Python中的list / tuple切片语法中是否有某种表达式求值？

Question

with numpy arrays, you can use some kind of inequality within the square bracket slicing syntax:

>>>arr = numpy.array([1,2,3])
>>>arr[arr>=2]
array([2, 3])

is there some kind of equivalent syntax within regular python data structures? I expected to get an error when I tried:

>>>lis = [1,2,3]
>>>lis[lis > 2]
2

but instead of an exception of some type, I get a returned value of 2, which doesn't make a lot of sense.

ps I couldn't find the documentation for this syntax at all, so if someone could point me to it for numpy and for regular python(if it exists) that would be great.

Answer 1

In Python 2.x lis > 2 returns True . This is because the operands have different types and there is no comparison operator defined for those two types, so it compares the class names in alphabetical order ( "list" > "int" ). Since True is the same as 1 , you get the item at index 1.

In Python 3.x this expression would give you an error (a much less surprising result).

TypeError: unorderable types: list() > int()

To do what you want you should use a list comprehension:

[x for x in lis if x > 2]

Answer 2

使用列表理解：

[a for a in lis if a>=2]