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Run-length decoding in MATLAB
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I have an array freq with frequencies and another one val with values.
val =[1 3 5 7];
freq=[2 3 3 2];
I want to get the array result.
result=[1 1 3 3 3 5 5 5 7 7];
One of the methods I've tried to get result is:
freq=[2 3 3 2];
val=[1 3 5 7];
result=[];
for i=1:length(val);
result=[result repmat(val(i),1,freq(i))];
end
It works, but with large arrays I expect some performance gain if I get rid of the for-loop. Is there any built in function for this? How would you calculate result for large arrays?
This can be made that way:
val = [1 3 5 7]
freq = [2 3 3 2]
res = repelem(val, freq)
res =
1 1 3 3 3 5 5 5 7 7
For large vectors, you could gain some performance by preallocating result and updating multiple cells at the same time.
result = zeros(sum(freq), 1);
j = 1;
for i=1:length(freq);
result(j:j+freq(i)-1) = val(i);
j = j + freq(i);
end;
Related
I have a Matrix A. I want to iterate over the inner part of the matrix (B), while also working with the rows and columns that are not part of B.
A = [1 4 5 6 7 1; B = [2 2 2 2;
8 2 2 2 2 1; 2 3 3 2;
9 2 3 3 2 1; 2 8 2 2];
0 2 8 2 2 1;
1 1 1 1 1 1];
I know it is possible to select the part of A like this:
[rows,columns] = size(A);
B = A([2:1:rows-1],[2:1:columns-1]);
for i = 1:(rows*columns)
%do loop stuff
endfor
This however wont work because I also need the outer rows and columns for my calculations. How can I achieve a loop without altering A?
So, why do not use two indexes for the inner matrix?
%....
for i=2:rows-1
for j=2:cols-1
% here, A(i,j) are the B elements, but you
% can still access to A(i-1, j+1) if you want.
end
end
%....
I want to change the individual elements in a matrix by different values simultaneously.
How do I do that?
For example: I want to change the first element in matrix A by certain amount and the second element by a different amount simultaneously.
{ A = [1; 2]
% instead of doing A(1) = .....
A(2) = .....
}
You can access the elements of a vector or matrix and replace them.
For a vector this is intuitive.
octave:16> A = 1:9
A =
1 2 3 4 5 6 7 8 9
octave:17> A([1 3 5 7 9]) = 0
A =
0 2 0 4 0 6 0 8 0
This can be done for a matrix as well. The elements of a matrix are arranged in a column-first manner. You can use a single index to access the elements of a matrix.
octave:18> A = [1 2 3; 4 5 6; 7 8 9]
A =
1 2 3
4 5 6
7 8 9
The 2nd element of A is the same as A(2, 1). The 4th element of A is the same as A(1, 2).
octave:21> A(2)
ans = 4
octave:22> A(4)
ans = 2
So, you can set all the odd elements of A to 0 in one go like this:
octave:19> A([1 3 5 7 9]) = 0
A =
0 2 0
4 0 6
0 8 0
Just add a vector with the differences. A += [0.1; 0.2]
octave:1> A = [1; 2];
octave:2> A += [0.1; 0.2]
A =
1.1000
2.2000
Suppose I have an array X of size n by p by q. I would like to reshape it as a matrix with p rows, and in each row put the concatenation of the n rows of size q, resulting in a matrix of size p by nq.
I managed to do it with a loop but it takes a while say if n=1000, p=300, q=300.
F0=[];
for k=1:size(F,1)
F0=[F0,squeeze(X(k,:,:))];
end
Is there a faster way?
