| Re: Obtaining exponent from a scientific format number [message #30279] |
Thu, 18 April 2002 01:41 |
Nigel Wade
Messages: 286
Registered: March 1998
|
Senior Member |
|
|
Juan I. Cicuendez wrote:
> Hi all,
>
> I have a problem and I am looking for a faster solution:
>
> I have a scientific formatted number (e.g. 6.8977653e-18) and I have
> to split the exponent and number into two parts like this:
> long:68977653
> exp:-25
> where the first factor has to be a long number and the second the
> exponent.
> The exponential factors can algo change.
>
> The solution I came up is to turn the number into strings and then
> byte(mystring), obtaining the position of '.' and 'e' and then back to
> numbers. This seems to be quite slow and since I have a large number
> of data I don't think is very efficient.
>
> Thank you very much in advance, I would appreciate any hints.
>
> Juan
How about a mathematical solution?
IDL> number=6.8977653d-18
IDL> digits=8
IDL> exponent=floor(alog10(number))-(digits-1)
IDL> mantissa=round(number/10.0d0^exponent)
IDL> print,exponent
-25
IDL> print,mantissa
68977653
IDL>
It won't be entirely accurate due to the use of logs and exponentiation,
but it might be close enough (I've used double everywhere to increase
accuracy).
--
-----------------------------------------------------------
Nigel Wade, System Administrator, Space Plasma Physics Group,
University of Leicester, Leicester, LE1 7RH, UK
E-mail : nmw@ion.le.ac.uk
Phone : +44 (0)116 2523568, Fax : +44 (0)116 2523555
|
|
|
|
| Re: Obtaining exponent from a scientific format number [message #30287 is a reply to message #30279] |
Wed, 17 April 2002 15:21  |
Dick Jackson
Messages: 347
Registered: August 1998
|
Senior Member |
|
|
"David Fanning" <david@dfanning.com> wrote in message
news:MPG.172759ba2f12aa6198988f@news.frii.com...
> Dick Jackson (dick@d-jackson.com) writes:
>
>> Here's a way that you might like, but I know some others won't, as it
uses
>> the mysterious and cryptic "regular expressions" feature.
>> ; " * . * e *"
>> strPieces = StRegEx(str, '(.*)(\.)(.*)(e)(.*)', /Extract, /SubExpr)
>> ; strPieces now contains StrArr(6, n), columns 1-5 hold the pieces
>
> I just think anything that has a bunch of &%#@s in it
> is a real programmer's solution, using real programmer
> language! :-)
Just to clarify, I think David would be paying a Real Compliment if he
capitalized Real Programmer, but it seems he's been working on his backhand!
:-)
Hmm, upon seeing cool IDL stuff David has done, I've sometimes said
"Unreal!". Does that make him an Unreal Programmer? :-)
Cheers,
--
-Dick
Dick Jackson / dick@d-jackson.com
D-Jackson Software Consulting / http://www.d-jackson.com
Calgary, Alberta, Canada / +1-403-242-7398 / Fax: 241-7392
|
|
|
|
| Re: Obtaining exponent from a scientific format number [message #30306 is a reply to message #30287] |
Wed, 17 April 2002 10:07 |
David Fanning
Messages: 11724
Registered: August 2001
|
Senior Member |
|
|
Dick Jackson (dick@d-jackson.com) writes:
> Here's a way that you might like, but I know some others won't, as it uses
> the mysterious and cryptic "regular expressions" feature.
> ; " * . * e *"
> strPieces = StRegEx(str, '(.*)(\.)(.*)(e)(.*)', /Extract, /SubExpr)
> ; strPieces now contains StrArr(6, n), columns 1-5 hold the pieces
I just think anything that has a bunch of &%#@s in it
is a real programmer's solution, using real programmer
language! :-)
Cheers,
David
--
David W. Fanning, Ph.D.
