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"""Utilities for working with lists and sequences."""
def flatten(x):
"""
Returns a single, flat list which contains all elements retrieved
from the sequence and all recursively contained sub-sequences
(iterables).
Examples:
>>> [1, 2, [3, 4], (5, 6)]
[1, 2, [3, 4], (5, 6)]
From http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks
"""
result = []
for el in x:
if hasattr(el, '__iter__') and not isinstance(el, basestring):
result.extend(flatten(el))
else:
result.append(el)
return result
def batch_size(items, size):
"""
Retrieves items in batches of the given size.
>>> l = range(1, 11)
>>> batch_size(l, 3)
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
>>> batch_size(l, 5)
[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
"""
return [items[i:i+size] for i in xrange(0, len(items), size)]
def batches(items, number):
"""
Retrieves items in the given number of batches.
>>> l = range(1, 11)
>>> batches(l, 1)
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]]
>>> batches(l, 2)
[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
>>> batches(l, 3)
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
>>> batches(l, 4)
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
>>> batches(l, 5)
[[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]
Initial batches will contain as many items as possible in cases where
there are not enough items to be distributed evenly.
>>> batches(l, 6)
[[1, 2], [3, 4], [5, 6], [7, 8], [9], [10]]
>>> batches(l, 7)
[[1, 2], [3, 4], [5, 6], [7], [8], [9], [10]]
>>> batches(l, 8)
[[1, 2], [3, 4], [5], [6], [7], [8], [9], [10]]
>>> batches(l, 9)
[[1, 2], [3], [4], [5], [6], [7], [8], [9], [10]]
>>> batches(l, 10)
[[1], [2], [3], [4], [5], [6], [7], [8], [9], [10]]
If there are more batches than items, empty batches will be appended
to the batch list.
>>> batches(l, 11)
[[1], [2], [3], [4], [5], [6], [7], [8], [9], [10], []]
>>> batches(l, 12)
[[1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [], []]
"""
div, mod= divmod(len(items), number)
if div > 1:
if mod:
div += 1
return batch_size(items, div)
else:
if not div:
return [[item] for item in items] + [[]] * (number - mod)
elif div == 1 and not mod:
return [[item] for item in items]
else:
# mod now tells you how many lists of 2 you can fit in
return ([items[i*2:(i*2)+2] for i in xrange(0, mod)] +
[[item] for item in items[mod*2:]])
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