using linspace in matlab - matlab

I want to create a vector which runs from 1 to 260 with increments of 360 between every whole number.
I can do this manually by: y=linspace(1,2,360); y1=linspace(2,3,360);... and so on.
By combining these I would have a vector which was 260*360=93600 long. However, there must be a easier way of doing this? preferably without a loop.

Maybe you can just do:
n=261;
linspace(1,n,(n-1)*360);

And what about y=(1:1/360:260) ?
Or if you want to have exactly 360 elements between 1 and 2 (included) as it seems from your use of linspace(1,2,360) you could do y=(1:1/359:260).
Also, your final vector would have less than 360*260 elements as you have to account for duplicates.

Related

Extract data using Matlab

How could I correlate the first (also second and third, etc) line of each function together? I meant I want to pick up the value of the first line of function 1, first line of function 2 and so on... Specially number of these lines (between two #...#) are not always equal.
Let's say the values between (# -#) is one matrix. I see that the problem is I need to break them down into different matrix and then equalize all these matrix to the same size by adding NaN values and then reconstruct them again. That is what I think but I dont know how to ask Matlab do these tasks.
Would you please give me a help?
Thank you very much !!

Predictive coding without using loops in Matlab?

I have a set of numbers and I want to use predictive coding to get smaller values for this set of data as each value should not differ to much from the last. I was just starting very simply with a expected value that each value would be the same as the last, and then just store the error.
For some simple data:
1 2 -3 1
The values I should get are
1 1 -5 4
The way I was doing this compression was one line, but to decompress I need the last value, so I put this in a loop. Is there a way to do this, and possibly more complex(looking at more the just the last value) predictive coding without needing to use Matlab loops.
As suggested by #Divakar, constructing the new values can be done by
B = [A(1) diff(A)];
Obtaining the original from this result, the inverse of the above procedure, is done through
A = cumsum(B);

Matlab : How to extract elements in a uniform interval?

I have a (5160 X 4) matrix.
I want to extract just (1,1),(41,1),(81,1),(121,1)........ in a uniform interval, only from the first column of the matrix.
Assuming that data is your matrix, you can do this:
A = data(1:40:5160,1);
The 1:40:5160 will create an array such that it starts at 1, and goes up in increments of 40 as much as possible up until 5160. Once you create 1:40:5160, you can use this array and access the corresponding rows, and you are accessing the first column using the index of 1 for the second parameter. Actually, the last row that gets extracted is 5121. We aren't able to go up to 5161 due to the fact that your matrix has 5160 rows, and we have also specified 5160 as the ending of the indexing.
NB: This is very basic MATLAB syntax. Any standard MATLAB tutorial should teach you this.

Matlab: 'Matrix dimensions must agree' less than operator (<)?

A quick question because i fear there may already be an answer (although i cant find it)
i am getting the error: Matrix dimensions must agree.
because i am useing '<'
now with all the other operators there is a way around this either by putting '.' infront or by using a different formula. So what do people do about the less than operator????
i don't see why the greater than or equal to (>=) works but yet less than does not!?
am i being stupid and missed something really obvious??
code snippet
matrix 1 represents an array of 16 numbers
matrix 2 can represents anywhere between 10 and 20 numbers
idx = (matrix2 >= matrix1 * 0.1 & matrix2 < matrix1 * 1.5);
any help guidance or advice on the topic would be much appreciated! thank you!
EDIT
i know the matrices are different sizes but is there a way to use less then with different size arrays? as im not bothered about the size of the array but the numbers within
If you want to compare parts of matrices, like M(1:3,10:12)>A(5:7,1:3), you, probably, have to use the function squeeze():
squeeze(M(1:3,10:12))>squeeze(A(5:7,1:3))
This function remotes singleton dimensions and everything works fine after.

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.
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.