I have a matrix index in Matlab with size GxN and a matrix A with size MxN.
Let me provide an example before presenting my question.
clear
N=3;
G=2;
M=5;
index=[1 2 3;
13 14 15]; %GxN
A=[1 2 3;
5 6 7;
21 22 23;
1 2 3;
13 14 15]; %MxN
I would like your help to construct a matrix Response with size GxM with Response(g,m)=1 if the row A(m,:) is equal to index(g,:) and zero otherwise.
Continuing the example above
Response= [1 0 0 1 0;
0 0 0 0 1]; %GxM
This code does what I want (taken from a previous question of mine - just to clarify: the current question is different)
Response=permute(any(all(bsxfun(#eq, reshape(index.', N, [], G), permute(A, [2 3 4 1])), 1), 2), [3 4 1 2]);
However, the command is extremely slow for my real matrix sizes (N=19, M=500, G=524288). I understand that I will not be able to get huge speed but anything that can improve on this is welcome.
Aproach 1: computing distances
If you have the Statistics Toolbox:
Response = ~(pdist2(index, A));
or:
Response = ~(pdist2(index, A, 'hamming'));
This works because pdist2 computes the distance between each pair of rows. Equal rows have distance 0. The logical negation ~ gives 1 for those pairs of rows, and 0 otherwise.
Approach 2: reducing rows to unique integer labels
This approach is faster on my machine:
[~,~,u] = unique([index; A], 'rows');
Response = bsxfun(#eq, u(1:G), u(G+1:end).');
It works by reducing rows to unique integer labels (using the third output of unique), and comparing the latter instead of the former.
For your size values this takes approximately 1 second on my computer:
clear
N = 19; M = 500; G = 524288;
index = randi(5,G,N); A = randi(5,M,N);
tic
[~,~,u] = unique([index; A], 'rows');
Response = bsxfun(#eq, u(1:G), u(G+1:end).');
toc
gives
Elapsed time is 1.081043 seconds.
MATLAB has a multitude of functions for working with sets, including setdiff, intersect, union etc. In this case, you can use the ismember function:
[~, Loc] = ismember(A,index,'rows');
Which gives:
Loc =
1
0
0
1
2
And Response would be constructed as follows:
Response = (1:size(index,1) == Loc).';
Response =
2×5 logical array
1 0 0 1 0
0 0 0 0 1
You could reshape the matrices so that each row instead lies along the 3rd dimension. Then we can use implicit expansion (see bsxfun for R2016b or earlier) for equality of all elements, and all to aggregate on the rows (i.e. false if not all equal for a given row).
Response = all( reshape( index, [], 1, size(index,2) ) == reshape( A, 1, [], size(A,2) ), 3 );
You might even be able to avoid some reshaping by using all in another dimension, but it's easier for me to visualise it this way.
Related
I have a matrix suppX in Matlab with size GxN and a matrix A with size MxN. I would like your help to construct a matrix Xresponse with size GxM with Xresponse(g,m)=1 if the row A(m,:) is equal to the row suppX(g,:) and zero otherwise.
Let me explain better with an example.
suppX=[1 2 3 4;
5 6 7 8;
9 10 11 12]; %GxN
A=[1 2 3 4;
1 2 3 4;
9 10 11 12;
1 2 3 4]; %MxN
Xresponse=[1 1 0 1;
0 0 0 0;
0 0 1 0]; %GxM
I have written a code that does what I want.
Xresponsemy=zeros(size(suppX,1), size(A,1));
for x=1:size(suppX,1)
Xresponsemy(x,:)=ismember(A, suppX(x,:), 'rows').';
end
My code uses a loop. I would like to avoid this because in my real case this piece of code is part of another big loop. Do you have suggestions without looping?
One way to do this would be to treat each matrix as vectors in N dimensional space and you can find the L2 norm (or the Euclidean distance) of each vector. After, check if the distance is 0. If it is, then you have a match. Specifically, you can create a matrix such that element (i,j) in this matrix calculates the distance between row i in one matrix to row j in the other matrix.
You can treat your problem by modifying the distance matrix that results from this problem such that 1 means the two vectors completely similar and 0 otherwise.
This post should be of interest: Efficiently compute pairwise squared Euclidean distance in Matlab.
