This is the input matrix
7 9 6
8 7 9
7 6 7
Based on the frequency their appearance in the matrix (Note. these values are for explanation purpose. I didn't pre-calculate them in advance. That why I ask this question)
number frequency
6 2
7 4
8 1
9 2
and the output I expect is
4 2 2
1 4 2
4 2 4
Is there a simple way to do this?
Here's a three-line solution. First prepare the input:
X = [7 9 6;8 7 9;7 6 7];
Now do:
[a m n] = unique(X);
b = hist(X(:),a);
c = reshape(b(n),size(X));
Which gives this value for c:
4 2 2
1 4 2
4 2 4
If you also wanted the frequency matrix, you can get it with this code:
[a b']
Here is a code with for-loop (a is input matrix, freq - frequency matrix with 2 columns):
weight = zeros(size(a));
for k = 1:size(freq,1)
weight(a==freq(k,1)) = freq(k,2);
end
Maybe it can be solved without loops, but my code looks like:
M = [7 9 6 ;
8 7 9 ;
7 6 7 ;];
number = unique(M(:));
frequency = hist(M(:), number)';
map = containers.Map(number, frequency);
[height width] = size(M);
result = zeros(height, width); %allocate place
for i=1:height
for j=1:width
result(i,j) = map(M(i,j));
end
end
Related
I would like to align and count vectors with different time stamps to count the corresponding bins.
Let's assume I have 3 matrix from [N,edges] = histcounts in the following structure. The first row represents the edges, so the bins. The second row represents the values. I would like to sum all values with the same bin.
A = [0 1 2 3 4 5;
5 5 6 7 8 5]
B = [1 2 3 4 5 6;
2 5 7 8 5 4]
C = [2 3 4 5 6 7 8;
1 2 6 7 4 3 2]
Now I want to sum all the same bins. My final result should be:
result = [0 1 2 3 4 5 6 7 8;
5 7 12 16 ...]
I could loop over all numbers, but I would like to have it fast.
You can use accumarray:
H = [A B C].'; %//' Concatenate the histograms and make them column vectors
V = [unique(H(:,1)) accumarray(H(:,1)+1, H(:,2))].'; %//' Find unique values and accumulate
V =
0 1 2 3 4 5 6 7 8
5 7 12 16 22 17 8 3 2
Note: The H(:,1)+1 is to force the bin values to be positive, otherwise MATLAB will complain. We still use the actual bins in the output V. To avoid this, as #Daniel says in the comments, use the third output of unique (See: https://stackoverflow.com/a/27783568/2732801):
H = [A B C].'; %//' stupid syntax highlighting :/
[U, ~, IU] = unique(H(:,1));
V = [U accumarray(IU, H(:,2))].';
If you're only doing it with 3 variables as you've shown then there likely aren't going to be any performance hits with looping it.
But if you are really averse to the looping idea, then you can do it using arrayfun.
rng = 0:8;
output = arrayfun(#(x)sum([A(2,A(1,:) == x), B(2,B(1,:) == x), C(2,C(1,:) == x)]), rng);
output = cat(1, rng, output);
output =
0 1 2 3 4 5 6 7 8
5 7 12 16 22 17 8 3 2
This can be beneficial for particularly large A, B, and C variables as there is no copying of data.
Unfortunately my programming skills are not that advanced and I really need to vectorize some loops to finish my thesis.
I tried to make things really clear and simple and I have the following two questions in matlab:
1.
If we have a 5x5 matrix A and we want to set the diagonal elements of this matrix to the diagonal of a matrix B, apart from diag(A)=diag(B) we could use :
for i=1:5
B(i,i)=A(i,i)
end
Now if I want to vectorize this I can not use:
i=1:5
B(i,i)=A(i,i)
In that way we assign each combination from 1:5. So, in the end we asign each element of A equal to B and not the diagonal.
Is there some way that we could assign each identical pair of (i,i)?
I tried :
i=1:5
j=1:5
B(i,find(j==i))=A(i,find(j==i))
But still does not work. I repeat I know the diag property but Im only interested on the particular problem.
2.
A similar problem is the fillowing.
b=[ones(2,2) ones(2,2)*2 ones(2,2)*3 ones(2,2)*4] ;
a = zeros(8,12);
for i=1:4
a((i-1)*2+1:(i)*2,(i-1)*3+1:(i)*3) = [8*ones(2,1) b(:,[2*(i-1)+1 2*i])];
end
Thank you for your time and for your help.
Let's bring some mask magic, shall we!
Problem #1
mask = eye(size(A))==1
A(mask) = B(mask)
For creating the mask, you can also use bsxfun -
N = size(A,1)
bsxfun(#eq,[1:N]',1:N)
Or finally, you can use linear indexing -
N = size(A,1)
A(1:N+1:N^2) = B(1:N+1:N^2)
Sample run -
>> A
A =
5 2 9 6 5
9 1 6 2 2
9 7 5 3 9
4 5 8 8 7
7 5 8 1 8
>> B
B =
5 5 2 8 2
1 1 6 5 2
7 8 5 4 4
1 8 9 8 8
1 7 6 1 8
>> mask = eye(size(A))==1;
>> A(mask) = B(mask)
A =
5 2 9 6 5
9 1 6 2 2
9 7 5 3 9
4 5 8 8 7
7 5 8 1 8
Problem #2
%// Append 8's at the start of every (2,2) block in b
b1 = reshape([8*ones(2,4) ; reshape(b,4,[])],2,[])
%// Mask where b1 values are to be put in an otherwise zeros filled array
mask = kron(eye(4,4),ones(2,3))==1
%// Initialize output arraya and set values from b1 into masked places
out = zeros(size(mask))
out(mask) = b1
For your first problem. Use logical indexing:
index = diag(ones(1,size(B,1))
B(index) = A(index)
I have variable A of size m by n. I want to generate B of size m by m*n, such as below example.
