Matching the size of two matrices based on values of a column

I'm trying to make two matrices have the same dimensions, by matching up the values in the third column. The third column contains time in a day in decimal form. One of the matrix only runs for a certain time, and I want the 2nd matrix to match that length and line up the times. I'm currently using the code: where B is the matrix with the 24hr time frame and A has a smaller timeframe.

function [refined] = samesized( A, B)

[row,~] = size(A);

[rows,cols] = size(B);

re = zeros(rows,cols);

i = 1;

j = 1;

while i < row

while j < rows

if A(i,3) == B(j,3)

refined(i,3) = B(j,3);

refined(i,2) = B(j,2);

refined(i,1) = B(j,1);

end

j = j+1;

end

j = 1;

i = i+1;

end

refined( ~any(refined,2), : ) = [];

end

However, it sometimes skips values in the output and actually makes the returned matrix smaller, even though the values in column matched. I was wondering if anyone could help me figure out how to fix this, or know of a better way to do it.

Thanks!

Answer 1

[Peter Perkins]

"contains time in a day in decimal form": That would be your problem. Use timetables with datetimes (or maybe durations?), and use the [inner|outer]join function.

>> tt1 = timetable(rand(10,1),'RowTimes',datetime(2020,11,1:10)+hours(1:10));
>> tt2 = timetable(rand(5,1),'RowTimes',datetime(2020,11,1:2:10)+hours(1:2:10));
>> tt12 = innerjoin(tt1,tt2,'Key','Time')
tt12 =
  5×2 timetable
            Time            Var1_tt1    Var1_tt2
    ____________________    ________    ________
    01-Nov-2020 01:00:00    0.13755     0.29541 
    03-Nov-2020 03:00:00     0.6501     0.99014 
    05-Nov-2020 05:00:00    0.32355     0.16225 
    07-Nov-2020 07:00:00    0.68174     0.44102 
    09-Nov-2020 09:00:00    0.54227     0.55718 

[Bryan Ikeda]

Hmm, okay thanks. I'll need to look for a workaround because the date comes in decimal form due to it being used for a calculation for solar position, and the smaller matrix is coming from an xcel data sheet.

[Peter Perkins]

Bryan, you can store the timestamps as datetimes, and then convert to "fractional days" in the calculation. Something like

seconds(timeofday(tt12.Time)) / 86400

Answer 2

[ICR]

Hi,

This should work.

% Generate random integers
B = randi(24,[24 3]);
B(:,3) = 24:-1:1; % Replace these two your data

A = randi(15,[15 3]);
A(:,3) = randperm(15,15)'; % Replace these two your data

[r1,c1] = size(A)
outputMat = zeros(r1,c1); % Initialise

for i =1:1:r1
    [row,~] = find(B(:,3) == A(i,3))
    outputMat(i,:) = B(row,:);
end

[Bryan Ikeda]

Hmm, the test example you gave worked, but when I put the actual numbers in it gave the following error:

Unable to perform assignment because the size of the left side is 1-by-3 and the size of the right side is

0-by-3.

Error in samesized3 (line 6)

outputMat(i,:) = B(row,:);

I should also mention that the hours in column 3 are not all integers, they are divided into 5 minute intervals (ie: 12:05 pm is 12.083), I'm wondering if that is causing some of the problems.

Thanks!