I have a 3D matrix A (size m*n*k) where m=latitude, n*longitude and k=time.
I want only specific values from first and second dimension, specified by a logical matrix B (size m*n), and I want only the timesteps specified by vector C (size k).
In the end this should become a 2D matrix D, since the first two dimesions will colapse to one.
What is the most easy approach to do this?
And also is it possible to combine logical with linear indizes here? For example B is logical and C is linear?
Sample code with rand:
A=rand(10,10,10);
B=randi([0 1], 10,10);
C=randi([0 1], 10,1);
D=A(B,C) %This would be my approach which doesnt work. The size of D should be sum(B)*sum(c)
Another example without rand:
A=reshape([1:27],3,3,3);
B=logical([1,0,0;1,0,0;0,0,0]);
C=(1,3); %get data from timestep 1 and 5
D=A(B,C);%What I want to do, but doesnÄt work that way
D=[1,19;2,20];%Result should look like this! First dimension is now all data from dimesion 1 and 2. New dimesion 2 is now the time.
A = rand(4,4,4);
B = randi([0 1], 4,4)
B =
1 1 0 1
1 0 1 1
0 0 1 0
1 0 1 1
>> C = randi([0 1],1,1,4);
>> C(:)
ans =
0
1
1
0
Then use bsxfun or implicit expansion expansion whith .* if newer Matlab version to generate a matrix of logical for you given coordinates.
>> idx = logical(bsxfun(#times,B,C))
idx(:,:,1) =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
idx(:,:,2) =
1 1 0 1
1 0 1 1
0 0 1 0
1 0 1 1
idx(:,:,3) =
1 1 0 1
1 0 1 1
0 0 1 0
1 0 1 1
idx(:,:,4) =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
Then your output is D = A(idx). However, note that this D is now an Nx1 array. Where N is number of true elements is B times number of true elements in C. 10x True in B and 2x True in C:
>> size(D)
ans =
20 1
An easy way to do it is to first reshape A into an m*n-by-k matrix, then do your indexing:
result = reshape(A, [], size(A, 3));
result = result(B, C);
In this case C can be either a logical vector or vector of indices.
Related
I need to create all possible permutation matrices for a matrix where every permutation matrix contains only one 1 in each column and each row, and 0 in all other places.
For example, below example in (1) is all possible permutation matrices for 2x2 matrix and in (2) is a all possible permutation matrices for 3x3 matrix and so on
So how can I get these matrices of a matrix NxN in MATLAB and store them into one three-dimensional matrix?
Here's my solution, using implicit expansion (tested with Octave 5.2.0 and MATLAB Online):
n = 3;
% Get all permutations of length n
p = perms(1:n);
% Number of permutations
n_p = size(p, 1);
% Set up indices, where to set elements to 1
p = p + (0:n:n^2-1) + (0:n^2:n^2*n_p-1).';
% Set up indices, where to set elements to 1 (for MATLAB R2016a and before)
%p = bsxfun(#plus, bsxfun(#plus, p, (0:n:n^2-1)), (0:n^2:n^2*n_p-1).');
% Initialize 3-dimensional matrix
a = zeros(n, n, n_p);
% Set proper elements to 1
a(p) = 1
The output for n = 3:
a =
ans(:,:,1) =
0 0 1
0 1 0
1 0 0
ans(:,:,2) =
0 1 0
0 0 1
1 0 0
ans(:,:,3) =
0 0 1
1 0 0
0 1 0
ans(:,:,4) =
0 1 0
1 0 0
0 0 1
ans(:,:,5) =
1 0 0
0 0 1
0 1 0
ans(:,:,6) =
1 0 0
0 1 0
0 0 1
Using repelem, perms and reshape:
n = 3; % matrix size
f = factorial(n); % number of permutation
rep = repelem(eye(n),1,1,f) % repeat n! time the diagonal matrix
res = reshape(rep(:,perms(1:n).'),n,n,f) % indexing and reshaping
Where res is:
res =
ans(:,:,1) =
0 0 1
0 1 0
1 0 0
ans(:,:,2) =
0 1 0
0 0 1
1 0 0
ans(:,:,3) =
0 0 1
1 0 0
0 1 0
ans(:,:,4) =
0 1 0
1 0 0
0 0 1
ans(:,:,5) =
1 0 0
0 0 1
0 1 0
ans(:,:,6) =
1 0 0
0 1 0
0 0 1
And according to your comment:
What I need to do is to multiply a matrix i.e Z with all possible
permutation matrices and choose that permutation matrix which
resulting a tr(Y) minimum; where Y is the results of multiplication of
Z with the permutation matrix. I Think I don't need to generate all
permutation matrices and store them in such variable, I can generate
them one by one and get the result of multiplication. Is that possible
?
You're trying to solve the assignment problem, you can use the well known hungarian algorithm to solve this task in polynomial time. No needs to generate a googleplex of permutation matrix.
