How to extract a 1D array from N-dimensional array?

[Kelvin Stott]

Hello,

I have a N-dimensional array called "Data", where the number of dimensions N is variable...

How can I extract a 1-dimensional array from this, comprising all elements along the Xth dimension, which are the first element along all the other dimensions?

For example, this would be equivalent to taking the values along one edge of a cubic 3D array, by specifying one of the 3 dimensions as X.

Thanks indeed for any help.

Kelvin

 

[shg]

You would also need to specify the values for all but one of the dimensions:

Function SuperDuperIndex(myArray, ParamArray aviDim as Variant)

... where aviDim is an N-element array that contains the fixed values of all but one dimension and one special value (Empty, maybe) that indicates the 'stripe'

 

[tusharm]

You cannot get a 1D array by specifying 1 dimension of a N dimension array. You get an array of dimension N-1.

For example, in a 3D array, if you specify the value of 1 dimension, you get a 2D matrix.

Hello,

I have a N-dimensional array called "Data", where the number of dimensions N is variable...

How can I extract a 1-dimensional array from this, comprising all elements along the Xth dimension, which are the first element along all the other dimensions?

For example, this would be equivalent to taking the values along one edge of a cubic 3D array, by specifying one of the 3 dimensions as X.

Thanks indeed for any help.

Kelvin

 

[Kelvin Stott]

Hi Tushar, I only want the first element of every other dimension, so I will end up with a 1D array, rather than N-1...

 

[Jerry Sullivan]

I'm trying to write a stand-alone function, where the input variable is an N-dimensional array:

Function GET1DARRAY(NArray, X)

I also have another function NDIM which can count the number of dimensions N in NArray, but from there I get stuck...

Kelvin, The function I suggested should do what you describe. Did you try it?

 

[Kelvin Stott]

Hi Jerry,

Sorry I missed your post (too many replies at once), let me take a closer look at this and see if I can get it to work, thanks.

 

[tusharm]

Maybe I am being particular dense but I still don't understand how "the first element of every other dimension" results in a 1D array. Take a 3D array. Suppose you specify element 2 of the 3rd dimension. This gives me a 2D array. What does "the first element of every dimension mean? I don't know what that means since you have to specify *two* indices to get any value!

Let's try this. Given that the 27 elements of a 3D array are indexed as below, which elements would constitute the 1D array of interest if I specify the index 3 value of 2?

0 0 0
0 0 1
0 0 2
0 1 0
0 1 1
0 1 2
0 2 0
0 2 1
0 2 2
1 0 0
1 0 1
1 0 2
1 1 0
1 1 1
1 1 2
1 2 0
1 2 1
1 2 2
2 0 0
2 0 1
2 0 2
2 1 0
2 1 1
2 1 2
2 2 0
2 2 1
2 2 2

Hi Tushar, I only want the first element of every other dimension, so I will end up with a 1D array, rather than N-1...

 

[Kelvin Stott]

Hi Tushar,

The easiest way to think of this is selecting one edge (1-dimensional array) of an N-dimensional polygon, which passes through the origin, by specifying one dimension X.

 

[shg]

Taking a 3-D array as an example, how would you specify which of the 8 edges to retrieve? One way (that has been suggested repeatedly) is to specify the fixed values of two of the dimensions, while the third dimension varies over its range. More generally, for an N-D cube, you would specify all but one dimension. That method is also constructive, since it's the way you index the array.

What's the alternative?

 

[Kelvin Stott]

For example:

For the 1st dimension (X=1) of a 3D array (N=3), the 1D array would be: [all, 1, 1]
For the 2nd dimension (X=2) of a 3D array (N=3), the 1D array would be: [1, all, 1]

For the 1st dimension (X=1) of a 4D array (N=4), the 1D array would be: [all, 1, 1, 1]
For the 3rd dimension (X=3) of a 4D array (N=4), the 1D array would be: [1, 1, all, 1]

For the 8th dimension (X=8) of a 24D array (N=24), the 1D array would be: [(1,)x7, all, (1,)x16]

Does that make sense now?

 

[Jerry Sullivan]

Taking a 3-D array as an example, how would you specify which of the 8 edges to retrieve? One way (that has been suggested repeatedly) is to specify the fixed values of two of the dimensions, while the third dimension varies over its range. More generally, for an N-D cube, you would specify all but one dimension. That method is also constructive, since it's the way you index the array.

What's the alternative?

That was my understanding of the OP question. The OP further said that each fixed index would be the first element (LBound) of that dimension.
For Tushar's example, the result of having X=2 would be a 1-Dim array having the values found at these indicies of the 3-Dim array.
0 1 0
0 2 0
0 3 0