Iterate through list two items at a time

[Dave Dean]

Hi all,
I'm looking for a way to iterate through a list, two (or more) items at a
time. Basically...

myList = [1,2,3,4,5,6]

I'd like to be able to pull out two items at a time - simple examples would
be:
Create this output:
1 2
3 4
5 6

Create this list:
[(1,2), (3,4), (5,6)]

I want the following syntax to work, but sadly it does not:
for x,y in myList:
print x, y

I can do this with a simple foreach statement in tcl, and if it's easy in
tcl it's probably not too hard in Python.

Thanks,
Dave

[Dan Bishop]

On Jan 2, 7:57 pm, "Dave Dean" <dave.d...@xili nx.comwrote:

Hi all,
I'm looking for a way to iterate through a list, two (or more) items at a
time. Basically...

myList = [1,2,3,4,5,6]

I'd like to be able to pull out two items at a time...

def pair_list(list_ ):
return[list_[i:i+2] for i in xrange(0, len(list_), 2)]

[bearophileHUGS]

Few alternative solutions (other are possible), I usually use a variant
of the first version, inside a partition function, the second variant
is shorter when you don't have a handy partition() function and you
don't want to import modules, and the forth one needs less memory when
the data is very long:

from itertools import izip, islice

data = [1,2,3,4,5,6,7]

for x1, x2 in (data[i:i+2] for i in xrange(0, len(data)/2*2, 2)):
print x1, x2

for x1, x2 in zip(data[::2], data[1::2]):
print x1, x2

for x1, x2 in izip(data[::2], data[1::2]):
print x1, x2

for x1, x2 in izip(islice(dat a,0,None,2), islice(data,1,N one,2)):
print x1, x2

Bye,
bearophile

[skip]

>I'm looking for a way to iterate through a list, two (or more) items
at a time. Basically...

myList = [1,2,3,4,5,6]

I'd like to be able to pull out two items at a time...

Dandef pair_list(list_ ):
Dan return[list_[i:i+2] for i in xrange(0, len(list_), 2)]

Here's another way (seems a bit clearer to me, but each person has their own
way of seeing things):

>>import string
string.letter s

'abcdefghijklmn opqrstuvwxyzABC DEFGHIJKLMNOPQR STUVWXYZ'

>>zip(string.le tters[::2], string.letters[1::2])

[('a', 'b'), ('c', 'd'), ..., ('W', 'X'), ('Y', 'Z')]

It extends readily to longer groupings:

>>zip(string.le tters[::3], string.letters[1::3], string.letters[2::3])

[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ...

Obviously, if your lists are long, you can substitute itertools.izip for
zip. There's probably some easy way to achieve the same result with
itertools.group by, but I'm out of my experience there...

Skip

[Gabriel Genellina]

At Tuesday 2/1/2007 22:57, Dave Dean wrote:

myList = [1,2,3,4,5,6]

I'd like to be able to pull out two items at a time - simple examples would
be:
Create this output:
1 2
3 4
5 6

b=iter(a)
for x in b:
y=b.next()
print x,y

b=iter(a)
for x,y in ((item, b.next()) for item in b):
print x,y

>Create this list:
[(1,2), (3,4), (5,6)]

b=iter(a)
[(item, b.next()) for item in b]

Note that they don't behave the same at the corner cases (empty list,
single item, odd length...)
--
Gabriel Genellina
Softlab SRL


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Todo lo que querías saber, y lo que ni imaginabas,
está en Yahoo! Respuestas (Beta).
¡Probalo ya!
http://www.yahoo.com.ar/respuestas

[Jeffrey Froman]

Dave Dean wrote:

I'm looking for a way to iterate through a list, two (or more) items at a
time.

Here's a solution, from the iterools documentation. It may not be the /most/
beautiful, but it is short, and scales well for larger groupings:

>>from itertools import izip
def groupn(iterable , n):

.... return izip(* [iter(iterable)] * n)
....

