Re: rfind bug ?
Chris Rebert
> With the following code, I would expect a result of 5 !!
>
>>>> a= 'word1 word2 word3'
>>>> a.rfind(' ',7)
> 11
>
> Is this a bug ?
No. Don't you think someone would have found such an obvious bug by now?
You want regular str.find(), which searches from left-to-right:
>>>> a= 'word1 word2 word3'
>>> a.find(' ')
5
str.rfind() is a variant of str.find() that searches from
right-to-left. The "r" is for "right".
Cheers,
Chris
--
http://blog.rebertia.com
James Mills
>
> With the following code, I would expect a result of 5 !!
>
>>>> a= 'word1 word2 word3'
>>>> a.rfind(' ',7)
> 11
>
> Is this a bug ?
Python's documentation states:
| rfind(...)
| S.rfind(sub [,start [,end]]) -> int
|
| Return the highest index in S where substring sub is found,
| such that sub is contained within s[start:end]. Optional
| arguments start and end are interpreted as in slice notation.
|
| Return -1 on failure.
"Return the highest index in S"
I haven't looked at python's source code for the
str object, but perhaps this is exactly what it's
algorithm does!
cheers
James
Peter Otten
[didn't see the original message]
> On Wed, Apr 21, 2010 at 1:51 AM, Stef Mientki <stef.m...@gmail.com>
> wrote:
>> With the following code, I would expect a result of 5 !!
>>
>>>>> a= 'word1 word2 word3'
>>>>> a.rfind(' ',7)
>> 11
>>
>> Is this a bug ?
>
> No. Don't you think someone would have found such an obvious bug by now?
Indeed.
OP: you may be looking for
>>> a = "a bb ccc"
>>> a[::-1].find(" ")
3
Peter
Paul Rudin
> OP: you may be looking for
>
>>>> a = "a bb ccc"
>>>> a[::-1].find(" ")
> 3
But you should be aware of the effeciency implications of doing
this. a[::-1] constructs a new list. It's probably faster to do e.g.:
len(a) - a.rfind(..) - 1
Peter Otten
> Peter Otten <__pet...@web.de> writes:
>
>
>> OP: you may be looking for
>>
>>>>> a = "a bb ccc"
>>>>> a[::-1].find(" ")
>> 3
>
>
> But you should be aware of the effeciency implications of doing
> this. a[::-1] constructs a new list.
A new string, yes.
> It's probably faster to do e.g.:
> len(a) - a.rfind(..) - 1
Yes, especially if the string is "long":
$ python -m timeit -s'a = "word1 word2 word3"' 'a[::-1].rfind(" ")'
1000000 loops, best of 3: 0.834 usec per loop
$ python -m timeit -s'a = "word1 word2 word3"*100' 'a[::-1].rfind(" ")'
100000 loops, best of 3: 5.04 usec per loop
$ python -m timeit -s'a = "word1 word2 word3"' 'len(a)-a.rfind(" ")-1'
1000000 loops, best of 3: 0.587 usec per loop
$ python -m timeit -s'a = "word1 word2 word3"*100' 'len(a)-a.rfind(" ")-1'
1000000 loops, best of 3: 0.592 usec per loop
But be aware of the following difference:
>>> a[::-1].rfind("not there")
-1
>>> len(a) - a.rfind("not there") -1
8
Peter
Stef Mientki
> On Wed, Apr 21, 2010 at 1:51 AM, Stef Mientki <stef.m...@gmail.com> wrote:
>
>> With the following code, I would expect a result of 5 !!
>>
>>
>>>>> a= 'word1 word2 word3'
>>>>> a.rfind(' ',7)
>>>>>
>> 11
>>
>> Is this a bug ?
>>
> No. Don't you think someone would have found such an obvious bug by now?
>
then the start index has no meaning ...
... and some would call that a bug.
cheers,
Stef
Jean-Michel Pichavant
Out[12]: -1
looks like the start index has a meaning ...
JM
Chris Rebert
> On 21-04-2010 10:56, Chris Rebert wrote:
>> On Wed, Apr 21, 2010 at 1:51 AM, Stef Mientki <stef.m...@gmail.com> wrote:
>>> With the following code, I would expect a result of 5 !!
>>>
>>>>>> a= 'word1 word2 word3'
>>>>>> a.rfind(' ',7)
>>>>>>
>>> 11
>>>
>>> Is this a bug ?
>>>
>> No. Don't you think someone would have found such an obvious bug by now?
>>
> if it's not a bug,
> then the start index has no meaning ...
> ... and some would call that a bug.
Ah, I neglected to take your use of .rfind()'s second parameter into account!
As can be interpolated from the part of the docs James quotes:
s.rfind(' ', 7) === s[7:].rfind(' ') + 7 # overlooking the 'not present' case
That is, the second parameter to .rfind(), namely `start`, is relative
to the *left* end of the string, not the right end. I can see how this
might be unintuitive, but it does make the API more uniform.
Cheers,
Chris
Alf P. Steinbach
It seems that the OP also thought it was relative to the left end of the string.
The difference is what it signifies: start of search, or end of search.
With rfind the "start" parameter signifies the end of the search, and
conversely, the third parameter "end" signifies where the search starts. :-)
Cheers,
- Alf
Stef Mientki
that's indeed what I was missing.
cheers,
Stef
>
> Cheers,
>
> - Alf
Chris Rebert
It really doesn't help that the example in question is kinda lousy
(particularly, it's symmetrical with respect to whitespace and has
only 2 spaces).
Case in point, my misinterpretation of the OP's misinterpretation
(i.e. having the indexing of `start` be relative to the end of the
string) still plausibly explains his confusion.
Cheers,
Chris
--
Excellent catch.
*Considers getting Alien Life-Form merch*