python equivalent of scala partition
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Chapters
00:00 Python Equivalent Of Scala Partition
00:28 Answer 1 Score 2
00:42 Accepted Answer Score 4
01:06 Answer 3 Score 1
01:36 Answer 4 Score 2
01:57 Thank you
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Full question
https://stackoverflow.com/questions/1869...
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Tags
#python #scala
#avk47
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Chapters
00:00 Python Equivalent Of Scala Partition
00:28 Answer 1 Score 2
00:42 Accepted Answer Score 4
01:06 Answer 3 Score 1
01:36 Answer 4 Score 2
01:57 Thank you
--
Full question
https://stackoverflow.com/questions/1869...
--
Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
--
Tags
#python #scala
#avk47
ACCEPTED ANSWER
Score 4
Using filter (require two iteration):
>>> items = [1,2,3,4,5]
>>> inGroup = filter(is_even, items) # list(filter(is_even, items)) in Python 3.x
>>> outGroup = filter(lambda n: not is_even(n), items)
>>> inGroup
[2, 4]
>>> outGroup
Simple loop:
def partition(item, filter_):
inGroup, outGroup = [], []
for n in items:
if filter_(n):
inGroup.append(n)
else:
outGroup.append(n)
return inGroup, outGroup
Example:
>>> items = [1,2,3,4,5]
>>> inGroup, outGroup = partition(items, is_even)
>>> inGroup
[2, 4]
>>> outGroup
[1, 3, 5]
ANSWER 2
Score 2
Scala
val (inGroup,outGroup) = items.partition(filter)
Python - Using List Comprehension
inGroup = [e for e in items if _filter(e)]
outGroup = [e for e in items if not _filter(e)]
ANSWER 3
Score 2
This version is lazy and doesn't apply the predicate twice to the same element:
def partition(it, pred):
buf = [[], []]
it = iter(it)
def get(t):
while True:
while buf[t]:
yield buf[t].pop(0)
x = next(it)
if t == bool(pred(x)):
yield x
else:
buf[not t].append(x)
return get(True), get(False)
Example:
even = lambda x: x % 2 == 0
e, o = partition([1,1,1,2,2,2,1,2], even)
print list(e)
print list(o)
ANSWER 4
Score 1
Scala has rich list processing api, and Python is much as that.
You should read document itertools. And you may find a receipe of partition.
from itertools import ifilterfalse, ifilter, islice, tee, count
def partition(pred, iterable):
'''
>>> is_even = lambda i: i % 2 == 0
>>> even, no_even = partition(is_even, xrange(11))
>>> list(even)
[0, 2, 4, 6, 8, 10]
>>> list(no_even)
[1, 3, 5, 7, 9]
# Lazy evaluation
>>> infi_list = count(0)
>>> ingroup, outgroup = partition(is_even, infi_list)
>>> list(islice(ingroup, 5))
[0, 2, 4, 6, 8]
>>> list(islice(outgroup, 5))
[1, 3, 5, 7, 9]
'''
t1, t2 = tee(iterable)
return ifilter(pred, t1), ifilterfalse(pred, t2)