| size(/dimen) that automatically fills in extra dimensions [message #91241] |
Sat, 20 June 2015 20:38  |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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Before I write a quick routine that does this, it seems like someone must have done this already:
Does anyone have a drop-in replacement for SIZE(/DIMEN) that automatically fills in missing trailing dimensions with 1?
I.e. I have an array A that is always 3xN, but N could be 1, 2, or 3. I want to find out N, but
Size(A, /DIMEN)[1]
fails if N eq 1 because IDL drops the final dimension.
(even better: this would be a nice switch for the official SIZE function to have, if anyone is listening)
-Jeremy.
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| Re: size(/dimen) that automatically fills in extra dimensions [message #91242 is a reply to message #91241] |
Sun, 21 June 2015 00:42  |
Dick Jackson
Messages: 347
Registered: August 1998
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Senior Member |
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On Saturday, 20 June 2015 20:38:07 UTC-7, Jeremy Bailin wrote:
> Before I write a quick routine that does this, it seems like someone must have done this already:
>
> Does anyone have a drop-in replacement for SIZE(/DIMEN) that automatically fills in missing trailing dimensions with 1?
>
> I.e. I have an array A that is always 3xN, but N could be 1, 2, or 3. I want to find out N, but
>
> Size(A, /DIMEN)[1]
>
> fails if N eq 1 because IDL drops the final dimension.
>
> (even better: this would be a nice switch for the official SIZE function to have, if anyone is listening)
>
> -Jeremy.
Hi Jeremy,
I've been in your shoes...
In case this is helpful, there is a way to force the array to have a (3, N) shape, using Reform:
IDL> a=indgen(5,1)
IDL> help,a
A INT = Array[5] ; OK, the 1 has been dropped
IDL> a=reform(a,[5,1], /OVERWRITE)
IDL> help,a
A INT = Array[5, 1]
; Here's a handy routine when you want to ensure you have at least 'n' dimensions
;------
PRO EnsureNDims, x, nDims
IF Size(x, /N_Dimensions) GE nDims THEN RETURN
newDims = Replicate(1L, nDims)
newDims[0] = Size(x, /Dimensions) > 1 ; Will work even if x is scalar
x = Reform(x, newDims, /Overwrite)
END
;------
It can be interesting to see when this changes:
IDL> a=indgen([5,1])
IDL> help,a
A INT = Array[5]
IDL> ensurendims,a,2
IDL> help,a
A INT = Array[5, 1]
IDL> a=a
IDL> help,a
A INT = Array[5, 1]
; that was OK, didn't break it
IDL> b=a
IDL> help,b
B INT = Array[5]
; that broke it
IDL> a=b
IDL> help,a
A INT = Array[5]
; that broke 'a'
IDL> ensurendims,a,2
IDL> help,a
A INT = Array[5, 1]
IDL> a=a+1
IDL> help,a
A INT = Array[5]
; that broke it
IDL> a++
IDL> help,a
A INT = Array[5, 1]
; ... but that's OK!
Hope this helps!
Cheers,
-Dick
Dick Jackson Software Consulting Inc.
Victoria, BC, Canada --- http://www.d-jackson.com
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| Re: size(/dimen) that automatically fills in extra dimensions [message #91243 is a reply to message #91242] |
Mon, 22 June 2015 06:43 |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On Sunday, June 21, 2015 at 3:42:38 AM UTC-4, Dick Jackson wrote:
> On Saturday, 20 June 2015 20:38:07 UTC-7, Jeremy Bailin wrote:
>> Before I write a quick routine that does this, it seems like someone must have done this already:
>>
>> Does anyone have a drop-in replacement for SIZE(/DIMEN) that automatically fills in missing trailing dimensions with 1?
>>
>> I.e. I have an array A that is always 3xN, but N could be 1, 2, or 3. I want to find out N, but
>>
>> Size(A, /DIMEN)[1]
>>
>> fails if N eq 1 because IDL drops the final dimension.
>>
>> (even better: this would be a nice switch for the official SIZE function to have, if anyone is listening)
>>
>> -Jeremy.
>
> Hi Jeremy,
>
> I've been in your shoes...
>
> In case this is helpful, there is a way to force the array to have a (3, N) shape, using Reform:
>
> IDL> a=indgen(5,1)
> IDL> help,a
> A INT = Array[5] ; OK, the 1 has been dropped
>
> IDL> a=reform(a,[5,1], /OVERWRITE)
> IDL> help,a
> A INT = Array[5, 1]
>
> ; Here's a handy routine when you want to ensure you have at least 'n' dimensions
> ;------
> PRO EnsureNDims, x, nDims
> IF Size(x, /N_Dimensions) GE nDims THEN RETURN
> newDims = Replicate(1L, nDims)
> newDims[0] = Size(x, /Dimensions) > 1 ; Will work even if x is scalar
> x = Reform(x, newDims, /Overwrite)
> END
> ;------
>
> It can be interesting to see when this changes:
>
> IDL> a=indgen([5,1])
> IDL> help,a
> A INT = Array[5]
> IDL> ensurendims,a,2
> IDL> help,a
> A INT = Array[5, 1]
> IDL> a=a
> IDL> help,a
> A INT = Array[5, 1]
> ; that was OK, didn't break it
>
> IDL> b=a
> IDL> help,b
> B INT = Array[5]
> ; that broke it
>
> IDL> a=b
> IDL> help,a
> A INT = Array[5]
> ; that broke 'a'
>
> IDL> ensurendims,a,2
> IDL> help,a
> A INT = Array[5, 1]
> IDL> a=a+1
> IDL> help,a
> A INT = Array[5]
> ; that broke it
>
> IDL> a++
> IDL> help,a
> A INT = Array[5, 1]
> ; ... but that's OK!
