In one piece of my program I doubt if i use the comparison correctly. i want to make sure that ( u0 <= u < u0+step ) before do something.
if not (u0 <= u) and (u < u0+step):
u0 = u0+ step # change the condition until it is satisfied
else:
do something. # condition is satisfied
You can do:
if not (u0 <= u <= u0+step):
u0 = u0+ step # change the condition until it is satisfied
else:
do sth. # condition is satisfied
Using a loop:
while not (u0 <= u <= u0+step):
u0 = u0+ step # change the condition until it is satisfied
do sth. # condition is satisfied
Operator precedence in python
You can see that not X has higher precedence than and. Which means that the not only apply to the first part (u0 <= u).
Write:
if not (u0 <= u and u < u0+step):
or even
if not (u0 <= u < u0+step):
In this particular case the clearest solution is the S.Lott answer
But in some complex logical conditions I would prefer use some boolean algebra to get a clear solution.
Using De Morgan's law ¬(A^B) = ¬Av¬B
not (u0 <= u and u < u0+step)
(not u0 <= u) or (not u < u0+step)
u0 > u or u >= u0+step
then
if u0 > u or u >= u0+step:
pass
... in this case the «clear» solution is not more clear :P
Why think? If not confuses you, switch your if and else clauses around to avoid the negation.
i want to make sure that ( u0 <= u < u0+step ) before do sth.
Just write that.
if u0 <= u < u0+step:
"do sth" # What language is "sth"? No vowels. An odd-looking word.
else:
u0 = u0+ step
Why overthink it?
If you need an empty if -- and can't work out the logic -- use pass.
if some-condition-that's-too-complex-for-me-to-invert:
pass
else:
do real work here
There are two ways. In case of doubt, you can always just try it. If it does not work, you can add extra braces to make sure, like that:
if not ((u0 <= u) and (u < u0+step)):
