| Re: Please help me avoid loops and conditionals [message #36331] |
Tue, 09 September 2003 12:30  |
Craig Markwardt
Messages: 1869
Registered: November 1996
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Senior Member |
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pford@bcm.tmc.edu (Patrick Ford) writes:
> function elp2, a, b, box_dim, vval, e_a,e_b, I_ratio
> x_box = box_dim/2
> box = intarr(box_dim,box_dim)
> o_val = fix(vval / I_ratio)
> v = fix(vval)
> for i = 0, box_dim-1 do begin
> for j = 0, box_dim-1 do begin
> x = float(i - x_box)
> y = float(j - x_box)
> if( ((x/(a+e_a))^2 + (y/(b+e_b))^2) LE 1.0) then $
> if( ((x/a)^2 + (y/b)^2) LE 1.0) then box(i,j) = o_val $
> else box(i,j) = v
> endfor; j = 0, box_dim-1 do
> endfor; i = 0, box_dim-1 do
> return, box
> end
>
>
> So how do I go about converting this into a Boolean matrix operation
> that avoids all of this? Would it be faster to create a mask array
> such as:
>
> x = float((indgen(box_dim) � box_dim/2) # replicate(1, box_dim))
> y = (transpose(x) / b)^2
It's close to what I would do.
I would start by creating the X and Y arrays more or less as you have
done:
x = (fltarr(box_dim)+1) ## findgen(box_dim)
y = findgen(box_dim) ## (fltarr(box_dim)+1)
and then initializing the array to zeroes:
box = intarr(box_dim,box_dim)
then as you said, use the WHERE statement to pull out the values of
interest.
wh = where( (x/(a+e_a))^2 + (y/(b+e_b))^2 LE 1.0, ct)
... and fill them in. You appear to have a two stage process.
if ct GT 0 then begin
box(wh) = o_val
wh1 = where( (x(wh)/a)^2 + (y(wh)/b)^2 LE 1.0, ct1)
if ct1 GT 0 then box(wh(wh1)) = v
endif
Just as you, I didn't test this or nuthin'.
Good luck,
Craig
--
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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: Please help me avoid loops and conditionals [message #36411 is a reply to message #36331] |
Wed, 10 September 2003 09:33  |
pford
Messages: 33
Registered: September 1996
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Member |
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Well, yours did not work (as you stated that you had not tried it
yet), but I did not spend any time on it after I noticed that you
forgot to place the origin in the center of the array. Instead I
debugged mine. But I have 2 follow up questions. Is it faster to use
integers the way I did it then convert to real (floating point) or
start off in floating point as you did? My mindset is still that
"integer is faster", which may not be the case because of the
conversion and the improved floating point hardware, like Altevec on
the PPC.
The second is related to the same program. I want to take a string of
4 contiguous bytes and convert them into a long unsigned or signed
integer. I remember doing this in Fortran, Pascal and C with some
variations on a theme of having variables of type byte and long point
to the same memory location. Is there some clever way of doing the
same in IDL? My current klugy method is based on the following:
R = ary(p) + (16 * ary(p+1)) + (16^2 * ary(p+2)) + (16^3 * ary(p+3))
Where ary is a byte array. The endian is irrelevant to what I want to
do here.
Thanks
Patrick Ford, MD
Baylor College of Medicine
Department of Radiology
Craig Markwardt <craigmnet@cow.physics.wisc.edu> wrote in message news:<onbrttlpy5.fsf@cow.physics.wisc.edu>...
> pford@bcm.tmc.edu (Patrick Ford) writes:
>> function elp2, a, b, box_dim, vval, e_a,e_b, I_ratio
>> x_box = box_dim/2
>> box = intarr(box_dim,box_dim)
>> o_val = fix(vval / I_ratio)
>> v = fix(vval)
>> for i = 0, box_dim-1 do begin
>> for j = 0, box_dim-1 do begin
>> x = float(i - x_box)
>> y = float(j - x_box)
>> if( ((x/(a+e_a))^2 + (y/(b+e_b))^2) LE 1.0) then $
>> if( ((x/a)^2 + (y/b)^2) LE 1.0) then box(i,j) = o_val $
>> else box(i,j) = v
>> endfor; j = 0, box_dim-1 do
>> endfor; i = 0, box_dim-1 do
>> return, box
>> end
>>
>>
>> So how do I go about converting this into a Boolean matrix operation
>> that avoids all of this? Would it be faster to create a mask array
>> such as:
>>
>> x = float((indgen(box_dim) - box_dim/2) # replicate(1, box_dim))
>> y = (transpose(x) / b)^2
>
> It's close to what I would do.
>
> I would start by creating the X and Y arrays more or less as you have
> done:
> x = (fltarr(box_dim)+1) ## findgen(box_dim)
> y = findgen(box_dim) ## (fltarr(box_dim)+1)
> and then initializing the array to zeroes:
> box = intarr(box_dim,box_dim)
> then as you said, use the WHERE statement to pull out the values of
> interest.
> wh = where( (x/(a+e_a))^2 + (y/(b+e_b))^2 LE 1.0, ct)
>
> ... and fill them in. You appear to have a two stage process.
> if ct GT 0 then begin
> box(wh) = o_val
> wh1 = where( (x(wh)/a)^2 + (y(wh)/b)^2 LE 1.0, ct1)
> if ct1 GT 0 then box(wh(wh1)) = v
> endif
>
> Just as you, I didn't test this or nuthin'.
>
> Good luck,
>
> Craig
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