Problem Expanding a Matrix

Hi all,

How can I do the following:

Say that I start with a vector with the elements [5 3 4 9 10] - think of these numbers like daily stock prices. I want to transform this vector into (an approx.) of intraday stock prices - 1/10 of day.

Therefore my vector should look like [ 5 4.8 4.6 4.4 .4.2 4 3.8 3.6 3.4 3.2 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9 10]

THank you!!

Answer 1

[Charles Martineau]

I figured out some other way

x = 0:4; y = [5 3 4 9 10]; >> xnew = 0:.1:4; ynew = interp1(x,y,xnew,'linear');

Answer 2

[Walter Roberson]

NewV = interp1(1:length(V), V, V(1):.1:V(end));

[Charles Martineau]

Hi Walter,

Thanks for the help but why I am generating a vector of NaN NaN NaN NaN NaN....

I simply created a vector V (3X1) and I get this strange result.

Thanks!

Answer 3

[Ryan]

clear i j
elements = [5 3 4]; % Currently what you have

% Matrix containing intraday prices where each row corresponds to the intraday prices for each of the members of elements
intraday = [1 2 3 4 5 6 7 8;1 2 3 4 5 6 7 8;1 2 3 4 5 6 7 8]; 

j = length(elements);
for i = 1:j
      newelements(i,:) = [elements(i),intraday(i,:)];
end

[Charles Martineau]

Hi Ryan,

thanks for the help but your code generates a matrix

5 1 2 3 4 5 6 7 8

3 1 2 3 4 5 6 7 8

4 1 2 3 4 5 6 7 8

how can modify it to have a vector that looks like:

[ 5 4.8 4.6 4.4 .4.2 4 3.8 3.6 3.4 3.2 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4 ... ?

Also I don't have a intraday vecteur. The intraday numbers are computed using the gap between two numbers.

[Ryan]

clear i j

elements = [5 3 4]; % Currently what you have

% Matrix containing intraday prices where each row corresponds to the intraday prices for each of the members of elements

intraday = [1 2 3 4 5 6 7 8;1 2 3 4 5 6 7 8;1 2 3 4 5 6 7 8];

j = length(elements);

for i = 1:j

newelements(i) = [elements(i),intraday(i,:)];

end

[r,c] = size(newelements);

newestelements = reshape(newelements,1,r*c);

I understand that you answered your own question, but I believe that should work. More round about than your approach though!

[Charles Martineau]

Hi Ryan,

THanks for help! I'll keep your answer in mind. The answer that I got came from StackOverflow

Answer 4

[Ryan]

that should read newelements(i,:) = [elements(i),intraday(i,:)];

it is the same as before, it just reshapes it at the end.