How to find the value in one column based on a value of a another column.

Hello,

I have a table with time being the first column and Pressure A, B, and C being columns 2-4. I would like to find a way to display the time it takes for the first instance of Pressure A, B and C (which occur at different times) to show a value greater than .6.

Any help is greatly appreciated, I have just started using matlab and am quite confused.

[pragnan nadimatla]

As per my understanding you want to find the values of one column (time) based on pressure values which are on other columns when certain threshold has been crossed.

Here’s an example of how to code :

% Generate example data
time = (0:0.1:1)';
pressureA = sin(2*pi*time);
pressureB = cos(2*pi*time);
pressureC = sin(4*pi*time);

% Combine data into a table
data = table(time, pressureA, pressureB, pressureC);

% Find the time it takes for each pressure to exceed 0.6
threshold = 0.6;
timeA = data.time(find(data.pressureA > threshold, 1));
timeB = data.time(find(data.pressureB > threshold, 1));
timeC = data.time(find(data.pressureC > threshold, 1));

% Display the results
disp(['Time for Pressure A to exceed ' num2str(threshold) ': ' num2str(timeA)]);
disp(['Time for Pressure B to exceed ' num2str(threshold) ': ' num2str(timeB)]);
disp(['Time for Pressure C to exceed ' num2str(threshold) ': ' num2str(timeC)]);

I have create a table using time and 3 pressure columns A,B,C and I have used matlab inbuilt function find inorder to find the exact time after the value has crossed certain threshold which is 0.6 in this case.I hope it resolves your query.

Answer 1

[Matt J]

% Generate example data
time = (0:0.1:1)';
pressureA = 20*sin(time/3);
pressureB = 20*sin(pi*time/4);
pressureC = 20*sin(pi*time/5);

% Combine data into a table
data = table(time, pressureA, pressureB, pressureC)

data = 11×4 table
    time    pressureA    pressureB    pressureC
    ____    _________    _________    _________

      0            0           0            0  
    0.1      0.66654      1.5692       1.2558  
    0.2       1.3323      3.1287       2.5067  
    0.3       1.9967      4.6689       3.7476  
    0.4       2.6588      6.1803       4.9738  
    0.5       3.3179      7.6537       6.1803  
    0.6       3.9734      9.0798       7.3625  
    0.7       4.6244       10.45       8.5156  
    0.8       5.2703      11.756       9.6351  
    0.9       5.9104      12.989       10.717  
      1       6.5439      14.142       11.756  

[~,i]=max(data{:,2:end}>6);

result = time(i)'

result = 1×3

1.0000 0.4000 0.5000