Buntalan
Buntalan

Reputation: 183

matlab percentage change between cells

I'm a newbie to Matlab and just stumped how to do a simple task that can be easily performed in excel. I'm simply trying to get the percent change between cells in a matrix. I would like to create a for loop for this task. The data is setup in the following format:

                DAY1 DAY2 DAY3...DAY 100

SUBJECT RESULTS

I could only perform getting the percent change between two data points. How would I conduct it if across multiple days and multiple subjects? And please provide explanation

Thanks a bunch

FOR EXAMPLE, FOR DAY 1 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5), DAY 2 SUBJECT1(RESULT=2), SUBJECT2(RESULT=8), SUBJECT3(RESULT=10), DAY 3 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5). I WANT THE PERCENT CHANGE SO OUTPUT WILL BE DAY 2 SUBJECT1(RESULT=100%), SUBJECT2(RESULT=100%), SUBJECT3(RESULT=100%). DAY3 SUBJECT1(RESULT=50%), SUBJECT2(RESULT=50%), SUBJECT3(RESULT=50%)

updated:

Hi thanks for responding guys. sorry for the confusion. zebediah49 is pretty close to what I'm looking for. My data is for example a 10 x 10 double. I merely wanted to get the percentage change from column to column. For example, if I want the percentage change from rows 1 through 10 on all columns (from columns 2:10). I would like the code to function for any matrix dimension (e.g., 1000 x 1000 double) zebediah49 could you explain the code you posted? thanks

updated2: zebediah49,

(data(1:end,100)- data(1:end,99))./data(1:end,99)

output=[data(:,2:end)-data(:,1:end-1)]./data(:,1:end-1)*100;

Observing the code above, How would I go about modifying it so that column 100 is used as the index against all of the other columns(1-99)? If I change the code to the following:

(data(1:end,100)- data(1:end,:))./data(1:end,:)

matlab is unable because of exceeding matrix dimensions. How would I go about implementing that?

UPDATE 3

zebediah49,

Worked perfectly!!! Originally I created a new variable for the index and repmat the index to match the matrices which was not a good idea. It took forever to replicate when dealing with large numbers. Thanks for you contribution once again.

Thanks Chris for your contribution too!!! I was looking more on how to address and manipulate arrays within a matrix.

Upvotes: 1

Views: 3490

Answers (2)

zebediah49
zebediah49

Reputation: 7611

It's matlab; you don't actually want a loop.

output=input(2:end,:)./input(1:end-1,:)*100;

will probably do roughly what you want. Since you didn't give anything about your matlab structure, you may have to change index order, etc. in order to make it work.

If it's not obvious, that line defines output as a matrix consisting of the input matrix, divided by the input matrix shifted right by one element. The ./ operator is important, because it means that you will divide each element by its corresponding one, as opposed to doing matrix division.

EDIT: further explanation was requested: I assumed you wanted % change of the form 1->1->2->3->1 to be 100%, 200%, 150%, 33%. The other form can be obtained by subtracting 100%.

input(2:end,:) will grab a sub-matrix, where the first row is cut off. (I put the time along the first dimension... if you want it the other way it would be input(:,2:end). Matlab is 1-indexed, and lets you use the special value end to refer to the las element. Thus, end-1 is the second-last. The point here is that element (i) of this matrix is element (i+1) of the original.

input(1:end-1,:), like the above, will also grab a sub-matrix, except that that it's missing the last column.

I then divide element (i) by element (i+1). Because of how I picked out the sub-matrices, they now line up.

As a semi-graphical demonstration, using my above numbers:

input:          [1 1 2 3 1]
input(2,end):     [1 2 3 1]
input(1,end-1): [1 1 2 3]

When I do the division, it's first/first, second/second, etc.

input(2:end,:)./input(1:end-1,:):
   [1   2   3   1  ]
./ [1   1   2   3  ]
---------------------
== [1.0 2.0 1.5 0.3]

The extra index set to (:) means that it will do that procedure across all of the other dimension.

EDIT2: Revised question: How do I exclude a row, and keep it as an index. You say you tried something to the effect of (data(1:end,100)- data(1:end,:))./data(1:end,:). Matlab will not like this, because the element-by-element operators need them to be the same size. If you wanted it to only work on the 100th column, setting the second index to be 100 instead of : would do that. I would, instead, suggest setting the first to be the index, and the rest to be data. Thus, the data is processed by cutting off the first:

output=[data(2:end,2:end)-data(2:end,1:end-1)]./data(2:end,1:end-1)*100;

OR, (if you neglect the start, matlab assumes 1; neglect the end and it assumes end, making (:) shorthand for (1:end).

output=[data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100;

However, you will probably still want the indices back, in which case you will need to append that subarray back:

output=[data(1,1:end-1) data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100];

This is probably not how you should be doing it though-- keep data in one matrix, and time or whatever else in a separate array. That makes it much easier to do stuff like this to data, without having to worry about excluding time. It's especially nice when graphing.

Oh, and one more thing:

(data(:,2:end)-data(:,1:end-1))./data(:,1:end-1)*100;

is identically equivalent to

data(:,2:end)./data(:,1:end-1)*100-100;

Upvotes: 4

Christopher Creutzig
Christopher Creutzig

Reputation: 8774

Assuming zebediah49 guessed right in the comment above and you want

1  4  5
2  8 10
1  4  5

to turn into

   1    1    1
 -.5  -.5  -.5

then try this:

data = [1,4,5; 2,8,10; 1,4,5];
changes_absolute = diff(data);
changes_absolute./data(1:end-1,:)

ans =

    1.0000    1.0000    1.0000
   -0.5000   -0.5000   -0.5000

You don't need the intermediate variable, you can directly write diff(data)./data(1:end,:). I just thought the above might be easier to read. Getting from that result to percentage numbers is left as an exercise to the reader. :-)

Oh, and if you really want 50%, not -50%, just use abs around the final line.

Upvotes: 2

Related Questions