Example:
A = [1 2 3;
4 5 6;
7 8 9]
Should result with
B = [1 2 3 4 5 6 7 8 9;
1 2 3 4 5 6 7 8 9;
1 2 3 4 5 6 7 8 9]
Is there any way to do that without using loop? m and n is variable.
You should use the repmat Matlab funtion:
B = repmat(A,M,N) creates a large matrix B consisting of an M-by-N
tiling of copies of A. The size of B is [size(A,1)*M, size(A,2)*N].
The statement repmat(A,N) creates an N-by-N tiling.
For your specific case one solution may be:
A=A';
B=repmat(A(:)',3,1);
And for the general case one solution may be:
A_aux=reshape(A',1,size(A,1)*size(A,2));
B=repmat(A_aux,size(A,1),1);
Repmat is indeed the way to go here, as mentioned by #Nerea. This solution should give the same answer as his, but personally I consider it to be a bit more elegant:
B=repmat(reshape(A',1,[]),size(A,1),1);
To include a quite fast bsxfun solution:
A = [1 2 3 4;
5 6 7 8;
9 10 11 12]
A = A.' %'
B = bsxfun(#plus,zeros(size(A,2),1),A(:).')
or use kron, but its surely slower:
A = A.'
B = kron(A(:),ones(1,size(A,2))).'
B =
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12
No repmats
[m n] = size(A);
B = ones(m,1) * reshape( A.', 1, [] );
One approach
with repmat, reshape and permute combo!
out = repmat(reshape(permute(A,[3 2 1]),1,[]),size(A,1),1,1);
or Another approach without reshape but becomes a 2 liner
out1 = permute(A,[3 2 1]);
out = repmat(out1(:).',size(A,1),1,1);
Referring to Reshape row wise w/ different starting/ending elements number #Divakar came with a nice solution but, what if the number of columns is not always the same?
Sample run -
>> A'
ans =
4 9 8 9 6 1 8 9 7 7 7 4 6 2 7 1
>> out
out =
4 9 8 9 0 0
6 1 8 9 7 7
7 4 6 2 7 1
I took only the first 4 terms of A and put them in out, then fill the rest 2 empty cell with 0's. So the ncols = [4 6 6]. Unfortunately vet2mat doesn't allow vector as columns number.
Any suggestions?
You can employ bsxfun's masking capability here -
%// Random inputs
A = randi(9,1,15)
ncols = [4 6 5]
%// Initialize output arary of transposed size as compared to the desired
%// output arary size, as we need to insert values into it row-wise and MATLAB
%// follows column-major indexing
out = zeros(max(ncols),numel(ncols));
mask = bsxfun(#le,[1:max(ncols)]',ncols); %//'# valid positions mask for output
out(mask) = A; %// insert input array elements
out = out.' %//'# transpose output back to the desired output array size
Code run -
A =
5 3 7 2 7 2 4 6 8 1 9 7 5 4 5
ncols =
4 6 5
out =
5 3 7 2 0 0
7 2 4 6 8 1
9 7 5 4 5 0
You could use accumarray for that:
A = [4 9 8 9 6 1 8 9 7 7 7 4 6 2 7 1].'; %'// data
ncols = [4 6 6]; %// columns
n = max(ncols);
cs = cumsum(ncols);
ind = 1;
ind(cs+1) = 1;
ind = cumsum(ind(1:end-1)); %// `ind` tells the row for each element of A
result = accumarray(ind(:), A(:), [], #(x) {[x; zeros(n-numel(x),1)]}); %// split `A` as
%// dictated by `ind`, and fill with zeros. Each group is put into a cell.
result = [result{:}].'; %'// concatenate all cells
Say we have the following matrix
1 3 6
5 4 7
5 3 9
What I'm trying to do is for each row, I assign it the maximum value of the column. So, for instance, I'm expecting the following output:
x(1) = 6
x(2) = 7
x(3) = 9
I tried doing that using by writing the code below, but didn't get the expected putput:
x=[1 3 6;5 4 7; 5 3 9]
[rows, columns] = size(x);
for i=1:columns
for j=1:rows
[maximum, position] = max(x(j,:));
disp('MAXIMUM')
x(j)=maximum
end
end
What should I do to get the expected output?
You can use the built-in max function with a dimension specifier: max(x,[],dim).
In your case, assuming your matrix is called A:
>> x=max(A,[],2)
ans =
6
7
9
If I understood correctly your question, you can just use the max function. It naturally operates on columns, therfore, some transposition is necessary.
x=[1 3 6;5 4 7; 5 3 9]
y = max(x')'
y =
6
7
9
You can even reassing the values on the fly
x = max(x')'.