I got about ten thousand data points from the sensor. This data was extracted as an excel .csv file and loaded using readtable, resulting in a 10093x1 table.
However, the problem I am experiencing is that the 1st, 10th, 19th, etc., of these data are unnecessary values, so I want to exclude them.
I've found that these numbers have one thing in common: a row with a remainder of 1 when divided by 9.
How can I remove the rows where the divisor with 9 equals 1?
You're just removing every ninth row, so you can even use simple colon indexing
data(1:9:end, :) = [];
Alternatively, using your divisor theory, you can do this by using mod() and logical indexing:
idxRemove = mod(1:size(data,1),9) == 1; % True if remainder after division by 9 is 1
data(idxRemove,:) = []; % Remove unwanted rows
You can one-line it for efficiency, since idxRemove isn't stored in that case:
data(mod(1:size(data,1),9) == 1,:) = [];
In either way, using mod() will be slower and less readable than using the colon indexing.
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.
Starting wish a 7x4 binary matrix I need to change a random bit in each column to simulate error. Have been trying to no avail.
A very straightforward way to do this is to use a for loop. It might not be the most efficient approach in MATLAB, but it's probably good enough considering your data set is so small.
Iterate through each of the four columns. On each iteration, randomly chose a number from 1 to 7 to represent the row in that column that you have selected to change. Finally, flip the bit at that row/column. The following code does just this. Assume that "A" is a binary matrix with 7 rows and 4 columns
for col=1:4; %// Iterate through each column
row = ceil(7*rand()); %// Randomly chose a number from 1 to 7 to represent row
A(row,col) = ~A(row,col); %// Flip the bit at the specified row/col
end
Another possibility is to create 4 random numbers in one call, and assign in a vectorized fashion:
rowNumbers = randi(4,[1 4])
A(rowNumbers,:) = ~A(rowNumbers,:);
Suppose I have 121 elements and want to get all combinations of 4 elements taken at a time, i.e. 121c4.
Since combnk(1:121, 4) takes a lot of time, I want to go for 2% of that combination by providing:
z = 1:50:length(121c4(:, 1))
For example: 1st row, 5th row, 100th row and so on, up to 121c4, picking only those rows from a 121c4 matrix without generating the complete combination (it's consuming too much for large numbers like 625c4).
If you haven't defined an ordering on the combinations, why not just use
randi(121,p,4)
where p is the number of combinations you want in your set ? With this approach you may, or may not, want to replace duplicates.
If you have defined an ordering on the combinations, tell us what it is.
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.