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Re: Arrays of Structures [message #52586 is a reply to message #52447] Fri, 09 February 2007 02:13 Go to previous messageGo to previous message
Mick Brooks is currently offline  Mick Brooks
Messages: 4
Registered: February 2007
Junior Member
On Feb 8, 8:42 pm, JD Smith <jdsm...@as.arizona.edu> wrote:

> The problem is that structs.a is not an array until the ".a" part has been
> applied.

<snip>

> Creating "structs.a" would cause a large temporary
> array to be created, only to finally index a single element. Compare the
> memory usage of the following:
>
> IDL> struct=replicate({a:lindgen(100,100,100)},100)
> IDL> print,(m=memory(/HIGHWATER))/1024/1024.,"MB"
> 382.472MB
> IDL> val=struct[10].a[4]
> IDL> print,(memory(/HIGHWATER)-m)/1024/1024.,"MB extra"
> 0.00000MB extra
> IDL> val2=(struct.a)[10,4]
> IDL> print,(memory(/HIGHWATER)-m)/1024/1024.,"MB extra"
> 381.470MB extra
>
> So the latter method first creates a temporary variable of size
> 100,100,100,100, and then pulls a single element out of it. Not
> exactly good form.

It seems that this is a reason to prefer my first "workaround" to my
second one, but it doesn't tell us anything about my problem ("unholy
notation" - I like that), which here would be
represented by val3=struct.a[10,4] i.e. leaving off the temporary-
creating parentheses.

If I try, I get the following:

IDL> struct=replicate({a:lindgen(100,100,100)},100)
IDL> print,(m=memory(/HIGHWATER))/1024/1024.,"MB"
385.836MB
IDL> val=struct[10].a[4]
IDL> print,((n=memory(/HIGHWATER))-m)/1024/1024.,"MB Extra"
-3.81445MB Extra
IDL> val3=struct.a[10,4]
IDL> print,(memory(/HIGHWATER)-n)/1024/1024.,"MB Extra"
0.00000MB Extra
IDL> HELP, val, val3
VAL LONG = 4
VAL3 LONG = Array[100]
IDL> PRINT, val3
410 [+ another 99 of the same]

The problem case doesn't use any extra memory (great!), but it gives a
different result (boo!).

Bob's and Mike's posts made me think that my original "structs.a" has
a leading shallow dimension, but that IDL elides it when evaluating
it.
So,
IDL> structs = replicate({a:1},50)
IDL> HELP, structs.a[0]
<Expression> INT = Array[50]

works, because our array subscript is within bounds on our leading
shallow dimension, but

IDL> HELP, structs.a[1]
% Subscript range values of the form low:high must be >= 0, < size,
with low
<= high: <No name>.
% Execution halted at: $MAIN$

is out of range.

If we ask for everything from the array that is structs.a
IDL> HELP, structs.a[*]
<Expression> INT = Array[1, 50]
we see the entire thing, leading shallow dimension included.

However, if we simply evaluate structs.a, IDL drops the leading
dimension, like so:
IDL> HELP, structs.a
<Expression> INT = Array[50]

Creating a temporary with parentheses also causes IDL to drop the
leading dimension too:
IDL> HELP, (structs.a)[*]
<Expression> INT = Array[50]

Does this make any more sense to anyone? I still can't use this idea
to work out what's going on
with val3 above though...

Thanks again for everybody's help,

Mick

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