I am trying to figure out how to split a vector in matlab into subvectors.
I am solving a differential equation numerically using dde23. When You do this the length of the solution vector changes. Thus, I am finding it not so easy to use the mat2cell command that many people suggest.
All I am trying to do is split (as evenly as possible) a vector of length N into an arbitrary amount of sub-vectors whose length may vary depending on the length of the time vector. I am doing this so then I can find the maximum value of each vector on in each interval.
If I understand the question, maybe you can try to split it by using code below
dataset=[ 1 2 3 4 5 6 7 8 9 10]
splitpoint = randi[2 length(dataset)-1]
subset1 = dataset(1,1:splitpoint)
splitpoint = randi[length(subset1)+1 length(dataset)-1]
subset2 = dataset(1,length(subset1)+1:splitpoint)
After that you can choose where to finish and accept rest of it for last subset or you can define one list to hold each subset in the row of the list. So you can define while loop to handle it automatically by defining stop_criteria.
As the question says:
x = [1 2 3]
1 - 2 - 3 = -4
How do I get my -4?
Without a for-loop.
diff doesn't work. I don't get how Matlab has a way to SUM, but not a way to subtract.
It's an odd operation, I doubt there's a builtin function for that, but an easy way to do it would be:
2*x(1)-sum(x)
Another alternative is using dot product:
y = x*[1 ; -ones(numel(x)-1, 1)];
Here x is assumed a row vector.
Though probably not the best solution in your case, it can be nice say if you want other pattern for the summation, e.g. with weights - just replace the vector to the right by a vector of weights.
I want to reduce a two dimensional matrix to row vector.
But using reshape with large matrices is really slow. The other alternative is to use colon, but i want matrix's transpose to be colon and not the matrix itself.
e.g.
A=magic(3)
A =
8 1 6
3 5 7
4 9 2
A(:) will stack up all the columns one by one. but i am looking for something like this:
AA=A(2:3,:)';
and then reshape or colon AA instead of A.
The issue is i dont want to define additional variable like AA.
Is there anyway to reduce dimension of two dimensional matrix without reshape?
You can avoid the additional variable by linear indexing. For your example:
A([2 5 8 3 6 9])
which gives
3 5 7 4 9 2
What's happening here is that you treat A as if it was already transformed into a vector, and the elements of this one-dimensional array are accessed through indices 1 through 9. Using the colon is a special case of linear indexing, A(:) is the same as A(1 : end).
Figuring out the right linear indices can be tricky, but sub2ind can help with that.
It is possible that this slightly speeds up the code, mainly because (as #Shai wrote) you avoid writing data to an intermediate variable. I wouldn't expect too much, though.
Try looking at subsref. For your example, you could use it as follows:
subsref(A',struct('type','()','subs',{{2:3,':'}}))
Update: I misunderstood the original question; I thought the OP wanted to select rows from 2:3 from the transposed A matrix keeping the columns as is. I will keep the previous answer in case it could be useful to others.
I think he/she could use the following to slice and flatten a matrix:
subsref(A(2:3,:)', struct('type','()','subs',{{':'}}))
This would give as output:
[3 5 7 4 9 2]'
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.
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;
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.
I am using cross valind function on a very small data... However I observe that it gives me incorrect results for the same. Is this supposed to happen ?
I have Matlab R2012a and here is my output
crossvalind('KFold',1:1:11,5)
ans =
2
5
1
3
2
1
5
3
5
1
5
Notice the absence of set 4.. Is this a bug ? I expected atleast 2 elements per set but it gives me 0 in one... and it happens a lot that is the values are not uniformly distributed in the sets.
The help for crossvalind says that the form you are using is: crossvalind(METHOD, GROUP, ...). In this case, GROUP is the e.g. the class labels of your data. So 1:11 as the second argument is confusing here, because it suggests no two examples have the same label. I think this is sufficiently unusual that you shouldn't be surprised if the function does something strange.
I tried doing:
numel(unique(crossvalind('KFold', rand(11, 1) > 0.5, 5)))
and it reliably gave 5 as a result, which is what I would expect; my example would correspond to a two-class problem (I would guess that, as a general rule, you'd want something like numel(unique(group)) <= numel(group) / folds) - my hypothesis would be that it tries to have one example of each class in the Kth fold, and at least 2 examples in every other, with a difference between fold sizes of no more than 1 - but I haven't looked in the code to verify this.
It is possible that you mean to do:
crossvalind('KFold', 11, 5);
which would compute 5 folds for 11 data points - this doesn't attempt to do anything clever with labels, so you would be sure that there will be K folds.
However, in your problem, if you really have very few data points, then it is probably better to do leave-one-out cross validation, which you could do with:
crossvalind('LeaveMOut', 11, 1);
although a better method would be:
for leave_out=1:11
fold_number = (1:11) ~= leave_out;
<code here; where fold_number is 0, this is the leave-one-out example. fold_number = 1 means that the example is in the main fold.>
end