I have written some code that compresses a matrix to remove zero columns and rows, but I can't work out how to reconstruct the original matrix.
Say I have a matrix:
A = [ 0 3 0 2 1 0 6
3 0 0 4 8 0 5
0 0 0 0 0 0 0
2 4 0 0 2 0 1
1 8 0 2 0 0 7
0 0 0 0 0 0 0
6 5 0 1 7 0 0 ]
Here rows/columns 3 and 6 are empty, so my compression function will give the output:
A_dash = [ 0 3 2 1 6
3 0 4 8 5
2 4 0 2 1
1 8 2 0 7
6 5 1 7 0 ]
A_map = [ 1 2 4 5 7]
Where A_map is a vector mapping the indicies of the rows/columns of A_dash to A. This means that if A_map(3) = 4, then row/column 4 of A is the same as row/column 3 of A_dash - ie. a row/column of zeroes must be inserted between columns/rows 2 and 3 in A_dash
What is the easiest way people can suggest for me to recreate matrix A from A_dash, using the information in A_map?
Here is what I have got so far:
% orig_size is original number of columns/rows
c_count = size(A_dash,1);
A = zeros(c_count, orig_size); % c_count rows to avoid dimension mismatch
for ii = 1:c_count
A(:,A_map(ii)) == A_dash(:,ii);
end
This gives me the right result column-wise:
A = [ 0 3 0 2 1 0 6
3 0 0 4 8 0 5
2 4 0 0 2 0 1
1 8 0 2 0 0 7
6 5 0 1 7 0 0 ]
However, I'm not sure how i should go about inserting the rows, i suppose i could copy the first 1:i rows into one matrix, i:end rows to a second matrix and concatenate those with a zero row in between, but that feels like a bit of a
clunky solution, and probably not very efficient for large sized matrices..
Otherwise, is there a better way that people can suggest I store the map information? I was thinking instead of storing the mapping between column/row indices, that I just store the indices of the zero columns/rows and then insert columns/rows of zeros where appropriate. Would this be a better way?
You've got the indices of the valid rows/columns. Now all you've got to do is put them in a new matrix of zeros the same size as A:
B=zeros(size(A));
B(A_map,A_map)=A_dash
B =
0 3 0 2 1 0 6
3 0 0 4 8 0 5
0 0 0 0 0 0 0
2 4 0 0 2 0 1
1 8 0 2 0 0 7
0 0 0 0 0 0 0
6 5 0 1 7 0 0
Just to check...
>> A==B
ans =
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
A and B are equal everywhere, so we've reconstructed A.
If I have the matrix
1 0 0
0 0 1
0 0 0
and I want this form in MATLAB
1 2 3 1 2 3 1 2 3
1 1 1 2 2 2 3 3 3
1 0 0 0 0 0 0 1 0
also I want the values of third row in result. i.e. ans= [1 0 0 0 0 0 0 1 0]
Here you go -
[X,Y] = ndgrid(1:size(A,1),1:size(A,2));
out = [X(:).' ; Y(:).' ; A(:).']
For the last part of your question, use the last row of out : out(end,:) or A(:).'.
Sample run -
>> A
A =
1 0 0
0 0 1
0 0 0
>> [X,Y] = ndgrid(1:size(A,1),1:size(A,2));
>> out = [X(:).' ; Y(:).' ; A(:).']
out =
1 2 3 1 2 3 1 2 3
1 1 1 2 2 2 3 3 3
1 0 0 0 0 0 0 1 0
Let's have a M = [10 x 4 x 12] matrix. As example I take the M(:,:,4):
val(:,:,4) =
0 0 1 0
0 1 1 1
0 0 0 1
1 1 1 1
1 1 0 1
0 1 1 1
1 1 1 1
1 1 1 1
0 0 1 1
0 0 1 1
How can I obtain this:
val(:,:,4) =
0 0 3 0
0 2 2 2
0 0 0 4
1 1 1 1
1 1 0 1
0 2 2 2
1 1 1 1
1 1 1 1
0 0 3 3
0 0 3 3
If I have 1 in the first column then all the subsequent 1's should be 1.
If I have 0 in the first column but 1 in the second, all the subsequent 1's should be 2.
If I have 0 in the first and second column but 1 in the third then all the subsequent 1's should be 3.
If I have 0 in the first 3 columns but 1 in the forth then this one should be four.
Note: The logical matrix M is constructed:
Tab = [reshape(Avg_1step.',10,1,[]) reshape(Avg_2step.',10,1,[]) ...
reshape(Avg_4step.',10,1,[]) reshape(Avg_6step.',10,1,[])];
M = Tab>=repmat([20 40 60 80],10,1,size(Tab,3));
This is a very simple approach that works for both 2D and 3D matrices.
%// Find the column index of the first element in each "slice".
