linearly spacing a nX1 matrix

Hi!

I have an nx1 matrix. I want to linearly divide the value in each row into 16 columns thus giving me a nx16 linearly spaced matrix.

For a little example let us say I have a=[45; 50] now I want to creat a 2x4 matrix such that the values in consecutive clolumns are linearly spaced and their sum is equal to 45. Could someone help on this one please?

Answer 1

[the cyclist]

Your problem is still under-specified, and an infinite number of matrices will meet you conditions. Here is one possible solution:

a = [45; 50];

numberColumns = 4;

aMat = a + [0:numberColumns-1];

aMat = (aMat - mean(aMat,2) + a)/numberColumns;

[Z.khan]

Thank you very much for your response. I am sorry my question was under specified. What I meant was as follows

So, can I use linespace (or any other code) such that I produce a matrix 2x4 from 'a' such that the sum of each row in new matrix is equal to the coressponding row in a. so let us say we get b as follows by doing so

b=[i j k l m; n o p q] such that sum of i j k l m =45 and sum of n o p q=50. So actually the idea is to linearly divide 45 into a row of 4 columns and 50 into a row of linearly spaced 4 columns.

I hope it is better explained now. Thanks for your help.

[Stephen23]

"I hope it is better explained now."

The problem is still under-specified. For example, consider that both of these vectors add to 45 and are perfectly linearly spaced:

>> sum(7:1:11)
ans = 45
>> sum(5:2:13)
ans = 45

These are just two of infinite possible vectors that fulfill your requirements.

[the cyclist]

I edited my answer to try to do what I think you want. The result is

aMat =

   10.8750   11.1250   11.3750   11.6250
   12.1250   12.3750   12.6250   12.8750

Note that sum(aMat) is equal to [45; 50].

Your problem is still under-specified -- or I still don't understand -- meaning that there are an infinite number of output matrices that would meet the criteria of

• 4 columns
• each row sums to its corresponding number
• equal spacing

[Z.khan]

thank you very much again for your kindness. that is what I want but the problem is that I do not know what should be the appropriate step size given the start and end value and given the fact that the sum of new row should be a certain number?

so i know the start value (some positive number) and the end value (zero in my specific problem) and i know what should be the sum of the new row. How should i approach the problem then?

many thanks again.

[the cyclist]

I don't really understand your comment. Sorry.

Are you able to tell me if

aMat =
   10.8750   11.1250   11.3750   11.6250
   12.1250   12.3750   12.6250   12.8750

is a valid solution to your problem? It meets every requirement you have stated -- at least the ones I have understood.

If it is not a valid solution, are you able to say specifically what the problem is with it?

[Z.khan]

I am sorry this is so confusing on my part. Your solution is helpful but here is the problem.

I want to have the first row of aMat begin with 25 let's say and end with 0 and so the rest (45-10=35) should be equally spaced such that from 25 to zero everything is equally spaced.

Similarly, I want the second row to begin with 23 let's say and end with zero but all four columns should be linearly spaced.

Many thanks

[the cyclist]

So, now you've gone from an under-specified problem to an over-specified one.

If you require ...

• start value
• end value
• number of columns

then you cannot guarantee a particular sum.

This code will get you the listed three:

a = [45; 50];

start  = [25; 23];
finish = [ 0;  0];

numberRows = size(a,1);
numberColumns = 4;


output = zeros(numberRows,numberColumns);

for ir = 1:numberRows
    output(ir,:) = linspace(start(ir),finish(ir),4);
end

output =

   25.0000   16.6667    8.3333         0
   23.0000   15.3333    7.6667         0
   

but sum(output,2) = [50; 46]

[Z.khan]

Thank you very much. You have been extremely kind. I will work on adjusting the sum somehow. Thanks a lot for you help.