · Accepted Answer · 2008-12-22 16:32:41Z

1

import itertools as it

def ncycle(iterable, n=1):
    if n == 1:
        return it.cycle(iterable)
    return it.cycle(it.chain(*it.izip(*([iterable]*n))))

answered Dec 22, 2008 at 16:32

Wietse

Comments

sujeet · Accepted Answer · 2017-10-24 23:38:20Z

1

I do it this way:

from itertools import cycle, repeat, chain
flatten = chain.from_iterable # better name

def ncycle(xs, n):
    return flatten(repeat(x, n) for x in cycle(xs))

# example
for n,x in enumerate(ncycle('abcd', 2)):
    print(x, end=" ")
    if n > 9:
        print("")
        break
# output: a a b b c c d d a a b

answered Oct 24, 2017 at 23:38

sujeet

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Comments

PEZ · Accepted Answer · 2008-12-20 18:40:13Z

0

[ "%d, %d" % (i, i) for i in [1, 2, 3, 4] * 4]

The last 4 there is the number of cycles.

answered Dec 20, 2008 at 18:40

PEZ

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