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
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。
其他回答
所有提出的解决方案的问题是,它将扫描和应用过滤功能两次。我会做一个简单的小函数,像这样:
def split_into_two_lists(lst, f):
a = []
b = []
for elem in lst:
if f(elem):
a.append(elem)
else:
b.append(elem)
return a, b
这样你就不会重复处理任何东西,也不会重复代码。
good.append(x) if x in goodvals else bad.append(x)
来自@dansalmo的这个优雅简洁的回答被埋没在评论中,所以我只是把它作为一个答案转发到这里,这样它就能得到应有的重视,尤其是对新读者来说。
完整的例子:
good, bad = [], []
for x in my_list:
good.append(x) if x in goodvals else bad.append(x)
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。
如果你坚持聪明,你可以采用温登的解决方案,再加上一点虚假的聪明:
def splay(l, f, d=None):
d = d or {}
for x in l: d.setdefault(f(x), []).append(x)
return d
解决方案
from itertools import tee
def unpack_args(fn):
return lambda t: fn(*t)
def separate(fn, lx):
return map(
unpack_args(
lambda i, ly: filter(
lambda el: bool(i) == fn(el),
ly)),
enumerate(tee(lx, 2)))
test
[even, odd] = separate(
lambda x: bool(x % 2),
[1, 2, 3, 4, 5])
print(list(even) == [2, 4])
print(list(odd) == [1, 3, 5])