I have a 3D array. I need to remove any elements that are in the same row, column position but on the next page (3rd dimension), and only use the first occurrence at that position. So if all pages were to multiply the result would be 0.
Since the 3D array may be of any size, I can't hard code solutions like isMember. I also can't use unique because elements can be the same, just not share the same position.
For example, input:
A(:,:,1) = [ 1 0 2];
A(:,:,2) = [ 1 1 0];
A(:,:,3) = [ 0 1 0];
the desired output is:
A(:,:,1) = [ 1 0 2];
A(:,:,2) = [ 0 1 0];
A(:,:,3) = [ 0 0 0];
How can I accomplish this?
Not the most elegant, but at least it works.
A(:,:,1) = [ 1 0 2 ];
A(:,:,2) = [ 1 1 0 ];
A(:,:,3) = [ 0 1 0 ];
for ii = 1:size(A,1)
for jj = 1:size(A,2)
unique_el = unique(A(ii, jj, :)); % Grab unique elements
for kk = 1:numel(unique_el)
idx = find(A(ii,jj,:) == kk); % Contains indices of unique elements
if numel(idx) > 1 % If an element occurs more than once
A(ii, jj, idx(2:end)) = 0; % Set to 0
end
end
end
end
A
A(:,:,1) =
1 0 2
A(:,:,2) =
0 1 0
A(:,:,3) =
0 0 0
I loop over the first two dimensions of A (rows and columns), find any unique elements which occur on a certain row and column location through the third dimensions (pages). Then set all occurrences of a unique element after the first to 0.
Given a more elaborate 3D matrix this still works:
A(:,:,1) = [1 0 2 0; 2 1 3 0];
A(:,:,2) = [1 1 0 0; 2 2 1 0];
A(:,:,3) = [0 1 1 3; 1 2 2 4];
A(:,:,1) =
1 0 2 0
2 1 3 0
A(:,:,2) =
0 1 0 0
0 2 1 0
A(:,:,3) =
0 0 1 3
1 0 2 4
If you want the first non-zero element and discard any element occurring afterwards, simply get rid of the unique() call:
A(:,:,1) = [1 0 2 0; 2 1 3 0];
A(:,:,2) = [1 1 0 0; 2 2 1 0];
A(:,:,3) = [0 1 1 3; 1 2 2 4];
for ii = 1:size(A,1)
for jj = 1:size(A,2)
idx = find(A(ii,jj,:) ~= 0); % Contains indices of nonzero elements
if numel(idx) > 1 % If more than one element
A(ii, jj, idx(2:end)) = 0; % Set rest to 0
end
end
end
A(:,:,1) =
1 0 2 0
2 1 3 0
A(:,:,2) =
0 1 0 0
0 0 0 0
A(:,:,3) =
0 0 0 3
0 0 0 4
My solution assumes, that, for a given "position", EVERY value after the first occurence of any value is cleared. Some of the MATLAB regulars around here had some discussions on that, from there comes the "extended" example as also used in Adriaan's answer.
I use permute and reshape to rearrange the input, so that we have all "positions" as "page" columns in a 2D array. Then, we can use arrayfun to find the proper indices of the first occurence of a non-zero value (kudos to LuisMendo's answer here). Using this approach again, we find all indices to be set to 0.
Let's have a look at the following code:
A(:,:,1) = [1 0 2 0; 2 1 3 0];
A(:,:,2) = [1 1 0 0; 2 2 1 0];
A(:,:,3) = [0 1 1 3; 1 2 2 4]
[m, n, o] = size(A);
B = reshape(permute(A, [3 1 2]), o, m*n);
idx = arrayfun(#(x) find(B(:, x), 1, 'first'), 1:size(B, 2));
idx = arrayfun(#(x) find(B(idx(x)+1:end, x)) + idx(x) + 3*(x-1), 1:size(B, 2), 'UniformOutput', false);
idx = vertcat(idx{:});
B(idx) = 0;
B = permute(reshape(B, o, m , n), [2, 3, 1])
Definitely, it makes sense to have a look at the intermediate outputs to understand the functioning of my approach. (Of course, some lines can be combined, but I wanted to keep a certain degree of readability.)
And, here's the output:
A =
ans(:,:,1) =
1 0 2 0
2 1 3 0
ans(:,:,2) =
1 1 0 0
2 2 1 0
ans(:,:,3) =
0 1 1 3
1 2 2 4
B =
ans(:,:,1) =
1 0 2 0
2 1 3 0
ans(:,:,2) =
0 1 0 0
0 0 0 0
ans(:,:,3) =
0 0 0 3
0 0 0 4
As you can see, it's identical to Adriaan's second version.
Hope that helps!
A vectorized solution. You can use the second output of max to find the index of the first occurence of a nonzero value along the third dimension and then use sub2ind to convert that to linear index.