>>list(groupn(m yList, 2))

[(0, 1), (2, 3), (4, 5), (6, 7), (8, 9), (10, 11)]

>>list(groupn(m yList, 3))

[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11)]

>>list(groupn(m yList, 4))

[(0, 1, 2, 3), (4, 5, 6, 7), (8, 9, 10, 11)]

>>for a,b in groupn(myList, 2):

.... print a, b
....
0 1
2 3
4 5
6 7
8 9
10 11

>>>

Jeffrey

[Dave Dean]

Thanks for all the fast responses. I'm particularly a fan of the zip
method, followed closely by the xrange example. All, of course, are a lot
of help!
Thanks,
Dave

[Gerard Flanagan]

Dave Dean wrote:

Hi all,
I'm looking for a way to iterate through a list, two (or more) items at a
time. Basically...

myList = [1,2,3,4,5,6]

I'd like to be able to pull out two items at a time - simple examples would
be:
Create this output:
1 2
3 4
5 6

Create this list:
[(1,2), (3,4), (5,6)]

A "padding generator" version:

def chunk( seq, size, pad=None ):
'''
Slice a list into consecutive disjoint 'chunks' of
length equal to size. The last chunk is padded if necessary.

>>list(chunk(ra nge(1,10),3))

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

>>list(chunk(ra nge(1,9),3))

[[1, 2, 3], [4, 5, 6], [7, 8, None]]

>>list(chunk(ra nge(1,8),3))

[[1, 2, 3], [4, 5, 6], [7, None, None]]

>>list(chunk(ra nge(1,10),1))

[[1], [2], [3], [4], [5], [6], [7], [8], [9]]

>>list(chunk(ra nge(1,10),9))

[[1, 2, 3, 4, 5, 6, 7, 8, 9]]

>>for X in chunk([],3): print X

'''
n = len(seq)
mod = n % size
for i in xrange(0, n-mod, size):
yield seq[i:i+size]
if mod:
padding = [pad] * (size-mod)
yield seq[-mod:] + padding

------------------------------------------------------------------

Gerard

[J. Clifford Dyer]

Gabriel Genellina wrote:

b=iter(a)
for x in b:
y=b.next()
print x,y

So as not to choke on odd-length lists, you could try

a = [1,2,3,4,5,6,7]
b = iter(a)
for x in b:
try:
y=b.next()
except StopIteration:
y=None
print x,y

Substitute in whatever for y=None that you like.

Cheers,
Cliff

[Wade Leftwich]

Jeffrey Froman wrote:

Dave Dean wrote:

I'm looking for a way to iterate through a list, two (or more) items at a
time.

Here's a solution, from the iterools documentation. It may not be the /most/
beautiful, but it is short, and scales well for larger groupings:

>from itertools import izip
def groupn(iterable , n):

... return izip(* [iter(iterable)] * n)
...

>list(groupn(my List, 2))

[(0, 1), (2, 3), (4, 5), (6, 7), (8, 9), (10, 11)]

>list(groupn(my List, 3))

[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11)]

>list(groupn(my List, 4))

[(0, 1, 2, 3), (4, 5, 6, 7), (8, 9, 10, 11)]

>for a,b in groupn(myList, 2):

... print a, b
...
0 1
2 3
4 5
6 7
8 9
10 11

>>

Jeffrey

This works great except you lose any 'remainder' from myList:

>>list(groupn(r ange(10),3))

[(0, 1, 2), (3, 4, 5), (6, 7, 8)] # did not include (9,)

The following might be more complex than necessary but it solves the
problem, and like groupn()
it works on infinite lists.

from itertools import groupby, imap
def chunk(it, n=0):
if n == 0:
return iter([it])
grouped = groupby(enumera te(it), lambda x: int(x[0]/n))
counted = imap(lambda x:x[1], grouped)
return imap(lambda x: imap(lambda y: y[1], x), counted)

>>[list(x) for x in chunk(range(10) , 3)]

[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

Note the chunks are iterators, not tuples as in groupn():

>>[x for x in chunk(range(10) , 3)]

[<itertools.im ap object at 0xb78d4c4c>,
<itertools.im ap object at 0xb78d806c>,
<itertools.im ap object at 0xb78d808c>,
<itertools.im ap object at 0xb78d4c6c>]
-- Wade Leftwich
Ithaca, NY