I was able to list all the prime numbers up to an inputted number, however as I am inexperienced at coding on Scratch, I have had difficulty constructing a list of all the perfect squares below a number.
For example, if you input 17 the output should be 16,9,4.
The main way to detect if a number is a perfect square is to take the square root and round it and then multiply it by itself. If it is a perfect square the answer will be the number you started with.
Let's try with 4: The square root of 4 is 2 and when you round it you get 2 and when you multiply 2 times 2 you do get 4 so 4 is a perfect number.
Let's try with 5: The square root of 5 is 2.23606797749979 and when you round it you get 2 and when you multiply 2 times 2 you do NOT get 5.
There are two ways to write the loop: you can count up or you can count down. Either way, for each number you do the test and if the number is a perfect square you display it.
counting down:
counting up:
I'm sure this is a trivial question for a signals person. I need to find the function in Matlab that outputs averaging of contiguous segments of windowsize= l of a vector, e.g.
origSignal: [1 2 3 4 5 6 7 8 9];
windowSize = 3;
output = [2 5 8]; % i.e. [(1+2+3)/3 (4+5+6)/3 (7+8+9)/3]
EDIT: Neither one of the options presented in How can I (efficiently) compute a moving average of a vector? seems to work because I need that the window of size 3 slides, and doesnt include any of the previous elements... Maybe I'm missing it. Take a look at my example...
Thanks!
If the size of the original data is always a multiple of widowsize:
mean(reshape(origSignal,windowSize,[]));
Else, in one line:
mean(reshape(origSignal(1:end-mod(length(origSignal),windowSize)),windowSize,[]))
This is the same as before, but the signal is only taken to the end minus the extra values less than windowsize.
I am having problem with MATLAB's maximum function. What I am supposed to do is to replace the maximum value of an array with a number. However, when there are more than one maximal value, the program updates all of them simultaneously. Is there a way to make it do it one by one? The order of replacement is not important; it can be done arbitrarily. The only important thing is to have MATLAB doing it one by one.
Thank you in advance.
The second output of max returns one index:
a=[5,5];
[b,idx]=max(a)
c=b-2;
a(idx)=c
When you say more than 1 maximum value, I assume you are talking about a matrix where max function operates on every column?
You can do the following:
a = [1 1 2;5 5 7; 3 2 9]
Obviously, the maximum value is going to be 9, but if you do the following:
max(a)
The result will be:
5 5 9
Based on each column.
The following may work for you?
max(a(:)) % Maximum value from a matrix (rerranged into 1 column)
you can do the same for the min function.
I found the detail and implementation of Local Ternary Pattern (LTP) on Calculating the Local Ternary Pattern of an image?. I want to ask more details that what the best way to choose the threshold t and also I have confusion in understand the role of reorder_vector = [8 7 4 1 2 3 6 9];
Unfortunately there isn't a good way to figure out what the threshold is using LTPs. It's mostly trial and error or by experimentation. However, I could suggest to make the threshold adaptive. You can use Otsu's algorithm to dynamically determine the best threshold of your image. This is assuming that the distribution of your intensities in the image is bimodal. In other words, there is a clear separation between objects and background. MATLAB has an implementation of this by the graythresh function. However, this generates a threshold between 0 and 1, so you will need to multiply the result by 255, assuming that the type of your image is uint8.
Therefore, do:
t = 255*graythresh(im);
im is the image that you desire to compute the LTPs. Now, I can certainly provide insight on what the reorder_vector is doing. Look at the following figure on how to calculate LTPs:
(source: hindawi.com)
When we generate the ternary code matrix (matrix in the middle), we need to generate an 8 element sequence that doesn't include the middle of the neighbourhood. We start from the east most element (row 2, column 3), then traverse the elements in counter-clockwise order. The reorder_vector variable allows you to select those specific elements that respect that order. If you recall, MATLAB can access matrices using column-major linear indices. Specifically, given a 3 x 3 matrix, we can access an element using a number from 1 to 9 and the memory is laid out like so:
1 4 7
2 5 8
3 6 9
Therefore, the first element of reorder_vector is index 8, which is the east most element. Next is index 7, which is the top right element, then index 4 which is the north facing element, then 1, 2, 3, 6 and finally 9.
If you follow these numbers, you will determine how I got the reorder_vector:
reorder_vector = [8 7 4 1 2 3 6 9];
By using this variable for accessing each 3 x 3 local neighbourhood, we would thus generate the correct 8 element sequence that respects the ordering of the ternary code so that we can proceed with the next stage of the algorithm.
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.