Storing values calculated in a 'for' loop in a matrix / vector

I would very much appreciate some help storing values in a matrix. I am trying to do a growth curve (See: http://kellymom.com/images/growth/growthcharts.gif). For that reason I have to do quantile equations for a set of data (age and height). Currently, I've made a 'for' loop, which does calculate the value as intended, however it only stores the last result. How do I make it store the data as a matrix like:

Quantile 25 %

Age Height

• 6 123
• 7 129
• 8 135

and so forth.

Thank you very much!

Note: male.data consists of two columns. First column is age and second is height of two thousand male participants. The file: age.height.mat is my data.

[EDITED, Jan, contents of the M-file from dropbox:]

%%Assignment 1 - Growth curves for men and women
clear all ;
clc ;
load age.height.mat

% Sorting the data to be plotted
male.data = sortrows(male.data) ;

plot(male.data(:,1),male.data(:,2),'r+')

male.growth = zeros(1, 73);
t = zeros(1, 73); 
    for k = 6:78
      male.growth(k-5) = male.data(male.data(:,1) == k, 2);
      t(k-5) = quantile(male.growth(k-5), 0.5);
    end

% Note that t sometimes returns 'not a number', when a specific age is not
% present. That won't be a problem in my final script, as I am not allowed
% to publish all my data for copyright reasons.

[Jan]

This forum does not have a strict text limit and postimng the code would be a really good idea.

[Tobias]

Thank you for editing my post. I will just sort the layout a bit, as most of my comments in the script are probably not necessary.

[Tobias]

Anybody who can help me with fixing the dimension problem I am currently having?

"In an assignment A(I) = B, the number of elements in B and I must be the same."

male.growth(k-5) = male.data(male.data(:,1) == k, 2);

[Tobias]

I would just like to add that I have solved my problem! So thank you for inspirational input

This can be closed.

Answer 1

[Jan]

Please do not start your code with the cruel killer. There is no benefit in removing all loaded functions from the memory by clear all. If you have a good reason to clear all variables, use clear variables or even better use a function, because it has its own workspace.

For storing the results in a vector:

male.growth = zeros(1, 73);  % Pre-allocate!!! Optional but recommended
t = zeros(1, 73); 
for k = 6:78  % Avoid using "i" as a variable
  male.growth(k-5) = male.data(male.data(:,1) == k, 2); % Without FIND
  t(k-5) = quantile(male.growth(k-5), 0.5);
end

[Tobias]

Alright, I'll make sure to keep that in mind.

While I get what you're doing, I'm getting an error in the line:

    male.growth(k-5) = male.data(male.data(:,1) == k, 2); % Without FIND

"In an assignment A(I) = B, the number of elements in B and I must be the same."

As I'm seeing it, the number of elements correspond just fine.

[Jan]

So start the debugger:

dbstop if error

Then run the code again until it stops at the error. Then type in the command window:

a = male.data(male.data(:,1) == k, 2)
size(a)

[Tobias]

After having debugged and writing the code you suggested, I got the following:

K>> size(a)

ans =

     104     1

Which, as far as I understand, should correspond correctly with the 104 boys at the age of 6 included in my data. Even though the dimension of male.growth is 1 x 73 after the code:

male.growth = zeros(1, 73);

Isn't it supposed to automaticly change the dimensions of male.growth to the dimensions of the result in:

    male.growth(k-5) = male.data(male.data(:,1) == k, 2);

[Jan]

@Tobias: While for:

male.growth = male.data(male.data(:,1) == k, 2);

the dimensions on the left side are adjusted automatically, as for:

male.growth(100) == 8;

the missing elements are set to implicitely, a vector cannot assigned to a scalar:

a(1) = 1:10;  % ERROR