>
> Hope this helps!
>
> Cheers,
> -Dick
>
> Dick Jackson Software Consulting Inc.
> Victoria, BC, Canada --- http://www.d-jackson.com
That's interesting... I can kind of see why certain ones do vs. don't, but I'm not sure I could have predicted each case a priori!
In this case, I don't actually need to change the dimensions, since the rest of my code works fine even when there's no trailing dimension -- I just need to be able to access its size. I could use this and then run Size right afterwards, but I've ended up writing it as a quick single function instead:
; Return's the length of the D-th dimension (starting with 1) of A,
; returning 1 for any missing trailing dimensions.
function size_d, a, d
s = size(a, /dimen)
if d le n_elements(s) then return, s[d]
return, 1
end
-Jeremy.
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| Re: size(/dimen) that automatically fills in extra dimensions [message #91245 is a reply to message #91243] |
Mon, 22 June 2015 06:59 |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On Monday, June 22, 2015 at 9:43:33 AM UTC-4, Jeremy Bailin wrote:
> On Sunday, June 21, 2015 at 3:42:38 AM UTC-4, Dick Jackson wrote:
>> On Saturday, 20 June 2015 20:38:07 UTC-7, Jeremy Bailin wrote:
>>> Before I write a quick routine that does this, it seems like someone must have done this already:
>>>
>>> Does anyone have a drop-in replacement for SIZE(/DIMEN) that automatically fills in missing trailing dimensions with 1?
>>>
>>> I.e. I have an array A that is always 3xN, but N could be 1, 2, or 3. I want to find out N, but
>>>
>>> Size(A, /DIMEN)[1]
>>>
>>> fails if N eq 1 because IDL drops the final dimension.
>>>
>>> (even better: this would be a nice switch for the official SIZE function to have, if anyone is listening)
>>>
>>> -Jeremy.
>>
>> Hi Jeremy,
>>
>> I've been in your shoes...
>>
>> In case this is helpful, there is a way to force the array to have a (3, N) shape, using Reform:
>>
>> IDL> a=indgen(5,1)
>> IDL> help,a
>> A INT = Array[5] ; OK, the 1 has been dropped
>>
>> IDL> a=reform(a,[5,1], /OVERWRITE)
>> IDL> help,a
>> A INT = Array[5, 1]
>>
>> ; Here's a handy routine when you want to ensure you have at least 'n' dimensions
>> ;------
>> PRO EnsureNDims, x, nDims
>> IF Size(x, /N_Dimensions) GE nDims THEN RETURN
>> newDims = Replicate(1L, nDims)
>> newDims[0] = Size(x, /Dimensions) > 1 ; Will work even if x is scalar
>> x = Reform(x, newDims, /Overwrite)
>> END
>> ;------
>>
>> It can be interesting to see when this changes:
>>
>> IDL> a=indgen([5,1])
>> IDL> help,a
>> A INT = Array[5]
>> IDL> ensurendims,a,2
>> IDL> help,a
>> A INT = Array[5, 1]
>> IDL> a=a
>> IDL> help,a
>> A INT = Array[5, 1]
>> ; that was OK, didn't break it
>>
>> IDL> b=a
>> IDL> help,b
>> B INT = Array[5]
>> ; that broke it
>>
>> IDL> a=b
>> IDL> help,a
>> A INT = Array[5]
>> ; that broke 'a'
>>
>> IDL> ensurendims,a,2
>> IDL> help,a
>> A INT = Array[5, 1]
>> IDL> a=a+1
>> IDL> help,a
>> A INT = Array[5]
>> ; that broke it
>>
>> IDL> a++
>> IDL> help,a
>> A INT = Array[5, 1]
>> ; ... but that's OK!
>>
>> Hope this helps!
>>
>> Cheers,
>> -Dick
>>
>> Dick Jackson Software Consulting Inc.
>> Victoria, BC, Canada --- http://www.d-jackson.com
>
> That's interesting... I can kind of see why certain ones do vs. don't, but I'm not sure I could have predicted each case a priori!
>
> In this case, I don't actually need to change the dimensions, since the rest of my code works fine even when there's no trailing dimension -- I just need to be able to access its size. I could use this and then run Size right afterwards, but I've ended up writing it as a quick single function instead:
>
> ; Return's the length of the D-th dimension (starting with 1) of A,
> ; returning 1 for any missing trailing dimensions.
> function size_d, a, d
> s = size(a, /dimen)
> if d le n_elements(s) then return, s[d]
> return, 1
> end
>
> -Jeremy.
Er, no that's not right -- that should be:
; Return's the length of the D-th dimension (starting with 0) of A,
; returning 1 for any missing trailing dimensions.
function size_d, a, d
s = size(a, /dimen)
if d lt n_elements(s) then return, s[d]
return, 1
end
-Jeremy.
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