我有这样的代码:
good = [x for x in mylist if x in goodvals]
bad = [x for x in mylist if x not in goodvals]
目标是根据mylist的内容是否满足条件,将它们拆分为另外两个列表。
我怎样才能做得更优雅呢?我能避免在mylist上做两个单独的迭代吗?我可以通过这样做来提高性能吗?
我有这样的代码:
good = [x for x in mylist if x in goodvals]
bad = [x for x in mylist if x not in goodvals]
目标是根据mylist的内容是否满足条件,将它们拆分为另外两个列表。
我怎样才能做得更优雅呢?我能避免在mylist上做两个单独的迭代吗?我可以通过这样做来提高性能吗?
当前回答
受到@gnibbler伟大(但简洁!)回答的启发,我们可以应用该方法映射到多个分区:
from collections import defaultdict
def splitter(l, mapper):
"""Split an iterable into multiple partitions generated by a callable mapper."""
results = defaultdict(list)
for x in l:
results[mapper(x)] += [x]
return results
然后可以使用splitter,如下所示:
>>> l = [1, 2, 3, 4, 2, 3, 4, 5, 6, 4, 3, 2, 3]
>>> split = splitter(l, lambda x: x % 2 == 0) # partition l into odds and evens
>>> split.items()
>>> [(False, [1, 3, 3, 5, 3, 3]), (True, [2, 4, 2, 4, 6, 4, 2])]
这适用于有更复杂映射的两个以上分区(也适用于迭代器):
>>> import math
>>> l = xrange(1, 23)
>>> split = splitter(l, lambda x: int(math.log10(x) * 5))
>>> split.items()
[(0, [1]),
(1, [2]),
(2, [3]),
(3, [4, 5, 6]),
(4, [7, 8, 9]),
(5, [10, 11, 12, 13, 14, 15]),
(6, [16, 17, 18, 19, 20, 21, 22])]
或者用字典来映射:
>>> map = {'A': 1, 'X': 2, 'B': 3, 'Y': 1, 'C': 2, 'Z': 3}
>>> l = ['A', 'B', 'C', 'C', 'X', 'Y', 'Z', 'A', 'Z']
>>> split = splitter(l, map.get)
>>> split.items()
(1, ['A', 'Y', 'A']), (2, ['C', 'C', 'X']), (3, ['B', 'Z', 'Z'])]
其他回答
你可以在Python中进行惰性函数编程,像这样:
partition = lambda l, c: map(
lambda iii: (i for ii in iii for i in ii),
zip(*(([], [e]) if c(e) else ([e], []) for e in l)))
函数式编程很优雅,但在Python中不是这样。如果你知道你的列表中没有None值,也可以参考这个例子:
partition = lambda l, c: map(
filter(lambda x: x is not None, l),
zip(*((None, e) if c(e) else (e, None) for e in l)))
我认为基于N个条件来划分一个可迭代对象是很方便的
from collections import OrderedDict
def partition(iterable,*conditions):
'''Returns a list with the elements that satisfy each of condition.
Conditions are assumed to be exclusive'''
d= OrderedDict((i,list())for i in range(len(conditions)))
for e in iterable:
for i,condition in enumerate(conditions):
if condition(e):
d[i].append(e)
break
return d.values()
例如:
ints,floats,other = partition([2, 3.14, 1, 1.69, [], None],
lambda x: isinstance(x, int),
lambda x: isinstance(x, float),
lambda x: True)
print " ints: {}\n floats:{}\n other:{}".format(ints,floats,other)
ints: [2, 1]
floats:[3.14, 1.69]
other:[[], None]
如果元素可以满足多个条件,则删除断点。
Good = [x for x in mylist if x in goodvals] Bad = [x for x in mylist if x not in goodvals] 我怎样才能做得更优雅呢?
代码已经非常优雅了。
使用集合可能会有轻微的性能改进,但差异是微不足道的。基于集合的方法也会丢弃重复项,并且不会保留元素的顺序。我发现列表理解也更容易阅读。
事实上,我们甚至可以更简单地使用for循环:
good, bad = [], []
for x in mylist:
if x in goodvals:
good.append(f)
else:
bad.append(f)
这种方法可以更容易地添加额外的逻辑。例如,代码很容易被修改为丢弃None值:
good, bad = [], []
for x in mylist:
if x is None:
continue
if x in goodvals:
good.append(f)
else:
bad.append(f)
还有另一个答案,简短但“邪恶”(用于理解列表的副作用)。
digits = list(range(10))
odd = [x.pop(i) for i, x in enumerate(digits) if x % 2]
>>> odd
[1, 3, 5, 7, 9]
>>> digits
[0, 2, 4, 6, 8]
下面是惰性迭代器方法:
from itertools import tee
def split_on_condition(seq, condition):
l1, l2 = tee((condition(item), item) for item in seq)
return (i for p, i in l1 if p), (i for p, i in l2 if not p)
它对每个项计算一次条件,并返回两个生成器,第一个生成条件为真时序列中的值,另一个生成条件为假时序列中的值。
因为它是惰性的,你可以在任何迭代器上使用它,甚至是无限迭代器:
from itertools import count, islice
def is_prime(n):
return n > 1 and all(n % i for i in xrange(2, n))
primes, not_primes = split_on_condition(count(), is_prime)
print("First 10 primes", list(islice(primes, 10)))
print("First 10 non-primes", list(islice(not_primes, 10)))
通常情况下,非惰性列表返回方法会更好:
def split_on_condition(seq, condition):
a, b = [], []
for item in seq:
(a if condition(item) else b).append(item)
return a, b
编辑:对于您更具体的用例,将项目按某些键分割到不同的列表中,这里有一个通用函数:
DROP_VALUE = lambda _:_
def split_by_key(seq, resultmapping, keyfunc, default=DROP_VALUE):
"""Split a sequence into lists based on a key function.
seq - input sequence
resultmapping - a dictionary that maps from target lists to keys that go to that list
keyfunc - function to calculate the key of an input value
default - the target where items that don't have a corresponding key go, by default they are dropped
"""
result_lists = dict((key, []) for key in resultmapping)
appenders = dict((key, result_lists[target].append) for target, keys in resultmapping.items() for key in keys)
if default is not DROP_VALUE:
result_lists.setdefault(default, [])
default_action = result_lists[default].append
else:
default_action = DROP_VALUE
for item in seq:
appenders.get(keyfunc(item), default_action)(item)
return result_lists
用法:
def file_extension(f):
return f[2].lower()
split_files = split_by_key(files, {'images': IMAGE_TYPES}, keyfunc=file_extension, default='anims')
print split_files['images']
print split_files['anims']