finding unknown values in a column by using the indices (indexes) of known values.

Hello, I'm struggling a lot with some of the indexing methods. Right now I have the next matrix:

    1.5000    1.7024
    1.5000    1.3119
    1.5000    1.3122
    0.5000    0.8158
    0.5000    1.1760
(it's actually a 200 by 2 matrix, but I have shortened it to make things clearer)

The values in row 1 have been provided by linspace(0.5, 1.5, 5) and the values in row 2 have been entered by the user (they are the user's reaction times, and the latter [the values in col 1] are the stimuli durations).

What I'm trying to do is to find all the user's input when the duration was 1.5000 and find the mean of those values.

So far I used: [r,c]=find(mat==1.5000), and got:

r =

     1
     2
     3
    10
    17
 What I need now is the values next to those rows (that is the ones in col 2) and find their mean.

Hope I was clear and concise. Thanks in advance

Answer 1

[Walter Roberson]

mat(r, 2)

Note: be cautious on those floating point comparisons! http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

[Juan Pablo]

It worked but I didn't understand why should I be cautious. Anyways, thank you very much!

[Ryan]

He is warning you that find(variable == specific number) may not actually return "true" values (matches) that you're expecting because of the reasons outlined in that Matlab Wikia post. It may not apply here, but it's a good tidbit to know about to save yourself potentially from future headaches.

[Juan Pablo]

Interesting, I'll take that into account, thanks

Answer 2

[Sean de Wolski]

accumarray time! (happy time :) )

x =[1.5000    1.7024
  1.5000    1.3119
  1.5000    1.3122
  0.5000    0.8158
  0.5000    1.1760]

[y,~,idxu] = unique(x(:,1)); %indexes of each row into each unique value
y(:,2) = accumarray(idxu,x(:,2),[],@mean) %their mean

[Juan Pablo]

>> y(:,2) = accumarray(idxu,x(:,2),[],@mean) Index exceeds matrix dimensions.

[Juan Pablo]

Something didn't work :(

[Juan Pablo]

But thanks anyway, for your kind support

[Sean de Wolski]

What do you have defined as x? If you clear you workspace and copy the above in, it works.

[Juan Pablo]

Ok I'll try that

[Juan Pablo]

This helped me a lot more, wish I knew how to use all those functions!

[Sean de Wolski]

doc unique
doc accumarray

!

Answer 3

[Elizabeth]

Probably not as efficient as those above, but this is the simplest way I always tried to do it.

j= find(row1==1.5);
for k=1:length(j)
d(k)=c(j(k));
end
meanvalues=mean(d)

[Sean de Wolski]

why not just:

m = mean(x(x(:,1)==1.5,2))

[Juan Pablo]

row1 doesn't seem to be a legal function on version R2011B

[Juan Pablo]

but thanks for your support

[Walter Roberson]

row1 = mat(:,1);

It is a column, but Elizabeth has called it a row because you confused rows and columns when you phrased the question.

[Juan Pablo]

That is true, thanks for the observation