I think this is what you want:
Y = reshape(permute(X, [2 1 3]), size(X,2), []);
Example with n=2, p=3, q=4:
>> X
X(:,:,1) =
0 6 9
8 3 0
X(:,:,2) =
4 7 1
3 7 4
X(:,:,3) =
4 7 2
6 7 6
X(:,:,4) =
6 1 9
1 4 3
>> Y = reshape(permute(X, [2 1 3]), size(X,2), [])
Y =
0 8 4 3 4 6 6 1
6 3 7 7 7 7 1 4
9 0 1 4 2 6 9 3
Try this -
reshape(permute(X,[2 3 1]),p,[])
Thus, for code verification, one can look into a sample case run -
n = 2;
p = 3;
q = 4;
X = rand(n,p,q)
F0=[];
for k=1:n
F0=[F0,squeeze(X(k,:,:))];
end
F0
F0_noloop = reshape(permute(X,[2 3 1]),p,[])
Output is -
F0 =
0.4134 0.6938 0.3782 0.4775 0.2177 0.0098 0.7043 0.6237
0.1257 0.8432 0.7295 0.2364 0.3089 0.9223 0.2243 0.1771
0.7261 0.7710 0.2691 0.8296 0.7829 0.0427 0.6730 0.7669
F0_noloop =
0.4134 0.6938 0.3782 0.4775 0.2177 0.0098 0.7043 0.6237
0.1257 0.8432 0.7295 0.2364 0.3089 0.9223 0.2243 0.1771
0.7261 0.7710 0.2691 0.8296 0.7829 0.0427 0.6730 0.7669
Rather than using vectorization to solve the problem, you could look at the code to try and figure out what may improve performance. In this case, since you know the size of your output matrix F0 should be px(n*q), you could pre-allocate memory to F0 and avoid the constant resizing of the matrix at each iteration of the for loop
n=1000;
p=300;
q=300;
F0=zeros(p,n*q);
for k=1:size(F,1)
F0(:,(k-1)*q+1:k*q) = squeeze(F(k,:,:));
end
While probably not as efficient as the other two solutions, it is an alternative. Try the above and see what happens!
I have a problem with removing the rows when the columns are identical.
I have used a for and if loop but the run time is too long.
I was thinking if there are any more efficient and faster run time method.
say
A=[ 2 4 6 8;
3 9 7 9;
4 8 7 6;
8 5 4 6;
2 10 11 2]
I would want the result to be
A=[ 2 4 6 8;
4 8 7 6;
8 5 4 6]
eliminating the 2nd row because of the repeated '9' and remove the 5th row because of repeated '2'.
You can use sort and diff to identify the rows with repeated values
A = A(all(diff(sort(A'))),:)
returns
A =
2 4 6 8
4 8 7 6
8 5 4 6
The trick here is how to find the rows with repeated values in an efficient manner.
How about this:
% compare all-vs-all for each row using `bsxfun`
>> c = bsxfun( #eq, A, permute( A, [1 3 2] ) );
>> c = sum( c, 3 ); % count the number of matches each element has in the row
>> c = any( c > 1, 2 ); % indicates rows with repeated values - an element with more than one match
>> A = A( ~c, : )
A =
2 4 6 8
4 8 7 6
8 5 4 6
My data is a 2096x252 matrix of double values. I need a for loop or an equivalent which performs the following:
Each time the matrix is reproduced the first array is deleted and the second becomes the first. When the loop runs again, the remaining matrix is reproduced and the first array is deleted and the next becomes the first and so on.
I've tried using repmat but it is too slow and tedious when dealing with large matrices (2096x252).
Example input:
1 2 3 4
3 4 5 6
3 5 7 5
9 6 3 2
Desired output:
1 2 3 4
3 4 5 6
3 5 7 5
9 6 3 2
3 4 5 6
3 5 7 5
9 6 3 2
3 5 7 5
9 6 3 2
9 6 3 2
Generally with Matlab it is much faster to pre-allocate a large array than to build it incrementally. When you know in advance the final size of the large array there's no reason not to follow this general advice.
Something like the following should do what you want. Suppose you have an array in(nrows, ncols); then
indices = [0 nrows:-1:1];
out = zeros(sum(indices),ncols);
for ix = 1:nrows
out(1+sum(indices(1:ix)):sum(indices(1:ix+1)),:) = in(ix:end,:);
end
This worked on your small test input. I expect you can figure out what is going on.
Whether it is the fastest of all possible approaches I don't know, but I expect it to be much faster than building a large matrix incrementally.
Disclaimer:
You'll probably have memory issues with large matrices, but that is not the question.
Now, to the business:
For a given matrix A, the straightforward approach with the for loop would be:
[N, M] = size(A);
B = zeros(sum(1:N), M);
offset = 1;
for i = 1:N
B(offset:offset + N - i, :) = A(i:end, :);
offset = offset + size(A(i:end, :), 1);
end
B is the desired output matrix.
However, this solution is expected to be slow as well, because of the for loop.
Edit: preallocated B instead of dynamically changing size (this optimization should achieve a slight speedup).