Idiomatic way of repeating items in a sequence.

[alr]

I need to repeat each item in a list n times, like this function does:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
for i in range(repetitions):
newlist.append(item)
return newlist

Output:

repeatitems(['a', 'b', 'c'], 3)

['a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c']

Clear and simple. But i wonder if there is a more idiomatic way. Surely not this:

def repeatitems(sequence, repetitions):
return reduce(lambda l, i: l + i, [[item] * repetitions for item in sequence])

?

[Max M]

alr wrote:

I need to repeat each item in a list n times, like this function does:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
for i in range(repetitions):
newlist.append(item)
return newlist

I would make just a minor change:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
newlist += repetitions*[item]
return newlist
regards Max M

[Duncan Booth]

an***@wmdata.com (alr) wrote in
news:f5*************************@posting.google.co m:

I need to repeat each item in a list n times, like this function does:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
for i in range(repetitions):
newlist.append(item)
return newlist

Output:

>>> repeatitems(['a', 'b', 'c'], 3)

['a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c']

Clear and simple. But i wonder if there is a more idiomatic way.

The most obvious one that springs to mind is just a slight simplification
of your version:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
newlist.extend([item] * repetitions)
return newlist
--
Duncan Booth du****@rcp.co.uk
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?

[Peter Otten]

alr wrote:

I need to repeat each item in a list n times, like this function does:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
for i in range(repetitions):
newlist.append(item)
return newlist
.... But i wonder if there is a more idiomatic way.

How about this?

def repeatItems(sequence, repetitions):
return [[item]*repetitions for item in sequence]

[Richard Brodie]

"Peter Otten" <__*******@web.de> wrote in message news:bd*************@news.t-online.com...

How about this?

def repeatItems(sequence, repetitions):
return [[item]*repetitions for item in sequence]

Unfortunately that is equivalent to:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
newlist.append([item] * repetitions)
return newlist

and not:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
newlist.extend([item] * repetitions)
return newlist

[Duncan Booth]

Duncan Booth <du****@NOSPAMrcp.co.uk> wrote in
news:Xn***************************@127.0.0.1:

The most obvious one that springs to mind is just a slight
simplification of your version:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
newlist.extend([item] * repetitions)
return newlist

Or, if you are in a "I've got a new toy to play with" mood you could use
itertools from Python 2.3 to obfuscate it somewhat:

from itertools import chain, izip, repeat
def repeatiterate(sequence, repetitions):
return chain(*izip(*repeat(sequence, repetitions)))

This version returns an iterator, so you might want to throw in a call to
'list' if you want to do anything other than iterating over the result.

--
Duncan Booth du****@rcp.co.uk
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?

[Peter Otten]

Richard Brodie wrote:

Unfortunately that is equivalent to:

def repeatitems(sequence, repetitions):
newlist = []
for item in sequence:
newlist.append([item] * repetitions)
return newlist

.... but it looked so good. I should have tested it, though.
Je suis desole :-(

[Terry Reedy]

"alr" <an***@wmdata.com> wrote in message
news:f5*************************@posting.google.co m...

I need to repeat each item in a list n times, like this function

does:

Is this really the right question? Any code that requires such an
n-repeat list *could* be rewritten (if you own it) to use the replist
in condensed form: (items, n). If this is not possible, one could
also define a replist class with a __getitem__(self, index) that
divides index by self.n and an __iter__() that returns an appropriate
generator.

Terry J. Reedy

[John Hunter]

>>>>> "alr" == alr <an***@wmdata.com> writes:

alr> reduce(lambda l, i: l + i, [[item] * repetitions for item in

This doesn't look too bad to me, but perhaps list comprehensions are
clearer?

seq = ['a', 'b', 'c']
print [x for x in seq for x in seq]

[John Hunter]

>>>>> "alr" == alr <an***@wmdata.com> writes:

alr> reduce(lambda l, i: l + i, [[item] * repetitions for item in
alr> sequence])
Oops, premature hit of send key. What I meant to say was

seq = ['a', 'b', 'c']
print [x for x in seq for i in range(len(seq))]

JDH

[Mike C. Fletcher]

John Hunter wrote:

This doesn't look too bad to me, but perhaps list comprehensions are
clearer?

seq = ['a', 'b', 'c']
print [x for x in seq for x in seq]

def repeat3( sequence, count=1 ): .... return [x for x in sequence for i in range(count) ]
.... repeat3( [2,3,4], 3 ) [2, 2, 2, 3, 3, 3, 4, 4, 4]

I *think* is what you were suggesting, and is indeed very clear. For
those into generators, this is fun (but has no huge advantage if you
wind up using repeat instead of irepeat, or if you're using small
sequences):
from __future__ import generators
def irepeat( sequence, count=1 ): .... countSet = range(count)
.... for item in sequence:
.... for i in countSet:
.... yield item
.... def repeat( sequence, count = 1 ): .... return list(irepeat(sequence, count))
.... repeat( [2,3,4], 3 )

[2, 2, 2, 3, 3, 3, 4, 4, 4]

Enjoy,
Mike

_______________________________________
Mike C. Fletcher
Designer, VR Plumber, Coder
http://members.rogers.com/mcfletch/

[Jeff Epler]

Here's one:

def repeatitems(sequence, repetitions):
r = [None] * repetitions
return [i for i in sequence for j in r]

Here's another, using generator functions (so the return is an iterator,
not a list):

def repeatitems(sequence, repetitions):
r = [None] * repetitions
for item in sequence:
for i in r:
yield item

In both cases I've performed an "optimization" by precomputing a list
with len(repetitions) instead of computing it once for each item in
sequence. Whether this makes a difference, I don't know.

Jeff

[alr]

Wow, tanks for alle the replies. My favourite is John Hunters/Bob
Gailers solution ([x for x in seq for i in range(repetitions)]). I had
forgotten that you could have nested for statements in list
literals... Aahz's point is taken. I happen to need to repeat lists of
strings (which are immutable), but that's not what i asked about now
is it. :-)

--
Regards
André Risnes