how to split matrix by percentage

I have a matrix R(u,v) and i want to split into 2 new matrices M, N. M will be the first 10% of the rows of R and N will be the rest 90% of the rows

[babis]

thanks for the answers. you helped a lot. I tried this:

a = x(1:end*0.1, 1:end);

b = x(end*0.1:end, 1:end);

c = x(1:end, 1:end*0.1);

d = x(1:end, end*0.1:end);

it seems that it works well.

[Image Analyst]

That's what I said, except that my code was robust and correct while yours was not. What if end*0.1 is a fractional number? It will fail whereas I cast to integer so my code won't fail. My code gives N and M like you asked for whereas yours doesn't, and yours doesn't give the final 90% where if you do b(end*.1:end) you'll run into the same problem of fractional indexes. I'd recommend you use my more robust code.

Answer 1

[Image Analyst]

Try this:

% Create some sample data 100 rows by 101 columns.
R = rand(100, 11); 
% Determine how many rows 10% is.
[rows, columns] = size(R);
% Determine the last row number of the top  (upper) 10% of rows.
lastRow = int32(floor(0.1 * rows));
% Get first 10% into one array M:
M = R(1:lastRow, :);
% Get the rest into one array N:
N = R(lastRow+1:end, :);

[Cappriece Clarke]

Hi,

To aid my understanding, shouldn't there be a value of 10% in the second line below? When I ran it as is, it gave me the total number of rows in the dataset.

% Determine how many rows 10% is.

[rows, columns] = size(R);

[Image Analyst]

@Cappriece Clarke, to know how many rows 10% of the total number of rows is, you first must determine the total number of rows. This is precisely what

[rows, columns] = size(R);

does. Now once you have rows, you can get 10% of it from 0.1 * rows. Then I round down to the nearest integer since you can't have a fractional row. So I do

% Determine the last row number of the top  (upper) 10% of rows.
lastRow = int32(floor(0.1 * rows));

and this does what it says. It gives you the last row number of the upper 10% of rows. So if you have 1009 rows, 10% of that is 100.9. Since you can't have 0.9 of a row, I round down from 100.9 to 100.

[Cappriece Clarke]

@Image Analyst, Thank you so much! I fully understand now.

As a follow up, how do I then determine the threshold value for each array (arrays M and N)?

I am currently working on a research paper that looks at the asymmetric exchange pass-through to prices in my country. The threshold value that I have found is said to be near the lower bound, ergo a recommendation was made to trim the data to the first 10% of the data and then establish the threshold for each groups (the 10% and the 90%). I tried using the code: options.trim=10; to no avail.

Can you kindly assist me in establilshing a code to determine the threshold for the entire data set and for each arrays?

Thank you in advance.

[Image Analyst]

Not sure what your options structure is, but you might want to sort your data first. Sounds like they might be talking about the value distribution (the histogram).

[Cappriece Clarke]

Thank you for your quick response.

Okay. So once I have created the histogram what would be the next move and is there a way to obtain the threshold value for the two arrays?

[Image Analyst]

I have no idea how to segment your histograms into two classes (values you're interested in, like foreground, and values you're not interested in, like background). It totally depends on the signal and what you want. If it's an image, maybe you can try imbinarize(). Or just pick some value, like

indexes = signal > someThreshold;

[Cappriece Clarke]

Hi Image Analyst,

I would like to test the maximum likelihood (mle) function on the parameters of a vector in a loop, to see if each parameter is significant. I would also like to run the wald test on each parameter, could you kindly assist me with this please? Thanks in advance.

The vector is called Colvect.

[Image Analyst]

Not sure I could help. You can try pca to find out how much of the variation is explained by each PC. I think there is also a stepwise regression in MATLAB but I don't know the details. I suggest you explain it thoroughly in a new question, not here in @babis's 8 year old question. Explain it well and attach your data and code and someone may be familiar with the process you should use.

[Cappriece Clarke]

Good day, I totally understand. Thank you.

Answer 2

[Azzi Abdelmalek]

%----Example------- 
n=randi(100);
m=10;
R=rand(n,m)
%---------------------
n1=ceil(0.1*n)