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My goal was to determine the size of a single dimension index based on the
contents of the first column [row] of a matrix. For instance if I have Q=(1,0,2) I want
an index N=2, for two non-0 elements for Q1 and Q3. Q can be any size and any
number of 0 elements can exist. I attach a simple replica of the program I used
to calculate this index, using the "for" and "if" functions. It seems simple so I am
feeling pretty stupid right now. Can you see anything wrong with the attached
program? It was developed in MathCAD 2001i [V8].
Solved! Go to Solution.
I can't open your program, my computer doesn't recognise "zipx"
Does the attached help?
I can't open your program, my computer doesn't recognise "zipx"
Does the attached help?
Fred
The program and set up you sent works as I would have intended mine to work.
There is not much different in what I did than what you show. The only real differences
were that I have a one dimensional matrix [simpler] and I did not express the LHS
of the program as a function or the rows(A). Instead I simply embedded this number
in the for statement. My lines are generally as follows.
N= i set to 0
for k E 1 .... cols(Q)
i=i+1 if Q(k) <=> 0 |Q(k) is the column k of Q
[i=i otherwise, or no statement same effect]
i
This would give me N=3, total columns of Q, instead of 2, in my example, with non-zero
elements.
I will try your suggestion and I am sure it will cure my issue.
Thank you,
Paul
To me Fred's solution looks a little different to what you were after.
Please re-upload the worksheet.
Mike
I opened this file (ZIPX) using WinZip 15.5.
P.S. Unfortunately, at this moment I do not have access to Mathcad.
Cheers.
Mike
Mike
I have tried to re-upload a document to the current discussion
but being a relatively new participant I could not find a way to
do this. Ultimately I uploaded the document to the programming
group but it doesn't appear in the discussion thread. My problem
has been solved. It was a silly oversight on my part. In my single
dimension array Q=(1 0 2) say, my program was written to check
Q<k> = 0. But Q<k> only returns the array of that one cell, not its
value. Hence I needed to at least add indices Q<k>[1,1 to the program.
Better though is simply to use Q[1,k = 0 as the test. This works.
Thank you,
Paul
Hi Paul,
I have just commented on the document you posted. If you click on the 'use advanced editor' link in the top right hand corner you can then upload worksheet and attached files.
Mike
I forgot to add - While it is your prerogative which version to choose to adopt. Fred's example doesn't indicate the total amount of zero's in a Matrix, which is what you were after isn't it?
Mike
Andy.
If you change Q<k> to Q[1,k address the kth element of the first row rather than the whole kth column your program works.
Thanks, Vladimir, for unpacking it!
Fred Kohlhepp wrote:
...Thanks, Vladimir, for unpacking it!
No problem.
The problem appears to be that k<0> is returning an array & that isn't equal to a scalar.
There are a couple of alternate methods to achieve the function attached
Best regards
Andy
Another method?
Mike
Mike,
That is a clever function. Not sure what the expression in the upper limit of the first sum is to accomplish?
in the original file, Q<=>0, doesn't exclude real number. But I don't think that syntex is recoginized anyway.
Mike,
That is a clever function. Not sure what the expression in the upper limit of the first sum is to accomplish?
Good spot Wayne.
It wasn't meant to accomplish anything. It was a typo
Mike
Andy: You are absolutely right. I needed the scalar not the array. Thanks Paul