why "g".count('')==2 ?

[ygao]

my question is as title!
thanks!

[Fredrik Lundh]

Re: why "g".c ount('')==2 ?

"ygao" wrote:
[color=blue]
> my question is as title![/color]

my answer as code:
[color=blue][color=green][color=darkred]
>>> s = "g"
>>> t = ""
>>> s[0:0+len(t)] == t[/color][/color][/color]
True[color=blue][color=green][color=darkred]
>>> s[1:1+len(t)] == t[/color][/color][/color]
True

</F>

[Steven D'Aprano]

Re: why &quot;g&quot;.c ount('')==2 ?

On Sat, 11 Mar 2006 13:37:05 +0100, Fredrik Lundh wrote:
[color=blue]
> "ygao" wrote:
>[color=green]
>> my question is as title![/color]
>
> my answer as code:
>[color=green][color=darkred]
>>>> s = "g"
>>>> t = ""
>>>> s[0:0+len(t)] == t[/color][/color]
> True[color=green][color=darkred]
>>>> s[1:1+len(t)] == t[/color][/color]
> True[/color]


Or in other words, imagine that Python is walking the string looking to
match the target. The empty string matches the boundary of every character
with the next character, or in other words, for a string s of length N,
s.count('') will equal N+1.

I'm not sure what to describe this surprising result as. It isn't a bug;
it isn't even really a gotcha. I guess the best description is that it is
just a surprising, but logical, result.

(Well, I was surprised -- but I can't fault the logic.)


--
Steven.

[Terry Reedy]

Re: why &quot;g&quot;.c ount('')==2 ?

For the same reason as[color=blue][color=green][color=darkred]
>>> "".count("" )[/color][/color][/color]
1[color=blue][color=green][color=darkred]
>>> "ab".count( "")[/color][/color][/color]
3

This is counting slice positions, which is one more that the length of the
string.

[Felipe Almeida Lessa]

Re: why &quot;g&quot;.c ount('')==2 ?

Em Sáb, 2006-03-11 às 04:25 -0800, ygao escreveu:[color=blue]
> my question is as title!
> thanks![/color]

Forget it. Just look:

$ python2.4 -mtimeit '"g".count(" ")'
1000000 loops, best of 3: 0.516 usec per loop
$ python2.4 -mtimeit 'len("g")+1'
1000000 loops, best of 3: 0.26 usec per loop


--
"Quem excele em empregar a força militar subjulga os exércitos dos
outros povos sem travar batalha, toma cidades fortificadas dos outros
povos sem as atacar e destrói os estados dos outros povos sem lutas
prolongadas. Deve lutar sob o Céu com o propósito primordial da
'preservação' . Desse modo suas armas não se embotarão, e os ganhos
poderão ser preservados. Essa é a estratégia para planejar ofensivas."

-- Sun Tzu, em "A arte da guerra"

[James Stroud]

Re: why &quot;g&quot;.c ount('')==2 ?

Steven D'Aprano wrote:[color=blue]
> On Sat, 11 Mar 2006 13:37:05 +0100, Fredrik Lundh wrote:
>
>[color=green]
>>"ygao" wrote:
>>
>>[color=darkred]
>>>my question is as title![/color]
>>
>>my answer as code:
>>
>>[color=darkred]
>>>>>s = "g"
>>>>>t = ""
>>>>>s[0:0+len(t)] == t[/color]
>>
>>True
>>[color=darkred]
>>>>>s[1:1+len(t)] == t[/color]
>>
>>True[/color]
>
>
>
> Or in other words, imagine that Python is walking the string looking to
> match the target. The empty string matches the boundary of every character
> with the next character, or in other words, for a string s of length N,
> s.count('') will equal N+1.
>
> I'm not sure what to describe this surprising result as. It isn't a bug;
> it isn't even really a gotcha. I guess the best description is that it is
> just a surprising, but logical, result.[/color]

Yes, but:

py> "gab".split ("")
Traceback (most recent call last):
File "<stdin>", line 1, in ?
ValueError: empty separator

The idea is not consistent between string functions. I actually consider
this latter example a bug.

Jaems

--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095

http://www.jamesstroud.com/