lisp question

lisp question

[suvayu ali]

Hi Elisp users,

I was trying to write some simple elisp code to manipulate properties
from an org-mode entry. I can't figure out what kind of variable is
returned and how I can extract "somecategory" into a string from it.

(org-get-category (point)) ; evaluating gives me the output below
#("somecategory" 0 12 (fontified t font-lock-fontified t face
org-meta-line org-category "property-test"))

Thanks for any guidance.

-- 
Suvayu

Open source is the future. It sets us free.

Re: lisp question

[Jambunathan K]

Suvayu

> Hi Elisp users,
>
> I was trying to write some simple elisp code to manipulate properties
> from an org-mode entry. I can't figure out what kind of variable is
> returned and how I can extract "somecategory" into a string from it.
>
> (org-get-category (point)) ; evaluating gives me the output below
> #("somecategory" 0 12 (fontified t font-lock-fontified t face
> org-meta-line org-category "property-test"))

Return value is a text with properties. You can use

(substring-no-properties (org-get-category (point))) 

to get just the text part.

I think you will be better off using the property APIs. You can use
(org-entry-get (point) "CATEGORY")

to extract the desired information.

See "Using the property API" section of the manual.

> Thanks for any guidance.

-- 

Re: lisp question

[suvayu ali]

Hi Jambunathan,

On Thu, Aug 4, 2011 at 6:52 AM, Jambunathan K <kjambunathan@gmail.com> wrote:
> Return value is a text with properties. You can use
>
> (substring-no-properties (org-get-category (point)))
>
> to get just the text part.
>
> I think you will be better off using the property APIs. You can use
> (org-entry-get (point) "CATEGORY")
>
> to extract the desired information.
>
> See "Using the property API" section of the manual.

Thanks a lot for both the pointers! :)

-- 
Suvayu

Open source is the future. It sets us free.

Re: lisp question

[Maarten Bergvelt]

In article <mailman.55.1178045196.32220.help-gnu-emacs@gnu.org>, A Soare wrote:
> What determined the mathematicians to give the definition of x^0 = 1, not defined for x=0 ?

Discussion of this point at 
http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/

-- 
Maarten Bergvelt		

Re: lisp question

[Scott Frazer]

A Soare wrote:
> What determined the mathematicians to give the definition of x^0 = 1, not defined for x=0 ?
> 

y = x^0
ln(y) = ln(x^0) = ln(x) * 0 = 0

and since ln(1) = 0, y = 1 for any (non-zero) value of x

Re: lisp question

[A Soare]

What determined the mathematicians to give the definition of x^0 = 1, not defined for x=0 ?

Re: lisp question

[Jesper Harder]

A Soare <alinsoar@voila.fr> writes:

> What determined the mathematicians to give the definition of x^0 = 1,
> not defined for x=0 ?

Because 0/0 is undefined:

       x^1      x
x^0 =  ---- =  ---
       x^1      x

x=0 ->  0/0.

Re: lisp question

[Marco Almeida]

At Tue, 01 May 2007 21:10:44 +0200,
Jesper Harder wrote:
> 
> A Soare <alinsoar@voila.fr> writes:
> 
> > What determined the mathematicians to give the definition of x^0 = 1,
> > not defined for x=0 ?
> 
> Because 0/0 is undefined:
> 
>        x^1      x
> x^0 =  ---- =  ---
>        x^1      x
> 
> x=0 ->  0/0.
> 

As far as I know, 0^0 = 0 by convention (pretty much like 0 != 1)
Also, I think that your argument is wrong. 

0^y = 0 for any y
0 is the characteristic of R and you can not divid by it

This means that your first step :

       x^1   
x^0 =  ---- 
       x^1  

is not valid, regardeless of the value you chose as exponent because you will be trying a division by zero.


Also, from Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik): 

Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x , if the binomial theorem is to be valid when x=0 , y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.

> 
> 
> _______________________________________________
> help-gnu-emacs mailing list
> help-gnu-emacs@gnu.org
> http://lists.gnu.org/mailman/listinfo/help-gnu-emacs
> 

Re: lisp question

[Brendan Halpin]

A Soare <alinsoar@voila.fr> writes:

> Why e^0 evaluates to 1?

x^0 is defined as 1 for all real values of x.

Brendan
-- 
Brendan Halpin,  Department of Sociology,  University of Limerick,  Ireland
Tel: w +353-61-213147 f +353-61-202569 h +353-61-338562; Room F2-025 x 3147
mailto:brendan.halpin@ul.ie  http://www.ul.ie/sociology/brendan.halpin.html

Re: lisp question

[Hartmut Figge]

Brendan Halpin schribselte
>A Soare <alinsoar@voila.fr> writes:

>> Why e^0 evaluates to 1?
>
>x^0 is defined as 1 for all real values of x.

Mhm, not if x is zero. In this case the result is not defined. IIRC. :)

Hartmut
-- 
Usenet-ABC-Wiki http://www.usenet-abc.de/wiki/
Von Usern fuer User  :-)

Re: lisp question

[Malte Spiess]

Hartmut Figge <h.figge@gmx.de> writes:

> Brendan Halpin schribselte
>>A Soare <alinsoar@voila.fr> writes:
>
>>> Why e^0 evaluates to 1?
>>
>>x^0 is defined as 1 for all real values of x.
>
> Mhm, not if x is zero. In this case the result is not defined. IIRC. :)

That is really a matter of how you define it. Normally it's easier to
say that 0^0=1, it makes calculations easier.

