Resizing an array - removing elements

I have an array of data, in this case 61 rows long. I need to reshape this into 3s - LSB1 = reshape(LSB,3,[]); but as 61 isn't divisible by 3, it wont work.

I want to be able to remove the last value of from the vector in this case, via an if statement (sometimes maybe to remove the last 2 values if there were 62 rows instead) i.e. an if statement to check divisibility by 3 and then to remove the correct number of elements from the end of the array.

Thank you

[Ahmad Kanzu Syauqi Firdaus]

you can also use imresize, especially when you want to decrease array dimension without changing array content generally. example:

X=rand(105,1); M=61; Xnew=imresize(X,[M,1]);

Answer 1

[Star Strider]

If you only want 60 rows in your array, just redefine a new array to have 60 rows:

LSB = rand(61,1);                               % Create Data
LSBnew = LSB(1:60,:);
LSB1 = reshape(LSBnew, [], 3);

[S]

Thank you!

[Star Strider]

My pleasure!

[S]

Hi again, How can I reshape a vector down so it is divisible by 3? Some of my data has 57 elements and therefore need to be brought down to 56 in this case. Is there a way to automate this for any data input?

Thanks!

[Star Strider]

My pleasure!

I don’t understand the 57 and 56 issue, though, since 57/3=19, and 56/3=18.666...

Yes. To do that, my code changes to:

LSB = rand(1, 56);                                        % Create Vector
vec_len = length(LSB) - rem(length(LSB),3);               % Trim To Have Length = Integer Multiple Of 3
LSB_new = reshape(LSB(1:vec_len), [], 3);                 % Reshape To Matrix

[S]

Thanks again!

One last thing, is there a way to reshape the vector so that elements 1,2 & 3 go across the new matrix (and then 3,4,5 etc) rather than going down the matrix?

so for a vector from 1-9, the reshaped one would read:

[1 2 3, 4 5 6, 7 8 9] rather than

[1 4 7, 2 5 8, 3 6 9]

[Star Strider]

My pleasure!

Since you are starting with a column vector, reverse the last two arguments in reshape and transpose the result:

x = [1:9]';
y = reshape(x, 3, [])';

This should do what you want.

Answer 2

[Stephen23]

Here is an example showing how this can be done automatically:

>> X = rand(4,3); % 4*3 = 12, not divisible by 5
>> R = 5;
>> reshape(X(1:end-mod(numel(X),R)),R,[])
ans =
      0.79221      0.93399
      0.95949      0.67874
      0.65574      0.75774
     0.035712      0.74313
      0.84913      0.39223