Inserting Zeros in a Matrix

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Ferd
Ferd 2012-3-20
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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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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采纳的回答

Dr. Seis
Dr. Seis 2012-3-20
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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更多回答（2 个）

Daniel Shub
Daniel Shub 2012-3-20
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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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
How about this:
startRow = 6;
NewDataCol = zeros(10,1);
NewDataCol(startRow-1 + (1:numel(DataCol))) = DataCol;

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