A(:,:,1) = [ 1 0 2];
A(:,:,2) = [ 1 1 0];
A(:,:,3) = [ 0 1 0];
[~, mi] =max(logical(A) ,[], 3);
sz=size(A) ;
[x, y] =ndgrid(1:sz(1),1:sz(2));
idx=sub2ind( sz, x,y,mi);
result=zeros(sz) ;
result(idx) =A(idx);
Related
I have a code for a 1D heat equation. Im trying to format a for loop so that the A matrix will follow a certain pattern of 1 -2 1 down the entire diagonal of a matrix that could be infinite. The pattern starts to take shape when I mess around with the initialized count at the beginning of the for loop but this changes the size of the matrix which fails the rest of the code.
My current code is below. The commented A matrix edits are what it should be.
N = 5;
%A(2,1:3) = [1 -2 1];
%A(3,2:4) = [1 -2 1];
%A(4,3:5) = [1 -2 1];
%A(5,4:6) = [1 -2 1];
A = zeros(N+1,N+1);
A(1,1) = 1;
for count=N:N+1
A(count+1,count:count+2) = [1 -2 1];
end
A(N+1,N+1) = 1;
In Matlab you can often avoid loops. In this case you can get the desired result with 2D convolution:
>> N = 6;
>> A = [1 zeros(1,N-1); conv2(eye(N-2), [1 -2 1]); zeros(1,N-1) 1]
A =
1 0 0 0 0 0
1 -2 1 0 0 0
0 1 -2 1 0 0
0 0 1 -2 1 0
0 0 0 1 -2 1
0 0 0 0 0 1
Or, depending on what you want,
>> A = conv2(eye(N), [1 -2 1], 'same')
A =
-2 1 0 0 0 0
1 -2 1 0 0 0
0 1 -2 1 0 0
0 0 1 -2 1 0
0 0 0 1 -2 1
0 0 0 0 1 -2
There are many simple ways of creating this matrix.
Your loop can be amended as follows:
N = 5;
A = zeros(N+1,N+1);
A(1,1) = 1;
for row = 2:N
A(row, row-1:row+1) = [1 -2 1];
end
A(N+1,N+1) = 1;
I've renamed count to row, we're indexing each row (from 2 to N, skipping the first and last rows), then finding with row-1:row+1 the three indices for that row that you want to address.
Directly indexing the diagonal and off-diagonal elements. Diagonal elements for an NxN matrix are 1:N+1:end. This is obviously more complex, I'd prefer the loop:
N = 6;
A = zeros(N,N);
A(1:N+1:end) = -2;
A(2:N+1:end-2*N) = 1; % skip last row
A(2*N+2:N+1:end) = 1; % skip first row
A(1,1) = 1;
A(N,N) = 1;
Using diag. We need to special-case the first and last rows:
N = 6;
A = diag(-2*ones(N,1),0) + diag(ones(N-1,1),1) + diag(ones(N-1,1),-1);
A(1,1:2) = [1,0];
A(end,end-1:end) = [0,1];
Consider an index vector consisting of ones and zeros:
I=[0 0 1 0 1 1 0 0 0];
How can I easily generate the following matrix in matlab:
J=[0 2;
1 1;
0 1;
1 2;
0 3];
Use diff:
I = [0 0 1 0 1 1 0 0 0];
d = diff(I);
ind = [1 find(d~=0)+1]; %// starting index of each new value
rep = diff([ind numel(I)+1]); %// number of repetitions of each new value
J = [ I(ind).' rep.' ];
Using strfind for a slightly bigger example -
I =[1 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0]
zero_pos = ['0' num2str(bsxfun(#eq,I,0),'%1d') '0']
ind3 = [ strfind(zero_pos,'01') ; strfind(zero_pos,'10')]
counts = diff(ind3(:))
var = zeros(numel(counts),1);
var(2:2:end)=1;
J = [var counts];
if ind3(1,1)-1>0
J = [1 ind3(1,1)-1;J];
end
Output
J =
1 2
0 2
1 1
0 1
1 2
0 3
1 4
0 2
0I have on matrix-
A=[1 2 2 3 5 5;
1 5 5 8 8 7;
2 9 9 3 3 5];
From matrix i need to count now many nonzero elements ,how any 1,how many 2 and how many 3 in each row of given matrix"A".For these i have written one code like:
[Ar Ac]=size(A);
for j=1:Ar
for k=1:Ac
count(:,j)=nnz(A(j,:));
d(:,j)=sum(A(j,:)== 1);
e(:,j)=sum(A(j,:)==2);
f(:,j)=sum(A(j,:)==3);
end
end
but i need to write these using on loop i.e. here i manually use sum(A(j,:)== 1),sum(A(j,:)== 2) and sum(A(j,:)== 3) but is there any option where i can only write sum(A(j,:)== 1:3) and store all the values in the different row i.e, the result will be like-
b=[1 2 1;
1 0 0;
0 1 2];
Matlab experts need your valuable suggestions
Sounds like you're looking for a histogram count:
U = unique(A);
counts = histc(A', U)';
b = counts(:, ismember(U, [1 2 3]));
Example
%// Input matrix and vector of values to count
A = [1 2 2 3 5 5; 1 5 5 8 8 7; 2 9 9 3 3 5];
vals = [1 2 3];
%// Count values
U = unique(A);
counts = histc(A', U)';
b = counts(:, ismember(U, vals));
The result is:
b =
1 2 1
1 0 0
0 1 2
Generalizing the sought values, as required by asker:
values = [ 1 2 3 ]; % or whichever values are sought
B = squeeze(sum(bsxfun(#(x,y) sum(x==y,2), A, shiftdim(values,-1)),2));
Here is a simple and general way. Just change n to however high you want to count. n=max(A(:)) is probably a good general value.