I would specifically look at the answer by Shai Bagon that uses matrix multiplication and broadcasting. You would then modify it so that you find distances that would be equal to 0:
nA = sum(A.^2, 2); % norm of A's elements
nB = sum(suppX.^2, 2); % norm of B's elements
Xresponse = bsxfun(#plus, nB, nA.') - 2 * suppX * A.';
Xresponse = Xresponse == 0;
We get:
Xresponse =
3×4 logical array
1 1 0 1
0 0 0 0
0 0 1 0
Note on floating-point efficiency
Because you are using ismember in your implementation, it's implicit to me that you expect all values to be integer. In this case, you can very much compare directly with the zero distance without loss of accuracy. If you intend to move to floating-point, you should always compare with some small threshold instead of 0, like Xresponse = Xresponse <= 1e-10; or something to that effect. I don't believe that is needed for your scenario.
Here's an alternative to #rayryeng's answer: reduce each row of the two matrices to a unique identifier using the third output of unique with the 'rows' input flag, and then compare the identifiers with singleton expansion (broadcast) using bsxfun:
[~, ~, w] = unique([A; suppX], 'rows');
Xresponse = bsxfun(#eq, w(1:size(A,1)).', w(size(A,1)+1:end));
New to MatLab here (R2015a, Mac OS 10.10.5), and hoping to find a solution to this indexing problem.
I want to change the values of a large 2D matrix, based on one vector of row indices and one of column indices. For a very simple example, if I have a 3 x 2 matrix of zeros:
A = zeros(3, 2)
0 0
0 0
0 0
I want to change A(1, 1) = 1, and A(2, 2) = 1, and A(3, 1) = 1, such that A is now
1 0
0 1
1 0
And I want to do this using vectors to indicate the row and column indices:
rows = [1 2 3];
cols = [1 2 1];
Is there a way to do this without looping? Remember, this is a toy example that needs to work on a very large 2D matrix. For extra credit, can I also include a vector that indicates which value to insert, instead of fixing it at 1?
My looping approach is easy, but slow:
for i = 1:length(rows)
A(rows(i), cols(i)) = 1;
end
sub2ind can help here,
A = zeros(3,2)
rows = [1 2 3];
cols = [1 2 1];
A(sub2ind(size(A),rows,cols))=1
A =
1 0
0 1
1 0
with a vector to 'insert'
b = [1,2,3];
A(sub2ind(size(A),rows,cols))=b
A =
1 0
0 2
3 0
I found this answer online when checking on the speed of sub2ind.
idx = rows + (cols - 1) * size(A, 1);
therefore
A(idx) = 1 % or b
5 tests on a big matrix (~ 5 second operations) shows it's 20% faster than sub2ind.
There is code for an n-dimensional problem here too.
What you have is basically a sparse definition of a matrix. Thus, an alternative to sub2ind is sparse. It will create a sparse matrix, use full to convert it to a full matrix.
A=full(sparse(rows,cols,1,3,2))
I'm trying to elegantly split a vector. For example,
vec = [1 2 3 4 5 6 7 8 9 10]
According to another vector of 0's and 1's of the same length where the 1's indicate where the vector should be split - or rather cut:
cut = [0 0 0 1 0 0 0 0 1 0]
Giving us a cell output similar to the following:
[1 2 3] [5 6 7 8] [10]
Solution code
You can use cumsum & accumarray for an efficient solution -
%// Create ID/labels for use with accumarray later on
id = cumsum(cut)+1
%// Mask to get valid values from cut and vec corresponding to ones in cut
mask = cut==0
%// Finally get the output with accumarray using masked IDs and vec values
out = accumarray(id(mask).',vec(mask).',[],#(x) {x})
Benchmarking
Here are some performance numbers when using a large input on the three most popular approaches listed to solve this problem -
N = 100000; %// Input Datasize
vec = randi(100,1,N); %// Random inputs
cut = randi(2,1,N)-1;
disp('-------------------- With CUMSUM + ACCUMARRAY')
tic
id = cumsum(cut)+1;
mask = cut==0;
out = accumarray(id(mask).',vec(mask).',[],#(x) {x});
toc
disp('-------------------- With FIND + ARRAYFUN')
tic
N = numel(vec);
ind = find(cut);
ind_before = [ind-1 N]; ind_before(ind_before < 1) = 1;
ind_after = [1 ind+1]; ind_after(ind_after > N) = N;
out = arrayfun(#(x,y) vec(x:y), ind_after, ind_before, 'uni', 0);
toc
disp('-------------------- With CUMSUM + ARRAYFUN')
tic
cutsum = cumsum(cut);
cutsum(cut == 1) = NaN; %Don't include the cut indices themselves
sumvals = unique(cutsum); % Find the values to use in indexing vec for the output
sumvals(isnan(sumvals)) = []; %Remove NaN values from sumvals
output = arrayfun(#(val) vec(cutsum == val), sumvals, 'UniformOutput', 0);
toc
Runtimes
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.068102 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.117953 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 12.560973 seconds.
