Is there any evaluation of the expression in slim syntax for a list / tuple inside Python?

Question

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

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

Is there any equivalent syntax in python regular data structures? I expected to get an error when I tried:

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

but instead of excluding some type, I get a return value of 2, which does not make much sense.

ps I could not find documentation for this syntax at all, so if someone could point me to numpy for it 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 these two types, so it compares class names in alphabetical order ( "list" > "int" ). Since True same as 1 , you get the item at index 1.

In Python 3.x, this expression will give you an error (a much less unexpected result).

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

To do what you want, you must use list comprehension:

 [x for x in lis if x > 2] 

Answer 2

Use list:

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