select corresponding values out of different columns of a matrix

Hi, We have a matrix (480 x 23) and we need to extract data from it. For example: in column 5 there are the numbers of 1 to 6 and we need to extract the numbers 1. At the same we need the corresponding reaction times of these values which can be found in column 14. Thus far, we have

[n,m]=size(trialList);
for k=1:n;
  r = trialList(k,5);
if  r == 1
 disp ('trailList14');
end
end

We are able to find the 80 rows where number 1 is mentioned. But instead showing the 80 corresponding reaction times, Matlab shows 80 times the word 'triallist14'.

How will we be able to select the corresponding reaction times with the corresponding values of column 5?

Answer 1

[Birdman]

 vals=trialList(find(trialList(:,5)==1),[5 14]);

[Guillaume]

find is not needed and a waste of time. Use the logical array returned by the comparison directly:

   vals = trialList(trialList(:, 5) == 1, [5 14]);

[Birdman]

Ups yes, even if I know that, I can not stop myself from using it.

[Birdman]

Julia's question moved here:

Thank you so much! we are not very common with Matlab so this platform is very useful :)

We have one other question... For another assignment, we want to extract more data. This is based on a selection of two columns. First, we need to make again a selection of the numbers 1 to 6 in column 5. Then when we have for example selected number 1 (in column 5), and we want to select from another column a number (1 to 5) in column 19, and then we want the corresponding reaction time of these values from column 14. Can we integrate this in the code we already got from you?

[Birdman]

Yes, try this(consider you select 1 from column 5 and 2 from column 19):

 vals = trialList(trialList(:,5)==1 & trialList(:,19)==2,14)

I put an AND operator, but you can change it to OR operator according to your need. That was just for an example.

[Dotje123]

thank you, the AND operator worked fine for us!

[Birdman]

You are welcome!

Answer 2

[Dotje123]

We were very content with the last solution you gave us. Now we are stuck again. We test more and more participants and every time we have to make new loops for a single participant. We were wondering if there is a way to get the output for all participants separately and only using one set of loops. thusfar, we have loaded a map with all the files of the participants.

load(['\\map...s' num2str(subjectNums(s)) '_2.mat'])

      for vals1_1_5 = trialList(trialList(:,5)==1 & trialList(:,19)==5,14);
          median1_1_5 = nanmedian(vals1_1_5)
      end

      for vals1_2_5 = trialList(trialList(:,5)==2 & trialList(:,19)==5,14);
          median1_2_5 = nanmedian(vals1_2_5)
      end

    for vals1_3_5 = trialList(trialList(:,5)==3 & trialList(:,19)==5,14);
        median1_3_5 = nanmedian(vals1_3_5)
    end

    for vals1_4_5 = trialList(trialList(:,5)==4 & trialList(:,19)==5,14);
        median1_4_5 = nanmedian(vals1_4_5)
    end

    for vals1_5_5 = trialList(trialList(:,5)==5 & trialList(:,19)==5,14);
        median1_5_5 = nanmedian(vals1_5_5)
    end

    for vals1_6_5 = trialList(trialList(:,5)==6 & trialList(:,19)==5,14);
        median1_6_5 = nanmedian(vals1_6_5)
    end

    participant1_5 = [median1_1_5 median1_2_5 median1_3_5 median1_4_5 median1_5_5 median1_6_5]

    %plot (overalmedian1, overalmedian2, overalmedian3, overalmedian4,
    %overalmedian5, overalmedian6, overalmedian7, overalmedian8, overalmedian9,
    %overalmedian10)

    load('s2_2');

    for vals2_1_5 = trialList(trialList(:,5)==1 & trialList(:,19)==5,14);
        median2_1_5 = nanmedian(vals2_1_5)
    end

    for vals2_2_5 = trialList(trialList(:,5)==2 & trialList(:,19)==5,14);
        median2_2_5 = nanmedian(vals2_2_5)
    end

    for vals2_3_5 = trialList(trialList(:,5)==3 & trialList(:,19)==5,14);
        median2_3_5 = nanmedian(vals2_3_5)
    end

