how to extend third dimension of matrix?

If I have a three dimensional matrix 180*360*5, suppose third dimension representing 5 months of a year (feb, march, may, aug, dec) third. I need to extend its third dimension and result should be 180*360*12. Now would I insert missing months. then I also want to assign zero on inserted months.

Answer 1

[the cyclist]

If A is your array:

A = cat(3,A,zeros(180,360,7));

[SONI SONI]

I understand but this code is extending only the length of array at right sight. as I mentioned I want to insert missing months(jan, apr.....). Sequence of month is discontinue(feb, march, may, aug, dec) . So I want to create a continue series (jan, feb, mar, apr, may, june, jul, aug, sep, oct, nov, dec). please suggest me how should I solve this problem?

[Stephen23]

This is not clear at all. You need to explain your requirements to us clearly. Your question stated:

• "result should be 180*360*12 180*360*12"
• "assign zero on inserted months"

Both The Cyclist's and my own answer provide you exactly this. Can you see how we tried to help you by giving you what you asked for ? If you want something else, then you will have to explain it in more detail, with examples. Sorry, but we don't read minds.

Answer 2

[Stephen23]

The fastest way to enlarge an existing array is to simply assign a value in one "corner" of your array:

A = [180x360x5] % define your array...
A(1,1,12) = 0;

This places the given value (zero) and fills the undefined values with zero. The final dimensions of this example will be 180x360x12. This method has numerous advantages over other methods (such as concatenation):

• very fast!
• no creation of intermediate variables
• even though the zero is class double, because it is a scalar double it will be cast to match the class of A (uint, int, and even character).
• by specifying the first indices as ones (1,1,...) it is robust to any changes in the array size, without rewriting.

[Stephen23]

Second attempt based on your comment:

B(180,360,12) = 0; % create array.
B(:,:,[2,3,5,8,12]) = A