array manipulation - a few basic questions

Hi, I have to do a bit of array manipulation and I really don't have a clue about it.

First I need to know how to take a double array, and turn it in to a 2046x1 array (twice as long), with 1024-2046 just being exactly the same as 1-1023.

Then I need to know how to chop off the first half of an array, in this case taking a 2046x1 array and disregarding the first 1-1023, leaving only a 1023x1 array made up from only the second half of the 2046x1 array.

I know it's simple stuff, but I just haven't done it before.

[Matt Fig]

Tom's question

Hi, I have to do a bit of array manipulation and I really don't have a clue about it.

First I need to know how to take a double array, and turn it in to a 2046x1 array (twice as long), with 1024-2046 just being exactly the same as 1-1023.

Then I need to know how to chop off the first half of an array, in this case taking a 2046x1 array and disregarding the first 1-1023, leaving only a 1023x1 array made up from only the second half of the 2046x1 array.

I know it's simple stuff, but I just haven't done it before.

Answer 1

[Rick Rosson]

Please try:

   x = rand(1023,1);
   y = repmat(x,2,1);
   z = y(1024:2046);

HTH.

Rick

[Jan]

Or: y = [x; x];

or: y = vertcat(x, x);

[Rick Rosson]

Or: z = y(1024:end);

[Jan]

I avoid the END in indexing expressions in general. There have been several bugs in nested expressions and it is "slower" than "1024:numel(y)" - if one could claim that micro-seconds matter.

[Tom]

Many thanks guys - all useful.

[Tom]

I guessed that z = y(numel(x)+1:numel(y)); would work too. That make sense?

[Rick Rosson]

Yes, at least in this particular case, but not generally.