    for vals2_4_5 = trialList(trialList(:,5)==4 & trialList(:,19)==5,14);
        median2_4_5 = nanmedian(vals2_4_5)
    end

    for vals2_5_5 = trialList(trialList(:,5)==5 & trialList(:,19)==5,14);
        median2_5_5 = nanmedian(vals2_5_5)
    end

    for vals2_6_5 = trialList(trialList(:,5)==6 & trialList(:,19)==5,14);
        median2_6_5 = nanmedian(vals2_6_5)
    end

    participant2_5 = [median2_1_5 median2_2_5 median2_3_5 median2_4_5 median2_5_5 median2_6_5]

    load('s3_2');

    for vals3_1_5 = trialList(trialList(:,5)==1 & trialList(:,19)==5,14);
        median3_1_5 = nanmedian(vals3_1_5)
    end

    for vals3_2_5 = trialList(trialList(:,5)==2 & trialList(:,19)==5,14);
        median3_2_5 = nanmedian(vals3_2_5)
    end

    for vals3_3_5 = trialList(trialList(:,5)==3 & trialList(:,19)==5,14);
        median3_3_5 = nanmedian(vals3_3_5)
    end

    for vals3_4_5 = trialList(trialList(:,5)==4 & trialList(:,19)==5,14);
        median3_4_5 = nanmedian(vals3_4_5)
    end

    for vals3_5_5 = trialList(trialList(:,5)==5 & trialList(:,19)==5,14);
        median3_5_5 = nanmedian(vals3_5_5)
    end

    for vals3_6_5 = trialList(trialList(:,5)==6 & trialList(:,19)==5,14);
        median3_6_5 = nanmedian(vals3_6_5)
    end

    participant3_5 = [median3_1_5 median3_2_5 median3_3_5 median3_4_5 median3_5_5 median3_6_5]

To be clear, we now make a selection per participant for every condition. What we want, is to be able to load all the participants into one set of loops (shown below), and get an output of all the medians per participant.

   for vals1_1_5 = trialList(trialList(:,5)==1 & trialList(:,19)==5,14);
            median1_1_5 = nanmedian(vals1_1_5)
        end
        for vals1_2_5 = trialList(trialList(:,5)==2 & trialList(:,19)==5,14);
            median1_2_5 = nanmedian(vals1_2_5)
        end
      for vals1_3_5 = trialList(trialList(:,5)==3 & trialList(:,19)==5,14);
          median1_3_5 = nanmedian(vals1_3_5)
      end
      for vals1_4_5 = trialList(trialList(:,5)==4 & trialList(:,19)==5,14);
          median1_4_5 = nanmedian(vals1_4_5)
      end
      for vals1_5_5 = trialList(trialList(:,5)==5 & trialList(:,19)==5,14);
          median1_5_5 = nanmedian(vals1_5_5)
      end
      for vals1_6_5 = trialList(trialList(:,5)==6 & trialList(:,19)==5,14);
          median1_6_5 = nanmedian(vals1_6_5)
      end
      participant1_5 = [median1_1_5 median1_2_5 median1_3_5 median1_4_5 median1_5_5 median1_6_5]

Thank you in advance

[Guillaume]

Please start a new question.

Your design is flawed. A simple rule: if you start numbering variables (such as your valsx_y_z) you're doing it wrong.

We can discuss that in your new question.

[Stephen23]

@Julia Verhaegh: The problem is having lots of numbered variables, and then trying to access them. You would be much better of keeping your data together as much as possible, using vectors/matrices/arrays. When beginners use lots of numbered/sequential variable names then they force themselves into either copy-and-pasting code (huge waste of your time) or writing slow, complex, buggy code (huge waste of your computer's time). Read this to know more:

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval

Once you keep your data in just a few arrays then accessing it is trivial using indexing. Indexing is neat, very simple, and extremely efficient. It avoids all of the problems with numbered variables (which are just de-facto indices, so why not turn them into real indices?).

[Razvan Carbunescu]

While there might be a few more aspects to your question it seems like you are trying to compute medians for groups of values.

In R2018a you can accomplish this easier by keeping the data together and using groupsummary.

If using an earlier release can use the findgroups / splitapply model to achieve the same result.