Fanning Software Consulting
Phone: 970-221-0438, E-mail: david@dfanning.com
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Toll-Free IDL Book Orders: 1-888-461-0155
|
|
|
|
| Re: Obtaining exponent from a scientific format number [message #30307 is a reply to message #30306] |
Wed, 17 April 2002 09:56 |
Dick Jackson
Messages: 347
Registered: August 1998
|
Senior Member |
|
|
Hi Juan,
"Juan I. Cicuendez" <jicicuendez@gmv.es> wrote...
> I have a scientific formatted number (e.g. 6.8977653e-18) and I have
> to split the exponent and number into two parts like this:
> long:68977653
> exp:-25
> where the first factor has to be a long number and the second the
> exponent.
> The exponential factors can algo change.
>
> The solution I came up is to turn the number into strings and then
> byte(mystring), obtaining the position of '.' and 'e' and then back to
> numbers. This seems to be quite slow and since I have a large number
> of data I don't think is very efficient.
Here's a way that you might like, but I know some others won't, as it uses
the mysterious and cryptic "regular expressions" feature. I make a few
assumptions:
- you have an array of strings ready to process, for example:
str = ['6.8977653e-18', '-46.7654e-19']
- all of them have a '.' and an 'e'
If so, these 3 lines of IDL code should work for you:
; Match each string with a pattern like "*.*e*"
; " * . * e *"
strPieces = StRegEx(str, '(.*)(\.)(.*)(e)(.*)', /Extract, /SubExpr)
; strPieces now contains StrArr(6, n), columns 1-5 hold the pieces
; Put columns together to make desired 'long' values
longs = Reform(Long(strPieces[1,*] + strPieces[3,*]))
; Adjust values in last column to get correct exponents
exps = Reform(Long(strPieces[5,*]) + StrLen(strPieces[3,*]))
IDL> print, longs
68977653 -467654
IDL> print, exps
-11 -15
There may be methods that execute faster, on my system this takes 0.0003 s
per string.
Cheers,
--
-Dick
Dick Jackson / dick@d-jackson.com
D-Jackson Software Consulting / http://www.d-jackson.com
Calgary, Alberta, Canada / +1-403-242-7398 / Fax: 241-7392
|
|
|
|
| Re: Obtaining exponent from a scientific format number [message #30308 is a reply to message #30307] |
Wed, 17 April 2002 09:50 |
Michael Baca
Messages: 5
Registered: October 1999
|
Junior Member |
|
|
Well, part of this will depend on the format of the input number. The easy
way to get the exponent of the number is:
value=6.8977653d-18 ; the d will assign the value to a double
exp_num = floor(alog10(value)) ; works for both positive and negative
exponents
Then you just need to decide you starting position. In your message, you
moved the decimal place 7 positions to the right so the quick method would
be:
exp_num = exp_num - 7 ; moves the decimal 7 places to the right
base_val = floor(value/10.0d^exp_num)
Not fancy, but should do the job. The good news is that this will also work
on an array. So if you have an array of value, you can use the exact same
code and takes minimal effort. Much more efficient than looping over values
and counting positions in a string.
The biggest issue is handling the initial value. You may have problems
keeping track of the 7+ decimal places if not defined properly. My inital
attempts used 6.8977653e-18 and would consistently lose the 3 in the base
number unless I defined the number as above (use d not e). Besides,
recalling my days as a physics TA, do you really know the number that well.
:)
Mike
"Juan I. Cicuendez" <jicicuendez@gmv.es> wrote in message
news:4b04c513.0204170601.64440c58@posting.google.com...
> Hi all,
>
> I have a problem and I am looking for a faster solution:
>
> I have a scientific formatted number (e.g. 6.8977653e-18) and I have
> to split the exponent and number into two parts like this:
> long:68977653
> exp:-25
> where the first factor has to be a long number and the second the
> exponent.
> The exponential factors can algo change.
>
> The solution I came up is to turn the number into strings and then
> byte(mystring), obtaining the position of '.' and 'e' and then back to
> numbers. This seems to be quite slow and since I have a large number
> of data I don't think is very efficient.
>
> Thank you very much in advance, I would appreciate any hints.
>
> Juan
|
|
|
|
comp.lang.idl-pvwave archive
Members
Search
Help
Login
Home