result = [];
n = 3;
for col= 1:n
result = [result, sum(A==col, 2)];
end
result
e.g. for n = 10
result =
1 2 1 0 2 0 0 0 0 0
1 0 0 0 2 0 1 2 0 0
0 1 2 0 1 0 0 0 2 0
Why not use this?
B=[];
for x=1:size(A,1)
B=[B;sum(A(x,:)==1),sum(A(x,:)==2),sum(A(x,:)==3)];
end
I'd do this way:
B = [arrayfun(#(i) find(A(i,:) == 1) , 1:3 , 'UniformOutput', false)',arrayfun(#(i) find(A(i,:) == 2) , 1:3 , 'UniformOutput', false)',arrayfun(#(i) find(A(i,:) == 3) , 1:3 , 'UniformOutput', false)'];
res = cellfun(#numel, B);
Here is a compact one:
sum(bsxfun(#eq, permute(A, [1 3 2]), 1:3),3)
You can replace 1:3 with any array.
you can make an anonymous function for it
rowcnt = #(M, R) sum(bsxfun(#eq, permute(M, [1 3 2]), R),3);
then running it on your data returns
>> rowcnt(A,1:3)
ans =
1 2 1
1 0 0
0 1 2
and for more generalized case
>> rowcnt(A,[1 2 5 8])
ans =
1 2 2 0
1 0 2 2
0 1 1 0
I have a matrix A
1 1 0 0
0 1 0 0
1 0 0 1
0 0 1 0
0 0 0 0
0 1 1 1
1 1 0 0
1 0 0 0
0 0 0 1
I want this matrix to be split according to user's input say d = [1 2 3].
for i=2:length(d)
d(i) = d(i) + d(i-1); % d = [1 3 6]
end
This gives d = [1 (1+2) (1+2+3)] = d[1 3 6]. There are 9 rows in this matrix, calculate ceil of [(1/6)*9], [(3/6)*9] and [(6/6)*9]. Hence this gives [2 5 9]. First split up is first two rows , 2nd split up is next (5-2=3) 3 rows and third split is (9-5=4) 4 rows.
The output should be like:
The split up is: 1st split up->
1 1 0 0 % first 2 rows in matrix A
0 1 0 0
2nd split up->
1 0 0 1 % next 3 rows
0 0 1 0
0 0 0 0
3rd split up->
0 1 1 1 % next 4 rows
1 1 0 0
1 0 0 0
0 0 0 1
You can use mat2cell with input d = [1 2 3] to store the final splits in separate cell arrays
B = mat2cell(A, d+1, size(A,2));
or, to adapt it to your computation of the split row sizes:
d = [1 2 3];
c = cumsum(d); % [1, 3, 6]
s = ceil(size(A,1)*c/c(end)); % [2, 5, 9]
n = [s(1) diff(s)]; % [2, 3, 4]
B = mat2cell(A, n, size(A,2));
To display the splits you can add a command similar to:
cellfun(#disp, B)
I have a matrix A
A = [0 0 0 0 1; 0 0 0 0 2; 0 1 2 3 4];
and I would like to randomly permute the elements within each row. For example, matrix A2
A2 = [1 0 0 0 0; 0 0 0 2 0; 4 1 3 2 0]; % example of desired output
I can do this with a vector:
Av = [0 1 2 3 4];
Bv = Av(randperm(5));
But I am unsure how to do this a row at time for a matrix and to only permute the elements within a given row. Is this possible to do? I could construct a matrix from many permuted vectors, but I would prefer not to do it this way.
Thanks.
You can use sort on a random array of any size (which is what randperm does). After that, all you need to do is some index-trickery to properly reshuffle the array
A = [0 0 0 0 1; 0 0 0 0 2; 0 1 2 3 4];
[nRows,nCols] = size(A);
[~,idx] = sort(rand(nRows,nCols),2);
%# convert column indices into linear indices
idx = (idx-1)*nRows + ndgrid(1:nRows,1:nCols);
%# rearrange A
B = A;
B(:) = B(idx)
B =
0 0 1 0 0
0 2 0 0 0
2 1 3 4 0