I am creating a Matlab program to handle an array of data that contains finite integers (not necessary to be consecutive). Let says my data array is A, I need to find the unique values of A first, that forms a vector B, and I have to count the number of occurrence of each unique value in A, that gives another vector C. Finally, I need a vector E which has the same length of A but with the value of the ith element of E computed as E(i) = C(k) with k = A(i)
I have the code to create B and C as follows:
A=[5 5 1 1 3 1 3 11 9 6 -2];
[B, ia, ic]=unique(A);
C = accumarray(ic,1);
A could be a large vector and the elements of A could be very different and in a wide range, I wonder if there is any vectorize approach to generate the array E instead of the following loop method
E = zeros(size(A));
for n=1:length(E)
k=find(B==A(n));
E(n) = C(k);
end
As you say, to calculate the values for E, you need to calculate the indices in B which correspond to each element in A, informally where A == B. This obviously won't work because A and B are different sizes. But we can get what we want by thinking about things as a 2D grid. Transpose one of the vectors and equate them.
A == B.'
% or for older versions of MATLAB (won't implicitly expand the different sized variables)
bsxfun(#eq, A, B.')
ans =
7×11 logical array
0 0 0 0 0 0 0 0 0 0 1
0 0 1 1 0 1 0 0 0 0 0
0 0 0 0 1 0 1 0 0 0 0
1 1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0 0 0
% to help visualise
% A = [5 5 1 1 3 1 3 11 9 6 -2];
% B = [-2 1 3 5 6 9 11];
You can see that, for each column, the position of the 1 tells us the where we can find that element in A in vector B. We can use find to get the row indices of each 1, and use that to calculate E.
[indices, ~] = find(A == B.');
E = C(indices);
Final result
If you want to calculate this with one line of code, you can use do the following
A=[5 5 1 1 3 1 3 11 9 6 -2];
[B, ia, ic]=unique(A);
C = accumarray(ic,1);
E = C(mod(find(B.' == A)-1, length(B))+1).';
% or for older versions of MATLAB
E = C(mod(find(bsxfun(#eq, A, B.'))-1, length(B))+1).';
I want to obtain all the possible permutations of one vector elements by another vector elements. For example one vector is A=[0 0 0 0] and another is B=[1 1]. I want to replace the elements of A by B to obtain all the permutations in a matrix like this [1 1 0 0; 1 0 1 0; 1 0 0 1; 0 1 1 0; 0 1 0 1; 0 0 1 1]. The length of real A is big and I should be able to choose the length of B_max and to obtain all the permutations of A with B=[1], [1 1], [1 1 1],..., B_max.
Thanks a lot
Actually, since A and B are always defined, respectively, as a vector of zeros and a vector of ones, this computation is much easier than you may think. The only constraints you should respect concerns B, which shoud not be empty and it's elements cannot be greater than or equal to the number of elements in A... because after that threshold A will become a vector of ones and calculating its permutations will be just a waste of CPU cycles.
Here is the core function of the script, which undertakes the creation of the unique permutations of 0 and 1 given the target vector X:
function p = uperms(X)
n = numel(X);
k = sum(X);
c = nchoosek(1:n,k);
m = size(c,1);
p = zeros(m,n);
p(repmat((1-m:0)',1,k) + m*c) = 1;
end
And here is the full code:
clear();
clc();
% Define the main parameter: the number of elements in A...
A_len = 4;
% Compute the elements of B accordingly...
B_len = A_len - 1;
B_seq = 1:B_len;
% Compute the possible mixtures of A and B...
X = tril(ones(A_len));
X = X(B_seq,:);
% Compute the unique permutations...
p = [];
for i = B_seq
p = [p; uperms(X(i,:).')];
end
Output for A_len = 4:
p =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 1 0 0
1 0 1 0
1 0 0 1
0 1 1 0
0 1 0 1
0 0 1 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
Say I have a vector A of item IDs:
A=[50936
332680
107430
167940
185820
99732
198490
201250
27626
69375];
And I have a matrix B whose rows contains values of 8 parameters for each of the items in vector A:
B=[0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1 0 1 0 0 1 0 1 1 1
1 0 1 0 0 1 0 1 1 1
0 0 1 0 0 0 0 1 0 1
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1];
So, column 1 in matrix B represents data of item in row 1 of vector A, column 2 in matrix B represents data of item in row 2 of vector A, and so on. However, I want matrix B to contain the information in a different order of items stored in vector A2:
A2=[185820
198490
69375
167940
99732
332680
27626
107430
50936
201250];
How do I sort them, so that column 1 of matrix B contains data for item in row 1 of vector A2, column 2 of matrix B contains data for item in row 2 of vector A2, and so on?
My extremely crude solution to do this is the following:
A=A'; A2=A2';
for i=1:size(A,2)
A(2:size(B,1)+1,i)=B(:,i);
end
A2(2:size(B,1)+1,:)=zeros(size(B,1),size(B,2));
for i=size(A2,2)
for j=size(A,2)
if A2(1,i)==A(1,j)
A2(2:end,i)=A(2:end,j);
end
end
end
B2 = A2(2:end,:);
But I would like to know a cleaner, more elegant and less time consuming method to do this.
A possible solution
You can use second output of ismember function.
