Is there a cleaner and a shorter way to write this conditional in python?

Question

>>> def accept(d1, d2):
    if somefunc(d1,d2) > 32:
        h = 1
    else:
        h = 0
    return h

Does Python have a ternary conditional operator? doesn't give a solution for a case one want to return a value. A lambda based solution is preferable.

score 5 · Answer 1 · answered Oct 06 '13 at 21:03

The "return-value scenario" is no different than any other:

return 1 if somefunc(d1, d2) > 32 else 0

If for some reason you want a lambda:

lambda d1, d2: 1 if somefunc(d1, d2) > 32 else 0

Note that a lambda is no different than a function defined with def that returns the same thing. Lambdas are just regular functions.

score 3 · Accepted Answer · answered Oct 06 '13 at 21:06

3

Or, perhaps trickier,

return int(somefunc(d1, d2) > 32)

Note that int(True) == 1 and int(False) == 0.

answered Oct 06 '13 at 21:06

Tim Peters

Actually you could just return `somefunc(d1, d2) > 32` directly, since True and False already are `int`s. – BrenBarn Oct 06 '13 at 21:13
1

But that wouldn't return **exactly** what their original code returned. For all we know, someone on the receiving end is doing a `type()` check on the results ;-) – Tim Peters Oct 06 '13 at 21:14

score -1 · Answer 3 · answered Oct 06 '13 at 21:05

-1

Turn into a lambda (not a explicit function):

accept = lambda d1,d2: 1 if somefunc(d1, d2) > 32 else 0

answered Oct 06 '13 at 21:05

Nacib Neme

3 Answers3