I'm trying to figure out if there's a shorter way of writing this if statement (wrote a very basic, unrealistic if statement just so it's easy to see the point.
x = 3 if x == 1 or x == 2 or x == 3 or x==4: print x else: print "nope" I would like to write something along the lines of:
if x == or(1,2,3,4): is there any way of doing that? Or do I have to write out every option?
Thanks, Sean
you want:
if x in (1, 2, 3, 4): 
if x in {1, 2, 3, 4}: (using a set) is better. But for just four of them, it won't make a difference.
if statement.set((1,2,3,4)) isn't that terrible, certainly not the most terrible thing you'll face in Python 2.3…[1, 2, 3, 4].indexOf(x) + 1 to avoid the ugly != -1 although that is arguably even worse. :-)Use the in operator:
if x in range(1, 5) On Python2 you can use xrange.
On Python 3.2+ it is recommended to use set literals:
if x in {1, 2, 3, 4}:
Python’s peephole optimizer now recognizes patterns such x
in {1, 2, 3}as being a test for membership in a set of constants. The optimizer recasts thesetas afrozensetand stores the pre-built constant. Now that the speed penalty is gone, it is practical to start writing membership tests using set-notation. This style is both semantically clear and operationally fast.
[1, 2 ,3, 4].
xrange instead.in on an xrange is constant time in 2.x.%timeit 10000 in xrange(20000), %timeit 100000 in xrange(200000), %timeit 1000000 in xrange(2000000) and it's clearly linear.step and integer values, you can just write 1 <= x < 5…if x>=1 or x<=4: will do the job
if 1 <= x <= 4, which also implies and (as DSM points out).
x==10? (IOW, I think you mean and.) But this would also let x==1.5 pass too.
if 0 < x < 5 for that type of comparison.