[~ ,idx] = ismember(A2,A);
B2 = B(:,idx);
Update:I tested both my solution and another proposed by hbaderts
disp('-----ISMEMBER:-------')
tic
[~,idx]=ismember(A2,A);
toc
disp('-----SORT:-----------')
tic
[~,idx1] = sort(A);
[~,idx2] = sort(A2);
map = zeros(1,size(idx2));
map(idx2) = idx1;
toc
Here is the result in Octave:
-----ISMEMBER:-------
Elapsed time is 0.00157714 seconds.
-----SORT:-----------
Elapsed time is 4.41074e-05 seconds.
Conclusion: the sort method is more efficient!
As both A and A2 contain the exact same elements, just sorted differently, we can create a mapping from the A-sorting to the A2-sorting. For that, we run the sort function on both and save indexes (which are the second output).
[~,idx1] = sort(A);
[~,idx2] = sort(A2);
Now, the first element in idx1 corresponds to the first element in idx2, so A(idx1(1)) is the same as A2(idx2(1)) (which is 27626). To create a mapping idx1 -> idx2, we use matrix indexing as follows
map = zeros(size(idx2));
map(idx2) = idx1;
To sort B accordingly, all we need to do is
B2 = B(:, map);
[A2, sort_order] = sort(A);
B2 = B(:, sort_order)
MATLAB's sort function returns the order in which the items in A are sorted. You can use this to order the columns in B.
Transpose B so you can concatenate it with A:
C = [A B']
Now you have
C = [ 50936 0 0 1 1 0 0 0 0;
332680 0 0 0 0 0 0 0 0;
107430 0 0 1 1 1 0 0 0;
167940 0 0 0 0 0 0 0 0;
185820 0 0 0 0 0 0 0 0;
99732 0 0 1 1 0 0 0 0;
198490 0 0 0 0 0 0 0 0;
201250 0 0 1 1 1 1 0 0;
27626 0 0 1 1 0 0 0 0;
69375 0 0 1 1 1 0 0 1];
You can now sort the rows of the matrix however you want. For example, to sort by ID in ascending order, use sortrows:
C = sortrows(C)
To just swap rows around, use a permutation of 1:length(A):
C = C(perm, :)
where perm could be something like [4 5 6 3 2 1 8 7 9 10].
This way, your information is all contained in one structure and the data is always correctly matched to the proper ID.
Say I have a sparse non-rectangular matrix A:
>> A = round(rand(4,5))
A =
0 1 0 1 1
0 1 0 0 1
0 0 0 0 1
0 1 1 0 0
I would like to obtain the matrix B where the non-zero entries of A are replaced by their linear index in row-first order:
B =
0 2 0 4 5
0 7 0 0 10
0 0 0 0 15
0 17 18 0 0
and the matrix C that where the non-zero entries of A are replaced by the order in which they are found in a row-first search:
C =
0 1 0 2 3
0 4 0 0 5
0 0 0 0 6
0 7 8 0 0
I am looking for vectorized solutions for this problem that scale to large sparse matrices.
If I understand what you are asking, a couple of tranpositions should do the trick. The key is that find(A.') will do "row-first" indexing on A, where .' is the short hand for the transpose of a 2D matrix. So:
>> A = round(rand(4,5))
A =
0 1 0 1 1
0 1 0 0 1
0 0 0 0 1
0 1 1 0 0
then
B=A.';
B(find(B)) = find(B);
B=B.';
gives
B =
0 2 0 4 5
0 7 0 0 10
0 0 0 0 15
0 17 18 0 0
Here's a solution that doesn't require any transposing back and forth:
>> B = A; %# Initialize B
>> C = A; %# Initialize C
>> mask = logical(A); %# Create a logical mask using A
>> [r,c] = find(A); %# Find the row and column indices of non-zero values
>> index = c + (r - 1).*size(A,2); %# Compute the row-first linear index
>> [~,order] = sort(index); %# Compute the row-first order with
>> [~,order] = sort(order); %# two sorts
>> B(mask) = index %# Fill non-zero elements of B
B =
0 2 0 4 5
0 7 0 0 10
0 0 0 0 15
0 17 18 0 0
>> C(mask) = order %# Fill non-zero elements of C
C =
0 1 0 2 3
0 4 0 0 5
0 0 0 0 6
0 7 8 0 0
An outline (Matlab isn't on this machine, so verification is delayed):
You can use find() to get the coordinate list. Let T = A'; [r,c] = find(T)
From the coordinate list, you can create both B and C. Let valB = sub2ind([r,c],T) and valC = 1:length(r)
Use the sparse command to create B and C, e.g. B = sparse(r,c,valB), and then transpose, e.g. B = B' (or could do sparse(c,r,valB)).
Or, as #IanHincks suggests, let B = A'; B(find(B)) = find(B). (I'm not sure why .' is recommended, but, again, I don't have Matlab in front of to check.) For C, simply use C(find(C)) = 1:nnz(A). And transpose back, as he suggests.
Personally, I work with coordinate lists all the time, having migrated away from the sparse matrix representation, just to cut out the costs of index lookups.