Given a nXm matrix A and a mX2 matrix B and a matrix C of size mX1 containing 1s and 2s C=[1 2 1 2 1...], depending on which column, I want every row of A to be multiplied with. How can this be done? Or equivalently, given D = A*B how can I access only the values dictated by C. I tried D(:,C), but the result is not the expected.
Example a =[1 2; 3 4; 5 6] . c = [1 2 1] . a(?) = [1 4 5]
Any idea?
%example data
n=10;m=20;
A=rand(n,m)
B=rand(m,2)
C=round(rand(m,1))+1;
%solution:
B2=B(:,1); %multiplication vector
B2(C==2)=B(C==2,2) %change the ones where C==2
A*B2
You can run the following command for the last example:
a(sub2ind([3,2],1:3,c))'
In general case you can do like the following:
% n is the length of the D which is nx2 matrix
D(sub2ind([n,2],1:n,C))'
Related
Any idea how to formulate this sum other than using a loop?
sum(i) f(k(i),x) where k_i are some entries of a vector and x is a matrix.
Currently what I'm doing is this, but I'd rather have a general solution:
for ii=1:length(k)
psi=psi+f(k(ii),x)
end
If it's any concern:
f(k,x)=g(k)*exp(k*x)
Assuming g can take a vector input and returns a vector result of the same size, and that x is just a scalar
f=#(k,x) g(k).*exp(k*x);
psi=sum(f(k,x))
or if g can't be or isn't able to take vector input, you can do
g=#(k) arrayfun(g,k);
and then define f as before.
Do you mean that you want to sum only specific rows?
If so, this will do it:
a= [1 2 3 4;
5 2 7 2;
0 0 2 3];
k= [1 3]; %rows selection for sum
result= sum(a(k,:))
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 have multiple matrices of the same size and want to compare them.
As a result I need a matrix which gives me the biggest of the 3 for every value.
I will clarify what i mean with an example:
I have 3 matrices with data of 3 persons.
I would like to compare these 3 and get a matrix as result.
In that matrix every cell/value should be the name of the matrix who had the highest value for that cell. So if in the 3 matrices the first value (1 colum, 1 row) is accordingly 2, 5, 8 the first value of the result matrix should be 3 (or the name of the 3 matrix).
If the three matrices are A, B, C, do this:
[~, M] = max(cat(3,A,B,C),[],3);
It creates a 3D "matrix" and maximizes across the third dimension.
Concatenate them on the 3rd dimension, and the use the SECOND output from max to get exactly what you want
A = rand(3,3);
B = rand(3,3);
C = rand(3,3);
D = cat(3, A, B, C)
[~, Solution] = max(D, [], 3)
e.g.:
D =
ans(:,:,1) =
0.70101 0.31706 0.83874
0.89421 0.33783 0.55681
0.68520 0.11697 0.45631
ans(:,:,2) =
0.268715 0.213200 0.124450
0.869847 0.999649 0.153353
0.345447 0.023523 0.338099
ans(:,:,3) =
0.216665 0.297900 0.604734
0.103340 0.767206 0.660668
0.127052 0.430861 0.021584
Solution =
1 1 1
1 2 3
1 3 1
edit
As I did not know about the second argument of the max-function, here is what you should NOT use:
old
Well, quick&dirty:
x=[2 5 8];
w=max(x)
[~,loc] = ismember(w,x)
Suppose now I have two vectors of same length:
A = [1 2 2 1];
B = [2 1 2 2];
I would like to create a matrix C whose dim=m*n, m=max(A), n=max(B).
C = zeros(m,n);
for i = 1:length(A)
u = A(i);
v = B(i);
C(u,v)=C(u,v)+1;
end
and get
C =[0 2;
1 1]
More precisely, we treat the according indices in A and B as rows and columns in C, and C(u,v) is the number of elements in {k | A(i)=u and B(i)=v, i = 1,2,...,length(A)}
Is there a faster way to do that?
