Notimplemented GE function

Hi,
I am trying to sum alle positive value's in a vector.
Im trying this by;
A=([1;1;-1;-1])
For i=1:4
if A(i,1)>=0
sum(A(i,1))
end
end
But it doesnt work, GE is not implemented, i have a student version sow..
Anyone has an alternitive?
I will have a vector with appr. 8000 rows with data, is there an easy way to analyse this data?

Respuestas (6)

Andrew Newell
Andrew Newell el 17 de Feb. de 2011

1 voto

sum(A(A>=0))
EDIT: better still,
sum(A(A>0)
Walter Roberson
Walter Roberson el 17 de Feb. de 2011

1 voto

I suspect your actual code has
A = {[1;1;-1;-1]};
or
A = {1;1;-1;-1};
If so then A is a cell array, and A(i,1) is a cell array, and > is not defined for cell arrays. To access the content of A(i,1) in such a case, you would use A{i,1}
Andrew Newell
Andrew Newell el 17 de Feb. de 2011

0 votos

I should add that your code doesn't work because it has some errors in it, not because GE is not implemented. A version that is similar to yours but works is
A=([1;1;-1;-1]);
AposSum = 0;
for i=1:4
if A(i,1)>=0
AposSum = AposSum + A(i,1);
end
end
AposSum
Frank
Frank el 24 de Feb. de 2011

0 votos

How do i know if it is a cell/array/matrix or something?
It has in the workspace a Cube in front of it, re.. Could it be called " Tensile Data"?
I think i am doing something wrong with writing my data... I tried a lot..
Isnt it possible to export it to a Txt file and then read it in again?
Thanks in advance!

2 comentarios

Andrew Newell
Andrew Newell el 24 de Feb. de 2011
Look at the Value column, to the right of Name. It should say something like <8000x1 cell> or [8000x1 double]. If it's the latter, you should be able to use my method.
Andrew Newell
Andrew Newell el 24 de Feb. de 2011
If you're still not sure, type "whos A" and paste the answer in your question.

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Andrew Newell
Andrew Newell el 24 de Feb. de 2011

0 votos

If Walter is right and it's a cell array, you could do this:
ipos = [A{:}]>0;
if any(ipos)
Asum = sum(A{ipos});
else
Asum = 0;
end
That funny looking first line extracts the values of A using A{:} and then puts it in a double array using [].
Frank
Frank el 24 de Feb. de 2011

0 votos

I solved it!
B=double(A) makes the matrix numeric or something like that, now i can do everything i want with it!
Anyways, thank u all for the help!

3 comentarios

Andrew Newell
Andrew Newell el 24 de Feb. de 2011
So what type was it before?
Walter Roberson
Walter Roberson el 24 de Feb. de 2011
If double() worked to make it numeric, then it must have been symbolic.
Andrew Newell
Andrew Newell el 24 de Feb. de 2011
That would also explain the problem with GE.

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