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

I want to find the mean of the values in the second column only for the rows which have the value 3 in column 10 I have tried mean = mean(data(data(:,10)==3,8))

Answer 1

[Bruno Luong]

You command is correct (beside using 8th column instead of 2nd) BUT:

Don't name your (output) variable MEAN, you will have conflict later when using function MEAN.

[Catherine Branter]

thank you for this advice - yes, unfortunately it still doesn't work for me as it says Index exceeds the number of array elements (0). col2avg = mean(data(data(:,10)==3,2))

however, i manage to get the right answer with this code

column = data(data(:,10)==3,2)
[m,n] = size(column)
meancol = sum(column)/m

[Bruno Luong]

You have to make

>> clear mean

to undo the shadowing you made before

[Catherine Branter]

oh awesome, thank you!

[Catherine Branter]

that works!!

[Swazi Sharma]

Hey guys! If I want to do the same thing, but change the parameter from 3 to a word (for example: "January"). Can i just change the 3 to January, or do I have to change other bits in code?

[DGM]

Comparisons between words don't really work the same way, and the approach depends on the variable type.

For character vectors:

myword = 'apple';
myword == 'orange' % results in error if the words aren't the same size
strcmpi(myword,'orange') % this would be a valid comparison

If you don't need case-insensitivity, use strcmp() instead.

If the variables are actual strings (in double quotes), the behavior is a bit different.

myword = "apple"
myword == "orange" % this actually does work with strings
strcmpi(myword,"orange") % this still works with either

Similarly, the indexing expressions you use to apply this comparison would depend entirely on the type of array in which these variables exist. Practical arraying of strings and character vectors vary, so I can't guess what you have.

If I presume that you're using a cell array (note the case-insensitive behavior of strcmpi()):

A = {12 56 'orange';
     64 46 'apple';
     32 36 'Peach';
     64 16 'apple';
     32 26 'peAch';
     64 41 'apple';
     32 72 'peach';
     64 26 'apple';
     32 86 'peach';
     34 48 'banana'}

% mean of col2 for rows where col3 is 'peach'
mn = mean([A{strcmpi(A(:,3),'peach'),2}])

ans =
    55

Otherwise, you'll have to adapt this to whatever array type you're using.

Answer 2

[Matt J]

mean(data(data(:,10)==3,2))

[Catherine Branter]

that's the same as what i had except with a 2 - i had previously been trying to find it for the 8th column. this way isnt working!

[Matt J]

Nobody can know what you mean when you say it is not working when you do not show what you see and what you expect.

[madhan ravi]

Exactly I second @Matt ,+1 it works for me

[Catherine Branter]

this is the error i get:

Index exceeds the number of array elements (0).

Error in ML_Coursework (line 16) output = mean(data(data(:,10)==3,2))

[Bruno Luong]

Catherine Branter : look at my answer bellow