| Re: FOR statement [message #24722] |
Fri, 13 April 2001 13:17 |
Eli Beckerman
Messages: 3
Registered: April 2001
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Junior Member |
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> Eli,
>
> No matter how look at it, your example shows you *are* using a loop to
> create an evenly spaced mesh vector. I guess I was trying to say that
> using a loop is not the best way to do this in IDL. One of the wisest
> pieces of advice I ever received about IDL programming was
>
> "Try to think like an IDL programmer, not a Fortran or C programmer".
>
> In this spirit, I submit that the following method is preferable:
>
> nx = 1000 ; number of values required
> dx = 0.25 ; step size
> x1 = 0.0 ; start value
> radius = lindgen(nx) * dx + x1
>
> You might try changing the number of values to 10,000,000 and seeing
> which method is faster.
>
> Cheers,
> Liam.
Well, here's my FOR loop with some background info... The radius
vector was what was hanging me up -- it was not what the FOR loop was FOR!
I can't reason out any way of doing this without the loop, but I
imagine it can be done...
dist = sqrt((x-xc)^2 + (y-yc)^2) ;xc & yc are the center positions
bin = 0.25 ;of a distribution of points
psf = histogram(dist,binsize=bin) ;psf is the radial profile
norm = total(psf)
maxR = max(dist)
ee=fltarr((maxR/bin)+1)
rval=fltarr(maxR/bin)+1)
FOR i=0.0, maxR/bin DO BEGIN
radius = i * bin
insidecounts = n_elements(where(dist le radius))
rval(i) = radius
ee(i) = insidecounts/norm ;to establish the number of counts
;within a given radius
ENDFOR
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| Re: FOR statement [message #24744 is a reply to message #24722] |
Thu, 12 April 2001 14:04  |
Liam E. Gumley
Messages: 378
Registered: January 2000
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Senior Member |
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Eli Beckerman wrote:
> In defense of myself, I wasn't using the FOR loop to make an array
> (I just excluded all the unnecessary stuff from my posting, and
> seemingly confused everyone even further!)
>
> But clearly, I was mixing the semantics of indexing with assignment
> as Craig pointed out!
>
> FOR i=0, 999 do radius(i) = i * 0.25 is the way to go for what I
> wanted. Thanks for seeing through my morning foggyness.
Eli,
No matter how look at it, your example shows you *are* using a loop to
create an evenly spaced mesh vector. I guess I was trying to say that
using a loop is not the best way to do this in IDL. One of the wisest
pieces of advice I ever received about IDL programming was
"Try to think like an IDL programmer, not a Fortran or C programmer".
In this spirit, I submit that the following method is preferable:
nx = 1000 ; number of values required
dx = 0.25 ; step size
x1 = 0.0 ; start value
radius = lindgen(nx) * dx + x1
You might try changing the number of values to 10,000,000 and seeing
which method is faster.
Cheers,
Liam.
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| Re: FOR statement [message #24747 is a reply to message #24744] |
Thu, 12 April 2001 12:41 |
Eli Beckerman
Messages: 3
Registered: April 2001
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Junior Member |
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Thanks All.
In defense of myself, I wasn't using the FOR loop to make an array
(I just excluded all the unnecessary stuff from my posting, and
seemingly confused everyone even further!)
But clearly, I was mixing the semantics of indexing with assignment
as Craig pointed out!
FOR i=0, 999 do radius(i) = i * 0.25 is the way to go for what I
wanted. Thanks for seeing through my morning foggyness.
-Eli
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| Re: FOR statement [message #24752 is a reply to message #24747] |
Thu, 12 April 2001 09:18 |
Craig Markwardt
Messages: 1869
Registered: November 1996
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Senior Member |
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Eli Beckerman <ebeckerman@cfa.harvard.edu> writes:
> Hey,
>
> I just tried running a FOR loop in the hopes
> of incrementing the variable "i" by steps of 0.25 as follows:
>
> radius=fltarr(1000)
> FOR i=0.0, 100.0, 0.25 DO BEGIN
>
> radius(i)=i
>
> ENDFOR
>
>
> And what I end up with is an array that starts
> with the value 0.75 and is incremented by steps of 1.
>
> I'm following the convention of the FOR statement as
> presented in IDL's online help. What am I doing wrong?!
You appear to be mixing the semantics of indexing with assignment.
What do you think happens when you do radius(0.25) = 0.25, which is
followed by radius(0.5) = 0.5 and radius(0.75) = 0.75? IDL
automatically converts floating point array indices into integers by
rounding down, so all of the statements I just mentioned affect the
same array element, and 0.75 supercedes because it is last.
