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".
--
boB
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