Special case scenario: In cases where you might have runs of 1's, you need to modify few things as listed next -
%// Mask to get valid values from cut and vec corresponding to ones in cut
mask = cut==0
%// Setup IDs differently this time. The idea is to have successive IDs.
id = cumsum(cut)+1
[~,~,id] = unique(id(mask))
%// Finally get the output with accumarray using masked IDs and vec values
out = accumarray(id(:),vec(mask).',[],#(x) {x})
Sample run with such a case -
>> vec
vec =
1 2 3 4 5 6 7 8 9 10
>> cut
cut =
1 0 0 1 1 0 0 0 1 0
>> celldisp(out)
out{1} =
2
3
out{2} =
6
7
8
out{3} =
10
For this problem, a handy function is cumsum, which can create a cumulative sum of the cut array. The code that produces an output cell array is as follows:
vec = [1 2 3 4 5 6 7 8 9 10];
cut = [0 0 0 1 0 0 0 0 1 0];
cutsum = cumsum(cut);
cutsum(cut == 1) = NaN; %Don't include the cut indices themselves
sumvals = unique(cutsum); % Find the values to use in indexing vec for the output
sumvals(isnan(sumvals)) = []; %Remove NaN values from sumvals
output = {};
for i=1:numel(sumvals)
output{i} = vec(cutsum == sumvals(i)); %#ok<SAGROW>
end
As another answer shows, you can use arrayfun to create a cell array with the results. To apply that here, you'd replace the for loop (and the initialization of output) with the following line:
output = arrayfun(#(val) vec(cutsum == val), sumvals, 'UniformOutput', 0);
That's nice because it doesn't end up growing the output cell array.
The key feature of this routine is the variable cutsum, which ends up looking like this:
cutsum =
0 0 0 NaN 1 1 1 1 NaN 2
Then all we need to do is use it to create indices to pull the data out of the original vec array. We loop from zero to max and pull matching values. Notice that this routine handles some situations that may arise. For instance, it handles 1 values at the very beginning and very end of the cut array, and it gracefully handles repeated ones in the cut array without creating empty arrays in the output. This is because of the use of unique to create the set of values to search for in cutsum, and the fact that we throw out the NaN values in the sumvals array.
You could use -1 instead of NaN as the signal flag for the cut locations to not use, but I like NaN for readability. The -1 value would probably be more efficient, as all you'd have to do is truncate the first element from the sumvals array. It's just my preference to use NaN as a signal flag.
The output of this is a cell array with the results:
output{1} =
1 2 3
output{2} =
5 6 7 8
output{3} =
10
There are some odd conditions we need to handle. Consider the situation:
vec = [1 2 3 4 5 6 7 8 9 10 11 12 13 14];
cut = [1 0 0 1 1 0 0 0 0 1 0 0 0 1];
There are repeated 1's in there, as well as a 1 at the beginning and end. This routine properly handles all this without any empty sets:
output{1} =
2 3
output{2} =
6 7 8 9
output{3} =
11 12 13
You can do this with a combination of find and arrayfun:
vec = [1 2 3 4 5 6 7 8 9 10];
N = numel(vec);
cut = [0 0 0 1 0 0 0 0 1 0];
ind = find(cut);
ind_before = [ind-1 N]; ind_before(ind_before < 1) = 1;
ind_after = [1 ind+1]; ind_after(ind_after > N) = N;
out = arrayfun(#(x,y) vec(x:y), ind_after, ind_before, 'uni', 0);
We thus get:
>> celldisp(out)
out{1} =
1 2 3
out{2} =
5 6 7 8
out{3} =
10
So how does this work? Well, the first line defines your input vector, the second line finds how many elements are in this vector and the third line denotes your cut vector which defines where we need to cut in our vector. Next, we use find to determine the locations that are non-zero in cut which correspond to the split points in the vector. If you notice, the split points determine where we need to stop collecting elements and begin collecting elements.
However, we need to account for the beginning of the vector as well as the end. ind_after tells us the locations of where we need to start collecting values and ind_before tells us the locations of where we need to stop collecting values. To calculate these starting and ending positions, you simply take the result of find and add and subtract 1 respectively.
