[英]python equivalent of scala partition
我目前正在將一些Scala代碼移植到Python中,我想知道什么是最類似於Scala partition
pythonic方法? 特別是,在Scala代碼中我有一種情況,我根據是否從我傳入的某個過濾謂詞返回true或false來分區項目列表:
val (inGroup,outGroup) = items.partition(filter)
在Python中做這樣的事情的最佳方法是什么?
使用過濾器(需要兩次迭代):
>>> 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
簡單循環:
def partition(item, filter_):
inGroup, outGroup = [], []
for n in items:
if filter_(n):
inGroup.append(n)
else:
outGroup.append(n)
return inGroup, outGroup
例:
>>> items = [1,2,3,4,5]
>>> inGroup, outGroup = partition(items, is_even)
>>> inGroup
[2, 4]
>>> outGroup
[1, 3, 5]
斯卡拉
val (inGroup,outGroup) = items.partition(filter)
Python - 使用列表理解
inGroup = [e for e in items if _filter(e)]
outGroup = [e for e in items if not _filter(e)]
此版本是惰性的,不會將謂詞兩次應用於同一元素:
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)
例:
even = lambda x: x % 2 == 0
e, o = partition([1,1,1,2,2,2,1,2], even)
print list(e)
print list(o)
Scala有豐富的列表處理API,而Python也是如此。
你應該閱讀文檔itertools 。 你可能會發現一個分區。
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)
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