| Re: Principle of Least Surprise [message #52092] |
Tue, 09 January 2007 15:14  |
Jean H.
Messages: 472
Registered: July 2006
|
Senior Member |
|
|
... and
a[3:5].x=bindgen(3)
works perfectly!
It makes sense though... consider this case:
IDL> b=replicate({x:intarr(10)},10)
IDL> b[3].x = indgen(4)
IDL must understand that the specified array is the values of X in b[3].
If what you had written was working, IDL would not know if it is the
values of X or if it is the 1st value of 4 entries of b...
Jean
JD Smith wrote:
>
> Violated again. Consider:
>
> IDL> a=replicate(1b,10)
> IDL> a[3]=bindgen(3)
> IDL> print,a
> 1 1 1 0 1 2 1 1 1 1
>
> but now:
>
> IDL> a=replicate({x:1b},10)
> IDL> a[3].x=bindgen(3)
> % Expression must be a scalar in this context: <BYTE Array[3]>.
> % Execution halted at: $MAIN$
>
> This ought to work.
>
> JD
>
|
|
|
|
| Re: Principle of Least Surprise [message #52241 is a reply to message #52092] |
Tue, 09 January 2007 16:08  |
JD Smith
Messages: 850
Registered: December 1999
|
Senior Member |
|
|
On Tue, 09 Jan 2007 16:14:00 -0700, Jean H. wrote:
> ... and
> a[3:5].x=bindgen(3)
> works perfectly!
That doesn't prevent the first case I mentioned.
> It makes sense though... consider this case:
> IDL> b=replicate({x:intarr(10)},10)
> IDL> b[3].x = indgen(4)
>
> IDL must understand that the specified array is the values of X in b[3].
> If what you had written was working, IDL would not know if it is the
> values of X or if it is the 1st value of 4 entries of b...
Sure it would, easily. What is the dimension of b.x:
IDL> help,b.x
<Expression> INT = Array[10, 10]
IDL> help,b[3].x
<Expression> INT = Array[10]
I.e. the latter case has a 10 element array on its LHS, not a single index.
This is exactly analogous to:
IDL> b=intarr(10,10)
IDL> b[3,3]=indgen(4)
IDL> b[5,5]=indgen(4,2)
which IDL isn't confused about. I should equally be able to say:
IDL> b=replicate({x:intarr(10)},10)
IDL> b[3].x[2]=indgen(4,4)
and for it to do the right thing, without requiring index ranges to
make it happen. Why should implicit structure arrays be less capable
than their "normal" brethren?
IDL treats a LHS single index assigned to an array as a base from
which to fill in the entire array.... except that is for structure
dereferenced arrays. Certainly the form a[3:5] works, but it is
inefficient for large assignments, since it unnecessarily generates an
index list [3,4,5] before assigning.
Here's how to see how this hurts:
IDL> a=make_array(/LONG,long(100.e6))
IDL> print,memory(/HIGHWATER)/1024./1024.
382.465
382 MB for this array, OK. Now, let make another, one shorter:
IDL> b=make_array(/LONG,long(100.e6)-1L)
IDL> print,memory(/HIGHWATER)/1024./1024.
763.934
Now assign it using the "base index" trick:
IDL> a[1]=b
IDL> print,memory(/HIGHWATER)/1024./1024.
763.934
No additional memory used for the assignment from b to a, as
appropriate. Now, however, use an index range for the assignment, as
you are forced to do with structures:
IDL> a[1L:long(100.e6)-1]=b
IDL> print,memory(/HIGHWATER)/1024./1024.
1145.40
Whoops, we just used 382MB for the useless exercise of creating a large
index list [1,2,...,99999999] to consult while making the assignment.
Not only does this waste memory, it can really slow things down.
JD
|
|
|
|
comp.lang.idl-pvwave archive
Members
Search
Help
Login
Home
