I am trying to get M. To do this, it is necessary for Matlab to relate column 1 of A to column 1 of π΅^πand build a matrix M with 1 and 0 depending on whether in position π΄[π,π] and π΅^π[π,π] are equal to 1
A = [1 0 1; 0 1 1; 0 0 1 ];
B = [0 0 1 ; 0 1 0; 1 1 1];
for i = 1:3
for j =1:3
if A(i,j) == BT(i,j) && A(i,j)==1;
Z(i,j) = 1
end
end
end
When you use "if A(i,j) == BT(i,j) && A(i,j)==;" you are comparing individual elements. Instead you want to be comparing columns:
A(:, i) and BT(:, j).
Precisely, you want
for i = 1:3
for j = 1:3
M(i,j) = any( A(:,i) & BT(:,j) );
end
end
OR
You are comparing the columns of BT and the columns of A.
That is to say, the rows of B and the columns of A. You want to see if there are any occasions when both of the elements are 1. Thus you can compare the products of the terms in the rows of B and columns of A.
i.e. M = logical(B * A) should also give you the desired output.
NOTE that the data in B are different in your image examples and in your code.
Related
I want to find out row and column number of zeros in 3 dimensional space. Problem is I get output vectors(e.g row) of different length each time, hence dimension error occurs.
My attempt:
a (:,:,1)= [1 2 0; 2 0 1; 0 0 2]
a (:,:,2) = [0 2 8; 2 1 0; 0 0 0]
for i = 1 : 2
[row(:,i) colum(:,i)] = find(a(:,:,i)==0);
end
You can use linear indexing:
a (:,:,1) = [1 2 0; 2 0 1; 0 0 2];
a (:,:,2) = [0 2 8; 2 1 0; 0 0 0];
% Answer in linear indexing
idx = find(a == 0);
% Transforms linear indexing in rows-columns-3rd dimension
[rows , cols , third] = ind2sub(size(a) ,idx)
More on the topic can be found in Matlab's help
Lets assume your Matrix has the format N-by-M-by-P.
In your case
N = 3;
M = 3;
P = 2;
This would mean that the maximum length of rows and coloms from your search (if all entries are zero) is N*M=9
So one possible solution would be
%alloc output
row=zeros(size(a,1)*size(a,2),size(a,3));
colum=row;
%loop over third dimension
n=size(a,3);
for i = 1 : n
[row_t colum_t] = find(a(:,:,i)==0);
%copy your current result depending on it's length
row(1:length(row_t),i)=row_t;
colum(1:length(colum_t),i)=colum_t;
end
However, when you past the result to the next function / script you have to keep in mind to operate on the non-zero elements.
I would go for the vectorized solution of Zep. As for bigger matrices a it is more memory efficient and I am sure it must be way faster.
I am trying to solve this problem:
Write a function called cancel_middle that takes A, an n-by-m
matrix, as an input where both n and m are odd numbers and k, a positive
odd integer that is smaller than both m and n (the function does not have to
check the input). The function returns the input matrix with its center k-by-k
matrix zeroed out.
Check out the following run:
>> cancel_middle(ones(5),3)
ans =
1 1 1 1 1
1 0 0 0 1
1 0 0 0 1
1 0 0 0 1
1 1 1 1 1
My code works only when k=3. How can I generalize it for all odd values of k? Here's what I have so far:
function test(n,m,k)
A = ones(n,m);
B = zeros(k);
A((end+1)/2,(end+1)/2)=B((end+1)/2,(end+1)/2);
A(((end+1)/2)-1,((end+1)/2)-1)= B(1,1);
A(((end+1)/2)-1,((end+1)/2))= B(1,2);
A(((end+1)/2)-1,((end+1)/2)+1)= B(1,3);
A(((end+1)/2),((end+1)/2)-1)= B(2,1);
A(((end+1)/2),((end+1)/2)+1)= B(2,3);
A(((end+1)/2)+1,((end+1)/2)-1)= B(3,1);
A(((end+1)/2)+1,((end+1)/2))= B(3,2);
A((end+1)/2+1,(end+1)/2+1)=B(3,3)
end
You can simplify your code. Please have a look at
Matrix Indexing in MATLAB. "one or both of the row and column subscripts can be vectors", i.e. you can define a submatrix. Then you simply need to do the indexing correct: as you have odd numbers just subtract m-k and n-k and you have the number of elements left from your old matrix A. If you divide it by 2 you get the padding on the left/right, top/bottom. And another +1/-1 because of Matlab indexing.
