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I need to start with the highest value in the matrix (i,j). Then, go backwards to one of positions (i-1,j), (i, j-1), and (i-1, j-1) depending on the direction of movement used to construct the matrix. This method is used throughout until a matrix cell with zero value is reached.The above image shows what I want, I need to trace those marked in blue. I know there is a max function in matlab.
If I understand what you try to do...
Suppose M is your matrix:
result=[];
[~,ind]=max(M(:));
[r,c]=ind2sub(size(M),ind);
result=[result; r c];
while M(r,c)~=0
[~,f]=max([M(r-1,c),M(r,c-1),M(r-1,c-1)]);
switch f
case 1
r=r-1;
case 2
c=c-1;
case 3
r=r-1;
c=c-1;
end
result=[result; r c];
end
Then, result is a 2 x n matrix, where each row is the indexes of one step- from the upper to lower.
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In a sample code I have X(1:n, 1)
I don't understand how it works
n=3 in a sample
Is it a matrix? but isn't () used for indexing so how it's a double index?
I imagine that your X is a two-dimmensional matrix. Generalizing, it can be said that to access the elements of a matrix is done with: X(n,m).
The common case is to get one single number, in that case n and m are integer numbers. But you can also pass vectors to n and m positions and that way extract a submatrix from the original one.
As an example:
X = [1 2 3; 4 5 6; 7 8 9]
X(1:3,1) = [1; 4; 7]
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So for my probability class, my professor asked the following question on a homework problem:
A fair coin is flipped 10,000 times. Let X correspond to the difference between the number of heads and the number of tails. Using MATLAB, compute the expected value of X.
This is what I wrote to answer the question:
N = 10000;
i =0;
r=1/2;
Q=nchoosek(N,(X+N)/2);
X=(1,N);
for i=-N:N
P=Q*r.^(X+N)/2*(1-r)^(N-(X+N)/2) % probability_mass_function
E=sum(abs(X).*P); % expected value
end
However, is there faster and quicker way to compute this expected value? Any help would be greatly appreciated. Thanks
You can put all tests results in single matrix in one line and then calculate X per test, then average over X:
clear
TAIL=0; HEAD=1;
NumTests=121;
NumRollsPerTest=10*1000;
AllTestsRolls= rand(NumTests,NumRollsPerTest)>0.5 ; %head when rand>0.5
XperTest=sum(AllTestsRolls==HEAD,2)-sum(AllTestsRolls==TAIL,2);%every row is test so calc per test
ExpectedX=sum(XperTest)/length(XperTest)
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I would like to pick two different random elements from given array with their positions. Similar like with datasample, but with datasample there is a possibility of picking the same element twice.
I could use while loop or similar, but I suppose there is an easier way to do it.
Say you have a matrix A:n by m, you can choose two elements at random as following,
A=[2 7 8;5 4 6;8 3 11];%given array
[n m]=size(A);
x=2;%two different random elements
i=randperm(n,x)%row index for x elements
j=randperm(m,x)%column index for x elements
A(i(1),j(1)) %First random element
A(i(2),j(2)) %Second random element
If you try this you could get something like,
i =
2 3
j =
2 1
ans =
4
ans =
8
the code can be further simplified but just wanted to make it clear. Please let me know if you have any additional questions or require further clarification.
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I have 3 signals:
a = ecg(1000); // clean ecg with no noise
b = a + noise1; // with noise component on the ECG
c = a + noise2; // a,b,and c have the same dimension (1000x1)
now, i want to put the concatenated signal onto a single variable, x, so that the output of x would be:
x = a concatenated with b; b concatenated with c;
If by cascading you mean concatenating, simply do this:
x = [a b c];
This simply pieces a, b and c together so that they are one signal and the three signals are side-by-side each other. Doing ecg(1000) creates a 1 x 1000 vector. Assuming your noisy signals after that respect the same dimensions, then the above syntax should hold.
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How can I retrieve the last element of a vector only, provided that the length of that vector is unknown?
Use the special end keyword:
lastelement = myvector(end);
If the vector is called A, just use A(end).
In this case, use end, like #nispio and #David answered.
But it seems you think that not knowing the length can be a problem, but nope. That's because you can use length(v) if v is a column or row vector, or size(M) if M is a matrix.
Then, to get the last element of your vector, you could use (not recommended):
v(length(v)) if v is a row or a column vector
v(size(v,1)) if v is a column vector
v(size(v,2)) if v is a row vector
But if you use one of them, MATLAB will warn you:
The operation or expression <Indexing> has no evident effect.