> Hartmut

Greetings
Malte

Re: lisp question

[A Soare]

Why e^0 evaluates to 1?

Re: lisp question

[Peter Dyballa]

Am 30.04.2007 um 20:25 schrieb A Soare:

> Why e^0 evaluates to 1?

Because ln(e) is 1 and ln(1) is 0. The third reason I do not  
remember ...

--
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   Pete      <\
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lisp question

[Hadron]

from the lisp tutorial which comes with emacs 22:

,----
| 1.8.3 Variable Number of Arguments
| ----------------------------------
| 
| Some functions, such as `concat', `+' or `*', take any number of
| arguments.  (The `*' is the symbol for multiplication.)  This can be
| seen by evaluating each of the following expressions in the usual way.
| What you will see in the echo area is printed in this text after `=>',
| which you may read as `evaluates to'.
| 
|    In the first set, the functions have no arguments:
| 
|      (+)       => 0
| 
|      (*)       => 1
| 
|    In this set, the functions have one argument each:
| 
|      (+ 3)     => 3
| 
|      (* 3)     => 3
| 
|    In this set, the functions have three arguments each:
| 
|      (+ 3 4 5) => 12
| 
|      (* 3 4 5) => 60
`----

It kind of glosses over sections (1) and (2).

Why do (*) and (* 3) evaluate to 1?

Re: lisp question

[Barry Margolin]

In article <871wi385na.fsf@gmail.com>, Hadron <hadronquark@gmail.com> 
wrote:

> from the lisp tutorial which comes with emacs 22:
> 
> ,----
> | 1.8.3 Variable Number of Arguments
> | ----------------------------------
> | 
> | Some functions, such as `concat', `+' or `*', take any number of
> | arguments.  (The `*' is the symbol for multiplication.)  This can be
> | seen by evaluating each of the following expressions in the usual way.
> | What you will see in the echo area is printed in this text after `=>',
> | which you may read as `evaluates to'.
> | 
> |    In the first set, the functions have no arguments:
> | 
> |      (+)       => 0
> | 
> |      (*)       => 1
> | 
> |    In this set, the functions have one argument each:
> | 
> |      (+ 3)     => 3
> | 
> |      (* 3)     => 3
> | 
> |    In this set, the functions have three arguments each:
> | 
> |      (+ 3 4 5) => 12
> | 
> |      (* 3 4 5) => 60
> `----
> 
> It kind of glosses over sections (1) and (2).
> 
> Why do (*) and (* 3) evaluate to 1?

(* 3) evaluates to 3, not 1.

With associative functions, calling them with no arguments returns the 
identity value for that function.  This maintains the equivalence that

(<fun> <arguments>) == (<fun> (<fun> <part1>) (<fun> <part2>))

for any partitioning of the original arguments, including part1 or part2 
being empty.  E.g.

(* 3 4 5) = (* (* 3) (* 4 5)) = (* (*) (* 3 4 5))

-- 
Barry Margolin, barmar@alum.mit.edu
Arlington, MA
*** PLEASE post questions in newsgroups, not directly to me ***
*** PLEASE don't copy me on replies, I'll read them in the group ***

Re: lisp question

[Hadron]

Barry Margolin <barmar@alum.mit.edu> writes:

> In article <871wi385na.fsf@gmail.com>, Hadron <hadronquark@gmail.com> 
> wrote:
>
>> from the lisp tutorial which comes with emacs 22:
>> 
>> ,----
>> | 1.8.3 Variable Number of Arguments
>> | ----------------------------------
>> | 
>> | Some functions, such as `concat', `+' or `*', take any number of
>> | arguments.  (The `*' is the symbol for multiplication.)  This can be
>> | seen by evaluating each of the following expressions in the usual way.
>> | What you will see in the echo area is printed in this text after `=>',
>> | which you may read as `evaluates to'.
>> | 
>> |    In the first set, the functions have no arguments:
>> | 
>> |      (+)       => 0
>> | 
>> |      (*)       => 1
>> | 
>> |    In this set, the functions have one argument each:
>> | 
>> |      (+ 3)     => 3
>> | 
>> |      (* 3)     => 3
>> | 
>> |    In this set, the functions have three arguments each:
>> | 
>> |      (+ 3 4 5) => 12
>> | 
>> |      (* 3 4 5) => 60
>> `----
>> 
>> It kind of glosses over sections (1) and (2).
>> 
>> Why do (*) and (* 3) evaluate to 1?
>
> (* 3) evaluates to 3, not 1.

Sorry, yes.

>
> With associative functions, calling them with no arguments returns the 
> identity value for that function.  This maintains the equivalence that
>
> (<fun> <arguments>) == (<fun> (<fun> <part1>) (<fun> <part2>))
>
> for any partitioning of the original arguments, including part1 or part2 
> being empty.  E.g.
>
> (* 3 4 5) = (* (* 3) (* 4 5)) = (* (*) (* 3 4 5))

All clear. Thanks.