Re: can .range return a 1D array?

And if arr is a single row 2-D array, then

As will:

with application
arr = .transpose(.Transpose(arr))
end with

just another way...

Alan Beban wrote:

Bruce Bowler wrote:

Consider the example of implementing an algorithm that "works best" with a
1D array. Consider the case where sometimes you want to call that code
with row (or portion of a row) worth of data and the SAME code with a
column (or portion there of) worth of data. In 1 case the subscripts are
(I,1), in the other they're (1,I). Yes, I know I could implement it with
2 loops going from lbound(x,1) to ubound(x,1) and lbound(x,2) and ubound
(x,2). Now suppose I (or someone else who borrowed the code) wants to
call it with an array created via the ARRAY function. Yep, could code
that too, but the code is *MUCH* simpler to understand (and less likely to
contain errors) if it treats the input as a vector rather than an array.

I know moan and groan, it's not going to change. I'll just live with it.


Perhaps the following might be useful.

If arr is a single column 2-D array, then

arr = Application.Transpose(arr) will convert it to a 1-D array.

And if arr is a single row 2-D array, then

arr = Application.Index(arr,1,0) will convert it to a 1-D array.

And if the functions in the freely downloadable file at
http://home.pacbell.net/beban are available to your workbook, then

arr = OneD(arr) will convert either a single row 2-D array or a single
column 2-D array to a 1-D array, and will leave a 1-D array as a 1-D array.

Alan Beban

--

Dave Peterson
.

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