This question already has answers here:
shuffle matrix element in matlab
(2 answers)
Closed 5 years ago.
I have a matrix of values ranging from 0 to 3 in Matlab. I want to randomly shuffle the elements of the matrix, but only within the cells that have values in the range 1 -3 (so only within a subset of the whole matrix). Is there a way to do that? Thanks.
You can do this by getting an index of all the values of interest (such as a logical index), randomly permuting their order using randperm, then assigning them back into the matrix using the same index:
% Sample matrix with values from 0 to 3:
M = randi([0 3], 5)
M =
3 1 0 3 0
0 3 3 2 0
1 0 2 1 0
1 1 2 2 0
3 0 0 1 0
index = (M > 0); % Index of values from 1 to 3
values = M(index); % Vector of indexed values
M(index) = values(randperm(numel(values))) % Matrix with shuffled values
M =
2 3 0 2 0
0 3 1 1 0
2 0 3 3 0
1 1 2 1 0
3 0 0 1 0
Note that the zeroes are all still in the same place in the shuffled matrix. Note also that you still have the same number of ones, twos, and threes, since they are just shuffled to different spots.
Related
How can I count the number of values that are contiguous in an matrix? For example, if
A= [ 1 0 1 0 0 0 0 \
1 1 1 0 0 0 1\
1 1 0 1 1 1 1]
is a 7 by 3 matrix then the result should indicate that there are 12 contiguous values that are "1" (highlighted in bold), that there are 8 contiguous 0 values (highlighted in italics) and one lone 0 value. Code in IDL is preferred but MATLAB would also be helpful.
This is what you can do in MATLAB using the bwconncomp function. This is a function in the Image Processing Toolbox. I don't know a whole lot about IDL, it might have a similar function.
bwconncomp returns a struct with some information, one of the fields is PixelIdxList, which is a cell array with one element per connected component. Each of these elements is a vector with the indices to one of the array elements in the connected component. For the case of the 1 elements in your example, this cell array will have one vector with 12 values. For the case of the 0 elements, it will have two vectors, one with 1 value and one with 8:
>> A = [ 1 0 1 0 0 0 0 ; 1 1 1 0 0 0 1 ; 1 1 0 1 1 1 1 ];
>> CC = bwconncomp(A==1, 8);
>> cellfun(#numel, CC.PixelIdxList)
ans =
12
>> CC = bwconncomp(A==0, 8);
>> cellfun(#numel, CC.PixelIdxList)
ans =
1 8
The bwconncomp takes 4 or 8 as the second argument. This specifies what are considered connected elements (contiguous values, neighbors). 4 means only the 4 elements N,S,E,W are connected; 8 means also diagonal connections exist (8 neighbors).
I have a time-series matrix X whose first column contains user ID and second column contains the item ID they used at different times:
X=[1 4
2 1
4 2
2 3
3 4
1 1
4 2
5 3
2 1
4 2
5 4];
I want to find out which user used which item how many times, and save it in a matrix Y. The rows of Y represent users in ascending order of ID, and the columns represent items in ascending order of ID:
Y=[1 0 0 1
2 0 1 0
0 0 0 1
0 3 0 0
0 0 1 1]
The code I use to find matrix Y uses 2 for loops which is unwieldy for my large data:
no_of_users = size(unique(X(:,1)),1);
no_of_items = size(unique(X(:,2)),1);
users=unique(X(:,1));
Y=zeros(no_of_users,no_of_items);
for a=1:size(A,1)
for b=1:no_of_users
if X(a,1)==users(b,1)
Y(b,X(a,2)) = Y(b,X(a,2)) + 1;
end
end
end
Is there a more time efficient way to do it?
sparse creates a sparse matrix from row/column indices, conveniently accumulating the number of occurrences if you give a scalar value of 1. Just convert to a full matrix.
Y = full(sparse(X(:,1), X(:,2), 1))
Y =
1 0 0 1
2 0 1 0
0 0 0 1
0 3 0 0
0 0 1 1
But it's probably quicker to just use accumarray as suggested in the comments:
>> Y2 = accumarray(X, 1)
Y2 =
1 0 0 1
2 0 1 0
0 0 0 1
0 3 0 0
0 0 1 1
(In Octave, sparse seems to take about 50% longer than accumarray.)
