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
a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]
set(a) & set(xyz)
set([0, 9, 4, 6, 7])
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)
你可以像这样使用生成器表达式:
gen = (x for x in xyz if x not in a)
for x in gen:
print(x)
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)