[~, idx] = max(val,[],2);
%// Multiply the column index with each row of the initial matrix
bsxfun(#times, val, idx);
This could be one approach -
%// Concatenate input array along dim3 to create a 2D array for easy work ahead
M2d = reshape(permute(M,[1 3 2]),size(M,1)*size(M,3),[]);
%// Find matches for each case, index into each matching row and
%// elementwise multiply all elements with the corresponding multiplying
%// factor of 2 or 3 or 4 and thus obtain the desired output but as 2D array
%// NOTE: Case 1 would not change any value, so it was skipped.
case2m = all(bsxfun(#eq,M2d(:,1:2),[0 1]),2);
M2d(case2m,:) = bsxfun(#times,M2d(case2m,:),2);
case3m = all(bsxfun(#eq,M2d(:,1:3),[0 0 1]),2);
M2d(case3m,:) = bsxfun(#times,M2d(case3m,:),3);
case4m = all(bsxfun(#eq,M2d(:,1:4),[0 0 0 1]),2);
M2d(case4m,:) = bsxfun(#times,M2d(case4m,:),4);
%// Cut the 2D array thus obtained at every size(a,1) to give us back a 3D
%// array version of the expected values
Mout = permute(reshape(M2d,size(M,1),size(M,3),[]),[1 3 2])
Code run with a random 6 x 4 x 2 sized input array -
M(:,:,1) =
1 1 0 1
1 0 1 1
1 0 0 1
0 0 1 1
1 0 0 0
1 0 1 1
M(:,:,2) =
0 1 0 1
1 1 0 0
1 1 0 0
0 0 1 1
0 0 0 1
0 0 1 0
Mout(:,:,1) =
1 1 0 1
1 0 1 1
1 0 0 1
0 0 3 3
1 0 0 0
1 0 1 1
Mout(:,:,2) =
0 2 0 2
1 1 0 0
1 1 0 0
0 0 3 3
0 0 0 4
0 0 3 0
this is my matrix that displays a sample network graph
matrix =
0 1 1 1
1 0 1 0
0 0 0 1
1 1 1 0
where its a 4x4 matrix
1) 2) 3) 4)
1) 0 1 1 1
2) 1 0 1 0
3) 0 0 0 1
4) 1 1 1 0
i want to count this 4x4 matrix like
row 1 counts how many 1s i have and adds column 1 number of 1's to it and returns 1)=5 as total 1's in row 1 and col 1 = 5
i want my output to be like
1=5
2=4
3=4
4=5
This must be it -
out = sum([matrix matrix'],2)
Example run -
matrix =
1 1 1 1
1 0 0 0
0 1 0 1
0 0 1 1
out =
6
3
4
5
The above code would count 1s twice when they appear on the diagonal, which if you don't want, use this -
out1 = sum([matrix matrix'],2) - diag(matrix)
Example run -
matrix =
1 1 1 1
1 0 0 0
0 1 0 1
0 0 1 1
out1 =
5
3
4
4
I agree with the answer of Divakar, but once your graph gets larger and larger, you might not want to transpose the entire matrix. I suggest doing the sum first and then transposing afterwards:
sum(matrix,1)'+sum(matrix,2)-diag(matrix);
matrix =
0 1 1 1
1 0 1 0
0 0 0 1
1 1 1 0
degree=sum(matrix,1)'+sum(matrix,2)-diag(matrix)
degree =
5
4
4
5
This question already has an answer here:
Sort in ascending order, but keep zeros at last
(1 answer)
Closed 9 years ago.
I have a matrix like below-
x=[1 1 1 1 1;
2 1 1 1 0;
3 3 1 0 0;
3 2 2 0 0];
But i want to make this matrix like-
x=[2 2 3 0 0;
1 3 3 0 0;
1 1 1 2 0;
1 1 1 1 1];
I have already tried with "sort ascending" but then '0' will come first but i want to keep "0" on last and make one matrix where the number of '1's' will be from low to high throughout the matrix.I am trying but cannot do so.
I need Matlab experts help.
To get the zeros at the end, set them to infinity (inf), sort and then set them back to zero
x=[1 1 1 1 1;
2 1 1 1 0;
3 3 1 0 0;
3 2 2 0 0]
x(x == 0) = inf;
y = sort(x, 2, 'ascend');
y(y==inf) = 0;
Now count the number of 1s per row and reorder from least to most
[~, I] = sort(sum(y==1,2));
y(I, :)
ans =
2 2 3 0 0
1 3 3 0 0
1 1 1 2 0
1 1 1 1 1
EDIT:
The example data is ambiguous. Take this as the input:
x=[1 1 1 1 1;
2 1 1 1 0;
7 1 0 1 0;
3 3 1 0 0;
3 2 2 0 0];
Now the answers by Moshen (i.e. to just sort each column after sorting the rows) return
x =
2 3 7 0 0
1 2 3 0 0
1 1 3 0 0
1 1 1 2 0
1 1 1 1 1
Whereas mine returns
ans =
2 2 3 0 0
1 3 3 0 0
1 1 7 0 0
1 1 1 2 0
1 1 1 1 1
Mine preserves the row integrity. But it is not clear which answer the OP is after.