Inserting Zeros in a Matrix

2012 3 20
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Hi
Is there a way of inserting zeros in a data-matrix of 3x1(say). Only thing the zeros need to be inserted in those positions outside the data-matrix. So you would get a 10x1 matrix of zeros and datavalues.
For example,
DataCol = [x1;x2;x3] % Data column
NewDataCol = [0;0;0;0;0;x1;x2;x3;0;0] % New Data Column
The length of the DataCol changes each time.
Is there an easy way to insert the necessary zeros depending on the length of the data column?
Thanks
Ferd

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Daniel Shub
Daniel Shub 2012-3-20
You need to give more details. Do you always start with Nx1 and want to end up with Mx1 with M >= N? Assuming M = N+k, how do you distribute the k zeros (how many zeros go above and how many go below)?
Oleg Komarov
Oleg Komarov 2012-3-20
Yes, but you have to tell us on which position to start inserting DataCol each time into the colum of zeros.
Or phrased differently, what's the rule that decides how many zeros above and how many below.
Ferd
Ferd 2012-3-20
@Daniel - Hey! yea, I always start with a Nx1 data-matrix and the Mx1 is my universal matrix. I have a data-matrix(Nx1 = MainTiming channel in degrees from -12 to +10 TDC) 160x1 matrix long. And I also have a Universal matrix (Mx1 = Crank Angle Degrees from -180 to 180) 360x1 matrix long. I need to bring this data-matrix into a 360x1 matrix by inserting zeros outside these data values so that I can plot the MainTiming on a universal scale from -180 to +180 for ease.
@Oleg - Hey! the position to begin inserting 0's would be dependent on the data-matrix channel (in this case MainTiming in degrees).

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Dr. Seis
Dr. Seis 2012-3-20

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preZeros = 5;
postZeros = 2;
DataCol = rand(3,1);
%if really column vector
NewDataCol = zeros(length(DataCol)+preZeros+postZeros,1);
NewDataCol(preZeros+(1:length(DataCol)),1) = DataCol;

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Ferd
Ferd 2012-3-20
Hey Kewl thanks! This worked! I modified the preZeros and postZeros variable w.r.t the lengths of the parameters. Thanks again!

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Daniel Shub
Daniel Shub 2012-3-20

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Based on the additional information in your comment, you have x (160x1) and you want a function that will take in some offset b (0 <= b <= 200) and x and then create a matrix y (360x1) such that x(i) = y(i+b) for 1 <= i <= 160?
Without any error checking
f = @(b, x)([zeros(b, 1); x; zeros(200-b, 1)]);
NewDataCol = f(0, DataCol);
NewDataCol = f(200, DataCol);

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Ferd
Ferd 2012-3-20
Hey Daniel,
Technically, this works too based on how I questioned (my bad I didn't convey correctly). It does the question justice. The result is the first 200rows as 0's and the remainder is substituted with the DataCol. But then I would also need to have 0's post the datavalues. Ideally, if I plot these, it will give me the same output, but visually I would like to see the values in the form of 0's-Datavalues-0's matrix.
None-the-less, I appreciate the reply!
Thanks!
Daniel Shub
Daniel Shub 2012-3-20
I think that is what it does. I showed the edge conditions f(0, DataCol) which will have 0 preceding zeros and 200 following zeros and f(200, DataCol) which will have 200 preceding zeros and 0 following zeros. In general f(n, DataCol) will have n leading zeros and 200-n lagging zeros.
Daniel Shub
Daniel Shub 2012-3-20
Also, the best way to show appreciation on the answers is by up voting.
Ferd
Ferd 2012-3-20
Yup!
oh! will do... Didn't know about the voting thing. I am new to this Matlab environment.

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Geoff
Geoff 2012-3-20

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How about this:
startRow = 6;
NewDataCol = zeros(10,1);
NewDataCol(startRow-1 + (1:numel(DataCol))) = DataCol;

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