You probably want:
FOR i=0.0, 100.0, 0.25 DO radius(i*4)=i
Or better yet:
FOR i=0, 999 do radius(i) = i * 0.25
(which avoids the ambiguity of indexing by a floating point number)
Or even better:
radius = findgen(400)*0.25
(avoids FOR loops altogether)
Also, why do you have a 1000 element array when you only fill the
first 400 elements?
Craig
--
------------------------------------------------------------ --------------
Craig B. Markwardt, Ph.D. EMAIL: craigmnet@cow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
------------------------------------------------------------ --------------
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| Re: FOR statement [message #24753 is a reply to message #24752] |
Thu, 12 April 2001 09:15 |
Liam E. Gumley
Messages: 378
Registered: January 2000
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Senior Member |
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"Liam E. Gumley" wrote:
> Try this instead:
>
> nx = 1000 ; number of values required
What I meant to say was
nx = 401
but the rest of my post still applies.
Cheers,
Liam.
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| Re: FOR statement [message #24754 is a reply to message #24753] |
Thu, 12 April 2001 09:12 |
Liam E. Gumley
Messages: 378
Registered: January 2000
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Senior Member |
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Eli Beckerman wrote:
> I just tried running a FOR loop in the hopes
> of incrementing the variable "i" by steps of 0.25 as follows:
>
> radius=fltarr(1000)
> FOR i=0.0, 100.0, 0.25 DO BEGIN
>
> radius(i)=i
>
> ENDFOR
>
> And what I end up with is an array that starts
> with the value 0.75 and is incremented by steps of 1.
>
> I'm following the convention of the FOR statement as
> presented in IDL's online help. What am I doing wrong?!
You are using I as both an array index and a loop variable. This is not
a good idea.
Try this instead:
nx = 1000 ; number of values required
dx = 0.25 ; step size
x1 = 0.0 ; start value
radius = lindgen(nx) * dx + x1
print, radius[0:5]
Bottom line: Don't use loops to create mesh vectors or arrays.
Cheers,
Liam.
http://cimss.ssec.wisc.edu/~gumley/
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| Re: FOR statement [message #24755 is a reply to message #24754] |
Thu, 12 April 2001 08:57 |
Paul van Delst
Messages: 364
Registered: March 1997
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Senior Member |
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Eli Beckerman wrote:
>
> Hey,
>
> I just tried running a FOR loop in the hopes
> of incrementing the variable "i" by steps of 0.25 as follows:
>
> radius=fltarr(1000)
> FOR i=0.0, 100.0, 0.25 DO BEGIN
>
> radius(i)=i
>
> ENDFOR
>
> And what I end up with is an array that starts
> with the value 0.75 and is incremented by steps of 1.
>
> I'm following the convention of the FOR statement as
> presented in IDL's online help. What am I doing wrong?!
Try
i=-1L
FOR xi=0.0, 100.0, 0.25 DO BEGIN & i=i+1L & radius(i)=xi & endfor
IDL> print, radius[0:10]
0.00000 0.250000 0.500000 0.750000 1.00000 1.25000
1.50000
1.75000 2.00000 2.25000 2.50000
using i as the array subscript must be converting it to an integer type and screwing stuff
up somehow.
Using loop indices like this should be discouraged, I reckon.
paulv
--
Paul van Delst A little learning is a dangerous thing;
CIMSS @ NOAA/NCEP Drink deep, or taste not the Pierian spring;
Ph: (301)763-8000 x7274 There shallow draughts intoxicate the brain,
Fax:(301)763-8545 And drinking largely sobers us again.
paul.vandelst@noaa.gov Alexander Pope.
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| Re: FOR statement [message #24756 is a reply to message #24755] |
Thu, 12 April 2001 09:04 |
John-David T. Smith
Messages: 384
Registered: January 2000
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Senior Member |
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Eli Beckerman wrote:
>
> Hey,
>
> I just tried running a FOR loop in the hopes
> of incrementing the variable "i" by steps of 0.25 as follows:
>
> radius=fltarr(1000)
> FOR i=0.0, 100.0, 0.25 DO BEGIN
>
> radius(i)=i
>
> ENDFOR
>
> And what I end up with is an array that starts
> with the value 0.75 and is incremented by steps of 1.
>
> I'm following the convention of the FOR statement as
> presented in IDL's online help. What am I doing wrong?!
Indexing an array with a floating point number.
try:
FOR i=0.0, 100.0, 0.25 DO BEGIN radius[4*i]=i
or even better:
radius=findgen(401)/4.
JD
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