splitting a list into n groups

[Rajarshi Guha]

Hi,
is there an efficient (pythonic) way in which I could split a list into
say 5 groups? By split I mean the the first x members would be one group,
the next x members another group and so on 5 times. (Obviously x =
lengthof list/5)

I have done this by a simple for loop and using indexes into the list.
But it does'nt seemm very elegant

Thanks,

[Tim Hochberg]

Rajarshi Guha wrote:

Hi,
is there an efficient (pythonic) way in which I could split a list into
say 5 groups? By split I mean the the first x members would be one group,
the next x members another group and so on 5 times. (Obviously x =
lengthof list/5)

I have done this by a simple for loop and using indexes into the list.
But it does'nt seemm very elegant

Depending on what you're doing, this may or may not be appropriate, but
you might want to look at Numeric (or it's eventual successor numarray).
In this case, you could use reshape to map your list into a n by x
array, which you can treat like a nested list with respect to indexing.

import Numeric
data = range(25)
split = Numeric.reshape (data, (5,5))
split array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]]) split[0] array([0, 1, 2, 3, 4]) list(split[4])

[20, 21, 22, 23, 24]
You can sometimes use Numeric with nonnumeric data, but it tends to be
quircky and often is not worth the trouble, but if you've got numeric
data, try it out.

-tim

[Eddie Corns]

Rajarshi Guha <ra******@presi dency.com> writes:

Hi,
is there an efficient (pythonic) way in which I could split a list into
say 5 groups? By split I mean the the first x members would be one group,
the next x members another group and so on 5 times. (Obviously x =
lengthof list/5)

How about:

------------------------------------------------------------
def span (x, n):
return range(0, x, n)

def group (l, num):
n = (len(l)/num) + 1
return [l[s:s+n] for s in span (len(l), n)]

# test cases
l1 = range(100)
print group (l1, 5)
print group (l1, 6)
print group (l1, 4)
------------------------------------------------------------

Even though span could be folded in to make it shorter, abstracting it out
allows you to name and document what it does (return the start index of each
SPAN of n items). Also it doesn't worry too much about the last list being
the same size as the rest. I'm too lazy to check the boundary conditions
properly (an exercise for the reader). A completely non lazy person would
possibly define the more primitive group_by_n:

def group_by_n (l, n):
return [l[s:s+n] for s in span (len(l), n)]

and define group_into_num_ pieces in terms of that:

def group_into_num_ pieces (l, num):
return group_by_n (l,(len(l)/num) + 1)

giving you 3 useful functions for the price of 1, but I'm too lazy for that.

Eddie

[Terry Reedy]

"Rajarshi Guha" <ra******@presi dency.com> wrote in message
news:pa******** *************** ***********@pre sidency.com...

Hi,
is there an efficient (pythonic) way in which I could split a list into say 5 groups? By split I mean the the first x members would be one group, the next x members another group and so on 5 times. (Obviously x =
lengthof list/5)

I have done this by a simple for loop and using indexes into the list. But it does'nt seemm very elegant

Does it work correctly for all input cases? Is it readable? Is it
acceptibly fast? If so, it is probably 'Pythonic' enough.

TJR

[Paul Rubin]

Rajarshi Guha <ra******@presi dency.com> writes:

is there an efficient (pythonic) way in which I could split a list into
say 5 groups? By split I mean the the first x members would be one group,
the next x members another group and so on 5 times. (Obviously x =
lengthof list/5)

groups = [a[i:i+(len(a)//5)] for i in range(5)]

[Peter Otten]

Paul Rubin wrote:

Rajarshi Guha <ra******@presi dency.com> writes:

is there an efficient (pythonic) way in which I could split a list into
say 5 groups? By split I mean the the first x members would be one group,
the next x members another group and so on 5 times. (Obviously x =
lengthof list/5)

groups = [a[i:i+(len(a)//5)] for i in range(5)]

for k in range(10): .... a = range(k)
.... print [a[i:i+(len(a)//5)] for i in range(5)]
....
[[], [], [], [], []]
[[], [], [], [], []]
[[], [], [], [], []]
[[], [], [], [], []]
[[], [], [], [], []]
[[0], [1], [2], [3], [4]]
[[0], [1], [2], [3], [4]]
[[0], [1], [2], [3], [4]]
[[0], [1], [2], [3], [4]]
[[0], [1], [2], [3], [4]]

Is that what you expected?

