Prev: question about an exciting gotcha for unittests (and elsewhere)...
Next: shove does not store data as expected
From: Chris Rebert on 21 Apr 2010 04:56
On Wed, Apr 21, 2010 at 1:51 AM, Stef Mientki <stef.mientki(a)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?

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
From: James Mills on 21 Apr 2010 05:06
On Wed, Apr 21, 2010 at 6:51 PM, Stef Mientki <stef.mientki(a)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 ?

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
From: Peter Otten on 21 Apr 2010 05:15
Chris Rebert wrote:

[didn't see the original message]

> On Wed, Apr 21, 2010 at 1:51 AM, Stef Mientki <stef.mientki(a)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
From: Paul Rudin on 21 Apr 2010 05:42
Peter Otten <__peter__(a)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. It's probably faster to do e.g.:
len(a) - a.rfind(..) - 1
From: Peter Otten on 21 Apr 2010 05:54
Paul Rudin wrote:

> Peter Otten <__peter__(a)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
 |  Next  |  Last
Pages: 1 2 3
Prev: question about an exciting gotcha for unittests (and elsewhere)...
Next: shove does not store data as expected