The matrix <1x500> consists of different values, now I want to check if any of the values in the matrix occurs at least 3 times or more.
if (val occurs 3 times or more)
do
Help is very appreciated!
Another option from #KiW answer for when you need to know all the values that do appear at least 3 times is:
uniqA=unique(A);
counts=histcounts(A,[uniqA inf]);
vals_that_are_bigger=uniqA(counts>=3);
To check if any of them are bigger than 3, just
if any(counts>=3)
if numel(find(matrix)==val)>3
whatever you want to do
end
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 !!
I had previously wrote some code to split 3 columns into 4, however the code was very inefficient and time consuming. As I am working with millions of rows it wasn't suitable. (Below is my previous code)
tline = fgetl(fid);
ID=tline(1:4);
IDN = str2double(ID);
Day=tline(6:8);
DayN = str2double(Day);
HalfHour=tline(9:10);
HalfHourN = str2double(HalfHour);
Usage=tline(12:end);
UsageN = str2double(Usage);
There must be a more efficient and quicker way of doing this?
Going back to basics, I have produced a x by 3 matrix. but require an x by 4 matrix
To show what I am trying to do, examining one row -
I am trying to change
1001 36501 1005
to
1001 365 01 1005
Any help would be much appreciated!
Edit:
The second column I am trying to divide into two, is always composed of 5 characters. I am trying to get the first 3 characters into their own column, likewise for the remaining characters.
What might take time in your case is actually the use of the str2double function. It is known that this built-in function becomes very slow when the data set is large. You might try to get rid of it if possible.
you can use modulo
ans = (36501 - mod(36501,100))/100
This would give you 365
if you want the 1, it is mod(36501,100)
so this would effectively split your second column into 2 different numbers, you can then re name them etc.
hmmm on second thoughts, if all your numbers on your second column are 5 digits, this can be extremely efficient, since mod is computed in matlab by b = a - m.*floor(a./m);
check http://uk.mathworks.com/help/matlab/ref/mod.html it should work for vectors (i.e. your second column)
,A simple question but one I have yet to get a concrete answer. If I have two matrices, say A and B and I want to make them both the same size, say a 1x2 matrix of zeros. Is there a way to declare them both in one line of code? I ask because in my situation I will have over 10 matrices of the same size but I want an easier way to declare them.
So at first I might think it would look like the following (which is not valid):
A,B = zeros(1,2)
The deal()-function does exactly what you are looking for. You can distribute either one input to several output-variables or also distribute several input values to several output values. You need the first case:
[A, B] = deal(zeros(1,2));
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.