how to make a function that takes vector as input and generate matrix of its elements

Question

Muhammad Usman Saleem 2015-5-9

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评论： Geoff Hayes 2015-5-9

Hi everyone. I am going to make a function that take a vector as input and generate a matrix such that first column of matrix show vector elements, second colmn show square of the elements of vector and third column show cube of vector and so on..... Take a look on the example here

a=1:5;
m(a)=
   1     1   1    1
   4     8   16   32
   9    27   81  243
   16   64   256  1024
   25   125  625  3125

i am trying these codes but i unable to know how i make such matrix inside the second for loop

function H=hulk(v)
A=length(v);
H=zeros(A,A);
for i=1:A
    for j=1:A
        H(i,j)=i*j
    end
end
end

Kindly guide me how can i correct my code or any other alternate way to solve this.Thanks in advance for assistance...

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Answer 1

Geoff Hayes 2015-5-9

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Muhammad - a couple of things to consider. Try not to use i and j as indexing variables since both also represent the imaginary number. Note also how you are not using v as the first column of H. So first try

 numElems = length(v);
 H        = zeros(numElems,numElems);
 if ~iscolumn(v)
     v = v';
 end

The above assumes that v is a one-dimensional array. See how we convert v to a column vector to handle the case where it is a row vector.

Now you want each kth column of H to be v to the power of k. In your code, every element of your matrix H is just the row index multiplied by the column index which isn't what you want. Instead try

 for k=1:numElems
     H(:,k) = v.^k;
 end

We just loop over each column of H and set it to be v to the power of k. The colon in

H(:,k)

means that we consider all rows of H (instead specifying one for example) and so we are initializing this column of H to be

v^k

Try the above and see what happens!

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Geoff Hayes 2015-5-9

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Muhammad's answer moved here

@Geoff Hayes if i want to get only the matrix which will create powers till cube.It means this will not a square matrix. take an example below

output = hulk([1 2 3 4 5])
output =
   1     1
   4     8
   9    27
  16    64
  25   125

this will always created a matrix n*3(its 1st column is vector , second is square of elements of vector and 3rd column is cube). Where i need correction in your codes..Thanks in advance

Geoff Hayes 2015-5-9

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If you only want to calculate up to the nth power, then try the following

 nthPower = 3;
 numElems = length(v);
 H        = zeros(numElems, nthPower);
 if ~iscolumn(v)
     v = v';
 end
 for k=1:nthPower
     H(:,k) = v.^k;
 end

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Answer 2

Stephen23 2015-5-9

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编辑：Stephen23 2015-5-9

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bsxfun would be perfect for this, without requiring any loops (i.e vectorized code):

>> bsxfun(@power, (1:5).', 1:5)
ans =
         1           1           1           1           1
         2           4           8          16          32
         3           9          27          81         243
         4          16          64         256        1024
         5          25         125         625        3125