Very new to coding and Matlab here!
I have a cell array (16*1) (1 column, 16 rows)
In each of the cells, I have one 76*76 matrix.
In each of the matrices, I'd like to delete certain rows (e.g. 1 to 10, 22 and 76) and the same columns (1 to 10, 22, and 76).
I have no idea how to go about that. I've tried to read the getting started documentation, but it's still unclear for me.. I'd be very appreciative of any help and explanation!
Thank you :)
There are two ways to delete elements. One is to assign [] to the rows and columns you want to delete, the other is to select what you want.
%some example data
x={magic(76),magic(76)}
%First option assign []
x{1}([1:10,22,76],:)=[]
x{1}(:,[1:10,22,76])=[]
%Second option, index what you need:
%All values from 11 to 75 without 22
selector=setdiff(11:75,22)
x{2}=x{2}(selector,selector)
Obviously instead of the hard coded {1} or {2}, you would call these in a for loop.
Some advice, when you have a cell of equally sized matrices and you want to perform the same operation on each of them, it is likely that a (76,76,16) is better suited for you. To stick with above example:
x=ones(76,76,16);
x=x(selector,selector,:);
All done in a single line.
I have results which are 6 columns long however have been printed as 2 then 3 beneath then 1 beneath that! There are hundreds of lines and matlab will not except the structure of the matrix as it is now. Is there any way to tell matlab i want the first 5 results in their own columns then continuing down the rows after that?
My results appear as follows:
0.5 0
0.59095535915335684063 -0.59095535915335395405 -5.89791913085569763
33e-08
... repeated alot
thansk so much, em xx
I would just do a format shortE before you process the output, this will give you everything in scientific notation with 4 digits after the decimal. That 'should' allow you to fit your columns all in one line, so you don't have to deal with the botched output.
In general you should not want the output to be in a too specific format, but suppose you have this matrix:
M =[0.5 0 0.59095535915335684063 -0.59095535915335395405 -5.89791913085569763 33e-08];
To make it an actual matrix I will repeat it a bit:
M = repmat(M,10,1);
Now you can ensure that all six columns will fit on a normal screen by using the format.
format short
Try help format to find more options. Now simply showing the matrix will put all columns next to eachother. If you want one column below, the trick is to reduce your windows width untill it can only hold five columns. Matlab will now print the last column below the first.
M % Simply show the matrix
% Now reduce your window size
M % Simply show it again
This should help you display the numbers in matlab, if you want to process them further you can consider to write them to a file instead. Try help xlswrite for a simple solution.
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.
I have a data of 9 columns 14470 rows,
The first column is filled with 0 or 1. Zero means that there is no measurment and the whole row is not in my interest.... can some body help me in writing a loop which go through all lines and filter the data when in first column 1 exist?
You do not need a loop for this, remember Matlab is a matrix oriented programming language, loops should be avoided. I won't give you the answer, I think you can figure it out yourself, it's easy. This tutorial will help.
Have fun.
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.