I want to be able to take a sequence like:
my_sequence = ['foo', 'bar', 'baz', 'spam', 'eggs', 'cheese', 'yogurt']
Use a function like:
my_paginated_sequence = get_rows(my_sequence, 3)
To get:
[['foo', 'bar', 'baz'], ['spam', 'eggs', 'cheese'], ['yogurt']]
This is what I came up with by just thinking through it:
def get_rows(sequence, num):
count = 1
rows = list()
cols = list()
for item in sequence:
if count == num:
cols.append(item)
rows.append(cols)
cols = list()
count = 1
else:
cols.append(item)
count += 1
if count > 0:
rows.append(cols)
return rows
If you know you have a sliceable sequence (list or tuple),
def getrows_byslice(seq, rowlen):
for start in xrange(0, len(seq), rowlen):
yield seq[start:start+rowlen]
This of course is a generator, so if you absolutely need a list as the result, you'll use list(getrows_byslice(seq, 3))
or the like, of course.
If what you start with is a generic iterable, the itertools recipes offer help with the grouper
recipe...:
import itertools
def grouper(n, iterable, fillvalue=None):
"grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return itertools.izip_longest(fillvalue=fillvalue, *args)
(again, you'll need to call list
on this if a list is what you want, of course).
Since you actually want the last tuple to be truncated rather than filled up, you'll need to "trim" the trailing fill-values from the very last tuple.