Cleaning up conditionals
Erik
python at lucidity.plus.com
Fri Dec 30 18:22:02 EST 2016
On 30/12/16 23:00, Deborah Swanson wrote:
> Oops, indentation was messed up when I copied it into the email. Should
> be this:
>
> if len(l1[st]) == 0:
> if len(l2[st]) > 0:
> l1[st] = l2[st]
> elif len(l2[st]) == 0:
> if len(l1[st]) > 0:
> l2[st] = l1[st]
That's even worse!
Anyway, ignoring all that, if what you are trying to do is just do some
action based on a set of fixed comparisons that are known at the top of
the function (and do not mutate depending on the path through the
function), then you can just cache the comparisons and then compare the
resulting set of boolean results (note that my examples are NOT based on
your use-case, it's just pseudo-code which happens to use your expressions):
state = (len(l1[st]) == 0, len(l2[st]) > 0)
if state == (True, False):
pass
elif state == (False, True):
pass
... etc.
If the len() comparisons are tri-state (i.e., in some cases you want to
know if the length is <, ==, or > 0, depending on one of the other
comparisons) then you can do something like:
def clamp(foo):
return min(max(-1, foo), 1)
state = (clamp(cmp(len(l1[st]), 0), cmp(len(l2[st]), 0))
if state == (0, -1):
pass
elif state == (1, -1):
pass
... etc.
I'm not sure this makes it much more readable though - but if you make
the RHS of those comparisons a symbolic name you might be getting
somewhere -
ACTION1 = (0, -1)
ACTION2 = (1, -1)
if state == ACTION1:
pass
elif state == ACTION2:
pass
Hope that helps, E.
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