Each corresponding position in ind_after and ind_before tell us where we need to start and stop collecting values together. In order to accommodate for the beginning of the vector, ind_after needs to have the index of 1 inserted at the beginning because index 1 is where we should start collecting values at the beginning. Similarly, N needs to be inserted at the end of ind_before because this is where we need to stop collecting values at the end of the array.
Now for ind_after and ind_before, there is a degenerate case where the cut point may be at the end or beginning of the vector. If this is the case, then subtracting or adding by 1 will generate a start and stopping position that's out of bounds. We check for this in the 4th and 5th line of code and simply set these to 1 or N depending on whether we're at the beginning or end of the array.
The last line of code uses arrayfun and iterates through each pair of ind_after and ind_before to slice into our vector. Each result is placed into a cell array, and our output follows.
We can check for the degenerate case by placing a 1 at the beginning and end of cut and some values in between:
vec = [1 2 3 4 5 6 7 8 9 10];
cut = [1 0 0 1 0 0 0 1 0 1];
Using this example and the above code, we get:
>> celldisp(out)
out{1} =
1
out{2} =
2 3
out{3} =
5 6 7
out{4} =
9
out{5} =
10
Yet another way, but this time without any loops or accumulating at all...
lengths = diff(find([1 cut 1])) - 1; % assuming a row vector
lengths = lengths(lengths > 0);
data = vec(~cut);
result = mat2cell(data, 1, lengths); % also assuming a row vector
The diff(find(...)) construct gives us the distance from each marker to the next - we append boundary markers with [1 cut 1] to catch any runs of zeros which touch the ends. Each length is inclusive of its marker, though, so we subtract 1 to account for that, and remove any which just cover consecutive markers, so that we won't get any undesired empty cells in the output.
For the data, we mask out any elements corresponding to markers, so we just have the valid parts we want to partition up. Finally, with the data ready to split and the lengths into which to split it, that's precisely what mat2cell is for.
Also, using #Divakar's benchmark code;
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.272810 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.436276 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 17.112259 seconds.
-------------------- With mat2cell
Elapsed time is 0.084207 seconds.
...just sayin' ;)
Here's what you need:
function spl = Splitting(vec,cut)
n=1;
j=1;
for i=1:1:length(b)
if cut(i)==0
spl{n}(j)=vec(i);
j=j+1;
else
n=n+1;
j=1;
end
end
end
Despite how simple my method is, it's in 2nd place for performance:
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.264428 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.407963 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 18.337940 seconds.
-------------------- SIMPLE
Elapsed time is 0.271942 seconds.
Unfortunately there is no 'inverse concatenate' in MATLAB. If you wish to solve a question like this you can try the below code. It will give you what you looking for in the case where you have two split point to produce three vectors at the end. If you want more splits you will need to modify the code after the loop.
The results are in n vector form. To make them into cells, use num2cell on the results.
pos_of_one = 0;
% The loop finds the split points and puts their positions into a vector.
for kk = 1 : length(cut)
if cut(1,kk) == 1
pos_of_one = pos_of_one + 1;
A(1,one_pos) = kk;
end
end
F = vec(1 : A(1,1) - 1);
G = vec(A(1,1) + 1 : A(1,2) - 1);
H = vec(A(1,2) + 1 : end);
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)
I'm setting up a script and I want it to systematically go through ALL POSSIBLE 2x2, 3x3, and 4x4 matrices modulo 2, 3, 4, 5, 6, and 7. For example, for modulus 2 in a 2x2, there would be 16 possibilities (4^2 because there are 4 positions with 2 possibilities each). I'm having trouble getting MATLAB to not only form all these possibilities but to put them through my script one at a time. Any thoughts?
Thanks!
This solution uses allcomb from matlab file exchange.
%size
n=2
%maximum value
m=2
%generate input for allcomb
e=cell(1,n^2)
e(1:end)={[0:m-1]}
%generate all combinations.
F=reshape(allcomb(e{:}),[],n,n)
F is a 3D-Matrix, to get the first possibility use:
squeeze(F(1,:,:))
A slight generalization of this Q&A does the job in one line:
r = 2; %// number of rows
c = 2; %// number of columns
n = 2; %// considered values: 0, 1, ..., n-1
M = reshape(dec2base(0:n^(r*c)-1, n).' - '0', r,c,[]);
Result for r, c, n as above:
M(:,:,1) =
0 0
0 0
M(:,:,2) =
0 0
0 1
...
M(:,:,16) =
1 1
1 1