I have an adjacency matrix M, something like this:
[1 2 0 2 4
2 1 2 0 -1
0 3 1 2 3
2 0 2 1 0
4 -1 3 0 1]
I want to calculate the mean of all values below (but not including) the diagonal. The final output should be 1.5.
To get those values, I thought I'd use N = tril(M,-1). The issue is that I now have zeros in upper and lower part of the matrix N and therefore mean(sum(N)./sum(N~=0)) wouldn't work. Since I also have negative values, I can't just do the mean of values >=0 either. How can I do this?
In one line using logical indexing to extract just the values below the diagonal:
M = [ 1 2 0 2 4;
2 1 2 0 -1;
0 3 1 2 3;
2 0 2 1 0;
4 -1 3 0 1];
mean(M(tril(true(size(M)),-1)))
This returns 1.5 as #excaza indicated.
This question already has answers here:
How can I find indices of each row of a matrix which has a duplicate in matlab?
(3 answers)
Closed 8 years ago.
I have two matrices and I want to find the indices of rows in Matrix B which have the same row values in Matrix A. Let me give a simple example:
A=[1,2,3; 2,3,4; 3,5,7; 1,2,3; 1,2,3; 5,8,6];
B=[1,2,3; 29,3,4; 3,59,7; 1,29,3; 1,2,3; 5,8,6;1,2,3];
For example, for first row in matrix A, The row1, row5, and row7 in Matrix B are correspondences.
I have written below code but it doesn't return back all indices which have the same row value in matrix A and only one of them (row7) is backed !!
A_sorted = sort(A,2,'descend'); % sorting angles
B_sorted = sort(B,2,'descend'); % sorting angles
[~,indx]=ismember(A_sorted,B_sorted,'rows')
the result is
indx_2 =
7
0
0
7
7
6
It means for the first row in matrix A , only one row ( row 7) in Matrix B is available !! But as you can see for first row in matrix A there is three correspondent rows in matrix B (Row 1, row 5 and row 7)
I think the best strategy is to apply ismember to unique rows
%make matrix unique
[B_unique,B2,B3]=unique(B_sorted,'rows')
[~,indx]=ismember(A_sorted,B_unique,'rows')
%For each row in B_unique, get the corresponding indices in B_sorted
indx2=arrayfun(#(x)find(B3==x),indx,'uni',0)
If you want to compare all pairs of rows between A and B, use
E = squeeze(all(bsxfun(#eq, A, permute(B, [3 2 1])), 2));
or equivalently
E = pdist2(A,B)==0;
In your example, this gives
E =
1 0 0 0 1 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
1 0 0 0 1 0 1
1 0 0 0 1 0 1
0 0 0 0 0 1 0
The value E(ia,ib) tells you if the ia-th row of A equals the ib-th row of B.
If I have a matrix:
A = [1 2 3 4 5; 1 1 6 1 2; 0 0 9 0 1]
A =
1 2 3 4 5
1 1 6 1 2
0 0 9 0 1
How can I count the number of non-zero entries for each column? For example the desired output for this matrix would be:
2, 2, 3, 2, 3
I am not sure how to do this as size, length or numel do not appear to meet the requirements. Perhaps it would be best to remove zero entries first?
It's simply
> A ~= 0
ans =
1 1 1 1 1
1 1 1 1 1
0 0 1 0 1
> sum(A ~= 0, 1)
ans =
2 2 3 2 3
Here's another solution I can suggest that isn't very speed worthy for dense matrices but quite fast for sparse matrices (thanks #user1877862!). This also would mimic how one might do this in a compiled language, like C or Java, and perhaps for research purposes too. First find the row and column locations that are non zero, then do a histogram on just the column locations to count the frequency of how often you see a non-zero in each column. In other words:
[~,col] = find(A ~= 0);
counts = histc(col, 1:size(A,2));
find outputs the row and column locations of where a matrix satisfies some Boolean condition inside the argument of the function. We ignore the first output as we aren't concerned with the row locations.
The output we get is:
counts =
2
2
3
2
3