结合for-loop和if-statement的python方式

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

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)

其他回答

我可能会用:

for x in xyz: 
    if x not in a:
        print(x...)

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)

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

找到列表A和b的唯一公共元素的简单方法:

a = [1,2,3]
b = [3,6,2]
for both in set(a) & set(b):
    print(both)

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。

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