I trying to plot a function which has millions of points. Therefore it is not possible to plot the function as it would crash my computer. However, the entire plot consists of "elbows" in that more than 1 variable cannot change at a time.
Consider the following matrix
a = [1 2 3 4 4 4 4];
b = [1 1 1 1 2 3 4];
These points make the following figure when using plot(a,b)
However, upon closer inspection, I can reduce my plot vectors down to
a = [1 4 4];
b = [1 1 4];
and get the same plot.
My question is: what is a vectorized way to automatically remove every redundant point. I could easily loop through and search for points to remove but this would be expensive. I have tried using combinations of diff and cumsum to do this to no avail. I only have straight lines and right angles in the plot so I do not have to worry about anything of the form ax+b
You can use a diff(diff(a))~=0, or equivalently diff(a, 2)~=0, to get a perhaps cleaner looking solution. This will work whether you plug a or b into the diff function.
mat = [true, diff(a, 2)~=0, true];
a(mat)
ans =
1 4 4
b(mat)
ans =
1 1 4
Guess I'll answer my own question in case anyone else needs this. Basically what I'm doing is doing the first diff on each matrix to see if the values are changing between each element. The second diff looks for a change. Because we're looking for elbows, we end up switching from true on the x-axis to true on the y-axis. This is the point we're looking for. Then we add on the ends. From there we only keep the true elements for our original matrix.
mat = [true diff(logical(diff(a))-logical(diff(b))) true];
a(mat)
[1 4 4]
b(mat)
[1 1 4]
I'm open to alternative suggestions as this is very hacky.
Related
There is a downloadable function called CARTPROD which gives the cartesian product of given vectors
(link to CARTPROD function)
For example
cartprod(1:3,1:3)
ans =
1 1
2 1
3 1
1 2
2 2
3 2
1 3
2 3
3 3
However, is there a way in which I can specify how many times a given vector should be read in the cartesian product. I want something like this:
%If user chooses the vector to be used 4 times
cartprod(1:3,1:3,1:3, 1:3)
%If user chooses the vector to be used 2 times
cartprod(1:3,1:3)
I have tried thinking about it, but I can't think of doing it any way besides manually. Thanks!
What you're looking for is comma separated lists. Haven't tested this, but try
myvec={1:3,1:3,1:3,1:3};
cartprod(myvec{:}); %get cartprod of all vectors in the cell-array.
or as #Sardar_Usama pointed out, you can replace myvec={1:3,1:3,1:3,1:3} with this:
n=4; %number of repeated vectors
myvec=repmat({1:3},1,n); %repeat cell-array {1:3} 4 times
The other answer points out how you can use the same cartprod function from the FEX. However, there is another function named combvec (from the Neural Network Toolbox) which does exactly the same.
n = 4; %Number of times to be repeated
myvec = repmat({1:3},1,n); %Repeating the cell array vector
result = combvec(myvec{:}).'; %Converting to comma-separated list and applying combvec
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'm having troubles with randomly shuffling a vector without repeating numbers (ex. 1 1 is not acceptable but 1 2 is acceptable), given that each value is repeated equally.
More specifically, I would like to repeat the matrix [1:4] ten times (40 elements in total) so that 1, 2, 3 and 4 would all repeat 10 times without being consecutive.
If there is any clarification needed please let me know, I hope this question was clear.
This is what I have so far:
cond_order = repmat([1:4],10,1); %make matrix
cond_order = cond_order(:); %make sequence
I know randperm is quite relevant but I'm not sure how to use it with the one condition of non-repeating numbers.
EDIT: Thank you for all the responses.
I realize I was quite unclear. These are the examples I would like to reject [1 1 2 2 4 4 4...].
So it doesn't matter if [1 2 3 4] occurs in that order as long as individual values are not repeated. (so both [1 2 3 4 1 2 3 4...] and [4 3 1 2...] are acceptable)
Preferably I am looking for a shuffled vector meeting the criteria that
it is random
there are no consecutively repeating values (ex. 1 1 4 4)
all four values appear equal amount of times
Kind of working with the rejection sampling idea, just repeating with randperm until a sequence permutation is found that has no repeated values.
cond_order = repmat(1:4,10,1); %//make matrix
N = numel(cond_order); %//number of elements
sequence_found = false;
while ~sequence_found
candidate = cond_order(randperm(N));
if all(diff(candidate) ~= 0) %// check if no repeated values
sequence_found = true;
end
end
result = candidate;
The solution from mikkola got it methodically right, but I think there is a more efficient way:
He chose to sample based on equal quantities and check for the difference. I chose to do it the other way round and ended up with a solution requiering much less iterations.
n=4;
k=10;
d=42; %// random number to fail first check
while(~all(sum(bsxfun(#eq,d,(1:n).'),2)==k)) %' //Check all numbers to appear k times.
d=mod(cumsum([randi(n,1,1),randi(n-1,1,(n*k)-1)]),n)+1; %generate new random sample, enforcing a difference of at least 1.
end
A subtle but important distinction: does the author need an equal probability of picking any feasible sequence?
A number of people have mentioned answers of the form, "Let's use randperm and then rearrange the sequence so that it's feasible." That may not work. What will make this problem quite hard is if the author needs an equal chance of choosing any feasible sequence. Let me give an example to show the problem.
Imagine the set of numbers [1 2 2 3 4]. First lets enumerate the set of feasible sequences:
6 sequences beginning with 1: [1 2 3 2 4], [1 2 3 4 2], [1 2 4 2 3], [1 2 4 3 2], [1 3 2 4 2], [1 4 2 3 2].
