Dick Moores wrote:
> I have a bunch of functions I've collected in one script, "mycalc.py",
> which I use as a module, "mycalc". The last one I wrote, cmpSeq() is as
> follows:
>
[code]
>
> In a script, cmpSeq() works fine. For example followed by
>
[example]
>
> The output is:
>
[output]
>
> cmpSeq() is now copy-pasted into mycalc.py, but is not useable there:
>
[example]
>
> which produces:
>
[Traceback]
>
> However, other functions in mycalc work fine. For example:
>
[example]
>
> That may have been a bit long-winded; if so, I apologize.
Not at all! :)
I couldn't reproduce the error. The problem is not in the given code then.
Use PyChecker or PyLint to aid you find the problem.
Other comments:
Instead of:
for index in range(len(shorterOrEqualSequence)):
if seq1[index] != seq2[index]:
prints...
break
if index == len(shorterOrEqualSequence)-1:
print "sequences are identical thru end of shorter sequence at index",
index
this could be used:
for index in range(len(shorterOrEqualSequence)):
if seq1[index] != seq2[index]:
prints...
break
else:
print "sequences are identical thru end of shorter sequence at index",
index
And instead of:
if len(seq1) >= len(seq2):
shorterOrEqualSequence = seq2
else:
shorterOrEqualSequence = seq1
for index in range(len(shorterOrEqualSequence)):
etc.
this could be used:
for index in xrange(min(len(seq1), len(seq2))):
etc.
The main loop, I would write it like this:
from itertools import izip, count
def cmpSeq(seq1, seq2):
"""
find first index at which two sequences differ
"""
if seq1 == seq2:
print "Sequences are identical, and of length %d" % len(seq1)
return None
for i, ca, cb in izip(count(), seq1, seq2):
if ca != cb:
print "sequences first differ at index", index
print "seq1[%d] = %s" % (index, seq1[index])
print "seq2[%d] = %s" % (index, seq2[index])
break
else:
print "sequences are identical thru end of shorter sequence at
index", i
Javier
_______________________________________________
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