Peter

[Peter Otten]

Eddie Corns wrote:

Rajarshi Guha <ra******@presi dency.com> writes:

Hi,
is there an efficient (pythonic) way in which I could split a list into
say 5 groups? By split I mean the the first x members would be one group,
the next x members another group and so on 5 times. (Obviously x =
lengthof list/5)

How about:

------------------------------------------------------------
def span (x, n):
return range(0, x, n)

def group (l, num):
n = (len(l)/num) + 1
return [l[s:s+n] for s in span (len(l), n)]

# test cases
l1 = range(100)
print group (l1, 5)
print group (l1, 6)
print group (l1, 4)
------------------------------------------------------------

More test cases :-)

for k in range(20):
l1 = range(k)
lol = group(l1, 5)
print len(lol), lol

0 []
1 [[0]]
2 [[0], [1]]
3 [[0], [1], [2]]
4 [[0], [1], [2], [3]]
3 [[0, 1], [2, 3], [4]]
3 [[0, 1], [2, 3], [4, 5]]
4 [[0, 1], [2, 3], [4, 5], [6]]
4 [[0, 1], [2, 3], [4, 5], [6, 7]]
5 [[0, 1], [2, 3], [4, 5], [6, 7], [8]]
4 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
4 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]
4 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11]]
5 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12]]
5 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13]]
4 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14]]
4 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]
5 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16]]
5 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16, 17]]
5 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16, 17,
18]]

Peter

[Paul Rubin]

Peter Otten <__*******@web. de> writes:

groups = [a[i:i+(len(a)//5)] for i in range(5)]

for k in range(10):

... a = range(k)
... print [a[i:i+(len(a)//5)] for i in range(5)]
...
[[], [], [], [], []]

.... Is that what you expected?

Bah! Nope, got confused and mis-wrote the loop contents (started
thinking of donig it one way, then another, then got the two mixed up
while typing). Also, wasn't concerned about the case where the list
can't be chopped into equal length parts. Try this:

assert len(a) > 0 and len(a) % 5 == 0
d = len(a) // 5
groups = [a[d*i:d*(i+1)] for i in range(5)]

Handling the case where the pieces end up unequal because len(a) isn't
a multiple of 5 gets a little messy:

d = (len(a) + 4) // 5
groups = [a[d*i:d*(i+1)] for i in range(5)]

when a = range(6) gives

[[0, 1], [2, 3], [4, 5], [], []]

It's not clear to me whether that's good or bad: something like

[[0, 1], [2], [3], [4], [5]]

is probably better, but that would depend on the application.

[Cousin Stanley]

| ....
| More test cases
| ....

Peter ....

I use Outlook Express with the Quote-Fix add on
for reading news groups and got quite a chuckle
when I saw how your test-case code showed up here ....

http://fastq.com/~sckitching/Python/python_lol.png
[ 2 KB ]

--
Cousin Stanley
Human Being
Phoenix, Arizona

[Corey Coughlin]

Rajarshi Guha <ra******@presi dency.com> wrote in message news:<pa******* *************** ************@pr esidency.com>.. .

Hi,
is there an efficient (pythonic) way in which I could split a list into
say 5 groups? By split I mean the the first x members would be one group,
the next x members another group and so on 5 times. (Obviously x =
lengthof list/5)

I have done this by a simple for loop and using indexes into the list.
But it does'nt seemm very elegant

Thanks,

I had the same problem a while back. Here's the function I wrote:

def GroupList(inlis t, step=2):
outlist = []
ents = len(inlist)
if ents % step != 0:
print "In GroupList, the length of list ", inlist, " isn't
evenly"
print "divisible by step %i" % step
sys.exit(4)
maininds = filter(lambda x: x % step == 0, range(ents))
for i in maininds:
currlist = []
for j in range(step):
currlist.append (inlist[i+j])
outlist.append( currlist)
return outlist

As you can see, I made some assumptions (the default n is 2, the list
must be evenly divisible by the step, and so on) and I haven't heavily
debugged it, but it might help you out.