Then there are 6 sequences beginning with [2 1]: [2 1 2 3 4], [2 1 2 4 3], [2 1 3 2 4], [2 1 3 4 2], [2 1 4 2 3], [2 1 4 3 2]. By symmetry, there are 18 sequences beginning with 2 (i.e. 6 of [2 1], 6 of [2 3], 6 of [2 4]).
By symmetry there are 6 sequences beginning with 3 and another 6 starting with 4.
Hence there are 6 * 3 + 18 = 36 possible sequences.
Sampling uniformly from feasible sequences, the probability the first number is 2 is 18/36 = 50 percent! BUT if you just went with a random permutation, the probability the first digit is 2 would be 40 percent! (i.e. 2/5 numbers in set are 2)
If equal probability of any feasible sequence is required, you want 50 percent of a 2 as the first number, but naive use of randperm and then rejiggering numbers at 2:end to make sequence feasible would give you a 40 percent probability of the first digit being two.
Note that rejection sampling would get the probabilities right as every feasible sequence would have an equal probability of being accepted. (Of course rejection sampling becomes very slow as probability of being accepted goes towards 0.)
Following some of the discussion on here, I think that there is a trade-off between performance and the theoretical requirements of the application.
If a completely uniform draw from the set of all valid permutations is required, then pure rejection sampling method will probably be required. The problem with this of course is that as the size of the problem is increased, the rejection rate will become very high. To demonstrate this, if we consider the base example in the question being n multiples of [1 2 3 4] then we can see the number of samples rejected for each valid draw as follows (note the log y axis):
My alternative method is to randomly sort the array, and then if duplicates are detected then the remaining elements will again be randomly sorted:
cond_order = repmat(1:4,10,1); %make matrix
cond_order = reshape(cond_order, numel(cond_order), 1);
cond_order = cond_order(randperm(numel(cond_order)));
i = 2;
while i < numel(cond_order)
if cond_order(i) ~= cond_order(i - 1)
i = i + 1;
else
tmp = cond_order(i:end);
cond_order(i:end) = tmp(randperm(numel(tmp)));
end
end
cond_order
Note that there is no guarantee that this will converge, but in the case where is becomes clear that it will not converge, we can just start again and it will still be better that re-computing the whole sequence.
This definitely meets the second two requirements of the question:
B) there are no consecutive values
C) all 4 values appear equal amount of times
The question is whether it meets the first 'Random' requirement.
If we take the simplest version of the problem, with the input of [1 2 3 4 1 2 3 4] then there are 864 valid permutations (empirically determined!). If we run both methods over 100,000 runs, then we would expect a Gaussian distribution around 115.7 draws per permutation.
As expected, the pure rejection sampling method gives this:
However, my algorithm does not:
There is clearly a bias towards certain samples.
In the end, it depends on the requirements. Both methods sample over the whole distribution so both fill the core requirements of the problem. I have not included performance comparisons, but for anything other than the simplest of cases, I am confident that my algorithm would be much faster. However, the distribution of the draws is not perfectly uniform. Whether it is good enough is dependent on the application and the size of the actual problem.
I'm sure this is a very simple mistake by me somewhere! But when I use Matlab's graycomatrix function, I don't get the expected result. Instead of a matrix output I expect, I always get an 8 x 8 (nearly) zero matrix with one entry in the bottom right - usually equal to 16. I haven't changed the default settings or used 'offset', so I'm not too sure what the problem is.
That's because your image is not normalized!
Your image should be range 0-1, so:
I = [1 1 2; 2 2 3; 1 2 5]; %or any other I
glcm = graycomatrix(I/max(I(:))); % or I/255 , but it would not work for this example
should do the job.
In your case, Matlab interprets that everything avobe 1 is 1, therefore the co-occurrence matrix gives you a unique value in the max position.
This is a question about Matlab/Octave.
I am seeing some results of the medfilt1(1D Median filter command in matlab) computation by which I am confused.
EDIT:Sorry forgot to mention:I am using Octave for Windows 3.2.4. This is where i see this behavior.
Please see the questions below, and point if I am missing something.
1] I have a 1D data array b=[ 3 5 -8 6 0];
out=medfilt1(b,3);
I expected the output to be [3 3 5 0 0] but it is showing the output as [4 3 5 0 3]
How come? What is wrong here?
FYI-Help says it pads the data at boundaries by 0(zero).
2] How does medfilt2(2D median filter command in matlab) work.
Help says "Each output pixel contains the median value in the m-by-n neighborhood around the corresponding pixel in the input image".
For m=3,n=3, So does it calculate a 3x3 matrix MAT for each of input pixels placed at its center and do median(median(MAT)) to compute its median value in the m-by-n neighbourhood?
Any pointers will help.
thank you. -AD
I was not able to replicate your error with Matlab 7.11.0, but from the information in your question it seems like your version of medfilt1 does not differentiate between an odd or even n.
When finding the median in a vector of even length, one usually take the mean of the two median values,
median([1 3 4 5]) = (3+4)/2 = 3.5
This seems to be what happens in your case. Instead of treating n as odd, and setting the value to be 3, n is treated as even and your first out value is calculated to be
median([0 3 5]) = (3+5)/2 = 4
and so on.. EDIT: This only seems to happen in the endpoints, which suggest that the padding with zeros is not properly working in your Octave code.
For your second question, you are almost right, it calculates a 3x3 matrix in each center, but it does not do median(median(MAT)), but median(MAT(:)). There is a difference!
A = [1 2 3
14 5 33
11 7 13];
median(median(A)) = 11
median(A(:)) = 7