| Re: populating an array [message #48163] |
Thu, 30 March 2006 12:47  |
codepod
Messages: 19
Registered: March 2006
|
Junior Member |
|
|
Subir,
Are you trying to extract every other byte of data from your array and
then convert it to an int? If so, you could take the following
approach:
data = bytarr(202000 , /nozero)
;; Note: if you're filling the array later, /nozero will save a
;; little time (skips a bzero call).
;; Read your data as you mention in your message.
;; Here, just index into your data array using the array
;; stride syntax [0:*:2] Reform creates the array shape you need.
array = reform( fix( data[ 0:*:2 ] ), 1000, 101)
;; If your IDL version doesn't support strides, you can do this
;; with indgen, but it will probably be slightly slower.
array = reform( fix( data[ lindgen(101000)*2 ] ), 1000, 101)
;; Either of these will eleminate the costly for loop
;; And if your done with data at this point, you can just free it
data = 0b
Cheers - CP
subir.vasanth@gmail.com wrote:
> Greetings!
>
> I was wondering if there was a more efficient way to populate an array
> created using the MAKE_ARRAY function. This is how I am populating the
> array right now -
>
> data = BYTARR(202000L)
> ; populate byte array with valid data from some input source
> array = MAKE_ARRAY(1000,101, Type = 2)
> offset = 0L
> FOR k = 0L, 100999L DO BEGIN
> array(k) = FIX(data, offset)
> offset = offset + 2
> ENDFOR
>
> Is there a way I can populate 'array' without using a loop to populate
> each element, and instead do a array = FIX(data)??
>
> Thanks,
> subir
|
|
|
|
comp.lang.idl-pvwave archive
Members
Search
Help
Login
Home
