我知道如何在单独的行上使用for循环和if语句，例如:
>>> a = [2,3,4,5,6,7,8,9,0]
... xyz = [0,12,4,6,242,7,9]
... for x in xyz:
... if x in a:
... print(x)
0,4,6,7,9
我知道我可以使用一个列表理解来组合这些语句时很简单，例如:
print([x for x in xyz if x in a])
但是我在任何地方都找不到一个很好的例子(可以复制和学习)来演示一组复杂的命令(不仅仅是“print x”)，这些命令发生在for循环和一些if语句的组合之后。我期待的是这样的:
for x in xyz if x not in a:
print(x...)
这不是python应该工作的方式吗?
当前回答
使用intersection或intersection_update
十字路口: A = [2,3,4,5,6,7,8,9,0] Xyz = [0,12,4,6,242,7,9] Ans = sorted(set(a).intersection(set(xyz))) intersection_update: A = [2,3,4,5,6,7,8,9,0] Xyz = [0,12,4,6,242,7,9] B = set(a) b.intersection_update(某某) b是答案
其他回答
我个人认为这是最漂亮的版本:
a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]
for x in filter(lambda w: w in a, xyz):
print x
Edit
如果你非常热衷于避免使用lambda，你可以使用部分函数应用程序和使用operator模块(它提供了大多数操作符的函数)。
https://docs.python.org/2/library/operator.html#module-operator
from operator import contains
from functools import partial
print(list(filter(partial(contains, a), xyz)))
如果生成器表达式太复杂，你也可以使用生成器:
def gen():
for x in xyz:
if x in a:
yield x
for x in gen():
print x
根据The Zen of Python(如果你想知道你的代码是否“Pythonic”，可以去那里):
美总比丑好。 显性比隐性好。 简单比复杂好。 扁平比嵌套好。 可读性。
python获取两个集合的排序交集的方法是:
>>> sorted(set(a).intersection(xyz))
[0, 4, 6, 7, 9]
或者那些xyz但不在a中的元素:
>>> sorted(set(xyz).difference(a))
[12, 242]
但对于更复杂的循环，您可能希望通过遍历一个名称良好的生成器表达式和/或调用一个名称良好的函数来将其压平。试图将所有内容都放在一行上很少是“python式的”。
更新以下关于您的问题和接受的答案的其他评论
我不确定你想用枚举做什么，但如果a是字典，你可能想使用键，像这样:
>>> a = {
... 2: 'Turtle Doves',
... 3: 'French Hens',
... 4: 'Colly Birds',
... 5: 'Gold Rings',
... 6: 'Geese-a-Laying',
... 7: 'Swans-a-Swimming',
... 8: 'Maids-a-Milking',
... 9: 'Ladies Dancing',
... 0: 'Camel Books',
... }
>>>
>>> xyz = [0, 12, 4, 6, 242, 7, 9]
>>>
>>> known_things = sorted(set(a.iterkeys()).intersection(xyz))
>>> unknown_things = sorted(set(xyz).difference(a.iterkeys()))
>>>
>>> for thing in known_things:
... print 'I know about', a[thing]
...
I know about Camel Books
I know about Colly Birds
I know about Geese-a-Laying
I know about Swans-a-Swimming
I know about Ladies Dancing
>>> print '...but...'
...but...
>>>
>>> for thing in unknown_things:
... print "I don't know what happened on the {0}th day of Christmas".format(thing)
...
I don't know what happened on the 12th day of Christmas
I don't know what happened on the 242th day of Christmas
以下是对公认答案的简化/一行:
a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]
for x in (x for x in xyz if x not in a):
print(x)
12
242
注意，生成器是内联的。这是在python2.7和python3.6上测试的(注意print;))
即便如此，它还是很麻烦:x被提到了四次。
我喜欢Alex的回答，因为过滤器正是应用于列表的if，所以如果您想在给定条件下探索列表的子集，这似乎是最自然的方法
mylist = [1,2,3,4,5]
another_list = [2,3,4]
wanted = lambda x:x in another_list
for x in filter(wanted, mylist):
print(x)
这个方法对于分离关注点很有用，如果条件函数改变了，唯一需要修改的代码就是函数本身
mylist = [1,2,3,4,5]
wanted = lambda x:(x**0.5) > 10**0.3
for x in filter(wanted, mylist):
print(x)
当您不想要列表中的成员，而是修改这些成员时，生成器方法似乎更好，这似乎更适合生成器
mylist = [1,2,3,4,5]
wanted = lambda x:(x**0.5) > 10**0.3
generator = (x**0.5 for x in mylist if wanted(x))
for x in generator:
print(x)
此外，过滤器与生成器一起工作，尽管在这种情况下效率不高
mylist = [1,2,3,4,5]
wanted = lambda x:(x**0.5) > 10**0.3
generator = (x**0.9 for x in mylist)
for x in filter(wanted, generator):
print(x)
当然，这样写还是不错的:
mylist = [1,2,3,4,5]
wanted = lambda x:(x**0.5) > 10**0.3
# for x in filter(wanted, mylist):
for x in mylist if wanted(x):
print(x)