Yes. Use sparse. It assembles (i.e., sums up) the matrix values for repeating row-column pairs for you. You need an additional vector with the values that will be assembled into the matrix entries. If you use ones(size(A)), you will have exactly what you need - counting of repeated row-column pairs
spA=sparse(A, B, ones(size(A)));
full(spA)
ans =
0 2
1 1
The same can be obtained by simply passing scalar 1 to sparse function instead of a vector of values.
For matrices that have a large number of zero entries this is absolutely crucial that you use sparse storage. Another function you could use is accumarray. It can essentially do the same thing, but also works on dense matrix structure:
AA=accumarray([A;B]', 1);
AA =
0 2
1 1
You can pass size argument to accumarray if you want to create a matrix of specific size
AA=accumarray([A;B]', 1, [2 3]);
AA =
0 2 0
1 1 0
Note that you can actually also make it produce sparse matrices, and use a different operator in assembly (i.e., not necessarily a sum)
AA=accumarray([A;B]', 1, [2 3], #sum, 0, true)
will produce a sparse matrix (last parameter set to true) using sum for assembly and 0 as a fill value, i.e. a value which is used in cases a given row-column pair does not exist in A/B.
Assume I have two sets of matrix (A and B), inside each matrix contains few point coordinates, I want to find out point in B nearest to A and output a cell array C listed the nearest point pair coordinates accordingly and one cell array D register the unpaired spot, how should I do it?
To be more specific, here is what I want
Two sets of matrix contain spot xy coordinates;
A=[ 1 2; 3 4]; B=[1 3; 5 6; 2 1];
want to get C{1,1}=[1 2; 1 3]; C{2,1}= [3 4; 5 6]; D{1,1}=[2 1];
Thanks for the help.
There is not exactly one solution to this problem, take for example the (one-dimensional, but expandable to N-D) case:
A= [1; 3];
B= [2];
Then either A(1) or A(2) can be the leftover point. Which one your algorithm spits out, will depend on how it works, ie which point you take first to find the nearest point.
Such algorithm consists imo of
Finding distances between each combination of A(i) and B(j). If you have the statistics toolbox, pdist2 does this for you:
A=[ 1 2; 3 4];
B=[1 3; 5 6; 2 1];
dist = pdist2(A,B);
Looping over the smallest of A or B (I'll take A, cause it is smallest in your example) and finding for each point in A the closest point in the remaining set of B:
N = size(A,1);
matchAtoB=NaN(N,1);
for ii=1:N
dist(:,matchAtoB(1:ii-1))=Inf; % make sure that already picked points of B are not eligible to be new closest point
[~,matchAtoB(ii)]=min(dist(ii,:));
end
matchBtoA = NaN(size(B,1),1);
matchBtoA(matchAtoB)=1:N;
remaining_indices = find(isnan(matchBtoA));
Combine result to your desired output matrices C and D:
C=arrayfun(#(ii) [A(ii,:) ; B(matchAtoB(ii),:)],1:N,'uni',false);
D=mat2cell(B(remaining_indices,:),ones(numel(remaining_indices),1),size(B,2));
Note that this code will also work with 1D points or higher (N-D), the pdist2 flattens everything out to scalar distances.
Here's my take on the problem:
A=[1 2
3 4];
B=[1 3
5 6
2 1];
dists = pdist2(A,B);
[dists, I] = sort(dists,2);
c = NaN(size(A,1),1);
for ii = 1:size(A,1)
newC = find(~any(bsxfun(#eq, I(ii,:), c), 1));
c(ii) = I(ii,newC(1));
end
C = cellfun(#(x)reshape(x,2,2).',...
mat2cell([A B(c,:)], ones(size(A,1),1), 4), 'uni', false);
D = {B(setdiff(1:size(B,1),c), :)}
This solution assumes
all your vectors are 2D
stacked in rows of A and B
and A is always the source (i.e., everything is compared to A)
If these assumptions do not (always) hold, you'll have to take a more general approach, like the one suggested by #GuntherStruyf.