On Tue, Mar 31, 2015 at 3:37 PM, boB Stepp <robertvst...@gmail.com> wrote:
> On Tue, Mar 31, 2015 at 3:28 PM, Zachary Ware
> <zachary.ware+py...@gmail.com> wrote:
>> On Tue, Mar 31, 2015 at 3:23 PM, boB Stepp <robertvst...@gmail.com> wrote:
>>> The following behavior has me stumped:
>>>
>>> Python 2.7.8 (default, Jun 30 2014, 16:03:49) [MSC v.1500 32 bit
>>> (Intel)] on win32
>>> Type "copyright", "credits" or "license()" for more information.
>>>>>> L = ['#ROI:roi_0', '#TXT:text_0', '#1:one^two^three']
>>>>>> for i, item in enumerate(L):
>>> subitems = item.split(':')
>>> if subitems[0] == '#ROI':
>>> print subitems[1]
>>> if subitems[0] == '#TXT':
>>> print subitems[1]
>>> if subitems[0] == '#1' or '#2':
>>
>> Here's your problem: "#2" is always true. Try "if subitems[0] in
>> ['#1', '#2']:"
>
> Thanks, Zach! About the time your reply arrived I was starting to
> realize that my '#1' or '#2' might not be doing what I thought. In the
> "Python Pocket Reference" I was just looking at:
>
> X or Y If X is false then Y; else X.
>
> I forgot that different languages have different interpretations of "or".
In this case, the differing languages being Python and English :).
Also, not that since you aren't using the index for anything, you
don't need to use enumerate() to iterate over the list. Just do "for
item in L:". Of course, if you actually use the index in the real
code that I assume this was cut out of, keep enumerate; it's the right
tool for the job.
--
Zach
_______________________________________________
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
https://mail.python.org/mailman/listinfo/tutor