% Generate test data
n = 13;
m = 11;
A = reshape( 1:m*n, n, m )
k = 3;
% Do the calculations
start_row = (n-k)/2 + 1
start_col = (m-k)/2 + 1
A( start_row:start_row+k-1, start_col:start_col+k-1 ) = zeros( k )
function b = cancel_middle(a,k)
[n,m] = size(a);
start_row = (n-k)/2 + 1;
start_column = (m-k)/2 + 1;
end_row = (n-k)/2 + k;
end_column = (m-k)/2 + k;
a(start_row:end_row,start_column:end_column) = 0;
b = a;
end
I have made a function in an m file called cancel_middle and it basically converts the central k by k matrix as a zero matrix with the same dimensions i.e. k by k.
the rest of the matrix remains the same. It is a general function and you'll need to give 2 inputs i.e the matrix you want to convert and the order of submatrix, which is k.
I've a series of coordinates (i,j) and I want to loop through each one.
For example
A = ones(3,3);
i = [1 2 3];
j = [3 2 1];
I tried with this but it doesn't work:
for (i = i && j = j)
A(i,j) = 0;
end
I also tried this but it doens't work as expected:
for i = i
for j = j
A(i,j) = 0;
end
end
Desired result:
A =
1 1 0
1 0 1
0 1 1
Although A is a matrix in this example, I am working with table data.
The correct syntax to do what you want is:
A = ones(3,3);
i = [1 2 3];
j = [3 2 1];
for ii = 1:length( i )
A( i(ii) , j(ii) ) = 0;
end
Essentially you loop through each element and index i and j accordingly using ii. ii loops through 1..3 indexing each element.
This will give the a final result below.
>> A
A =
1 1 0
1 0 1
0 1 1
While this works and fixes your issue, I would recommend rayryeng's alternate solution with conversions if you don't have more complex operations involved.
Though this doesn't answer your question about for loops, I would avoid using loops all together and create column-major linear indices to access into your matrix. Use sub2ind to help facilitate that. sub2ind takes in the size of the matrix in question, the row locations and column locations. The output will be an array of values that specify the column-major locations to access in your matrix.
Therefore:
A = ones(3); i = [1 2 3]; j = [3 2 1]; %// Your code
%// New code
ind = sub2ind(size(A), i, j);
A(ind) = 0;
Given that you have a table, you can perhaps convert the table into an array, apply sub2ind on this array then convert the result back to a table when you're done. table2array and array2table are useful tools here. Given that your table is stored in A, you can try:
Atemp = table2array(A);
ind = sub2ind(size(Atemp), i, j);
Atemp(ind) = 0;
A = array2table(Atemp);
I need to replace the zeros (or NaNs) in a matrix with the previous element row-wise, so basically I need this Matrix X
[0,1,2,2,1,0;
5,6,3,0,0,2;
0,0,1,1,0,1]
To become like this:
[0,1,2,2,1,1;
5,6,3,3,3,2;
0,0,1,1,1,1],
please note that if the first row element is zero it will stay like that.
I know that this has been solved for a single row or column vector in a vectorized way and this is one of the nicest way of doing that:
id = find(X);
X(id(2:end)) = diff(X(id));
Y = cumsum(X)
The problem is that the indexing of a matrix in Matlab/Octave is consecutive and increments columnwise so it works for a single row or column but the same exact concept cannot be applied but needs to be modified with multiple rows 'cause each of raw/column starts fresh and must be regarded as independent. I've tried my best and googled the whole google but coukldnβt find a way out. If I apply that same very idea in a loop it gets too slow cause my matrices contain 3000 rows at least. Can anyone help me out of this please?
Special case when zeros are isolated in each row
You can do it using the two-output version of find to locate the zeros and NaN's in all columns except the first, and then using linear indexing to fill those entries with their row-wise preceding values:
[ii jj] = find( (X(:,2:end)==0) | isnan(X(:,2:end)) );
X(ii+jj*size(X,1)) = X(ii+(jj-1)*size(X,1));
General case (consecutive zeros are allowed on each row)
X(isnan(X)) = 0; %// handle NaN's and zeros in a unified way
aux = repmat(2.^(1:size(X,2)), size(X,1), 1) .* ...
[ones(size(X,1),1) logical(X(:,2:end))]; %// positive powers of 2 or 0
col = floor(log2(cumsum(aux,2))); %// col index
ind = bsxfun(#plus, (col-1)*size(X,1), (1:size(X,1)).'); %'// linear index
Y = X(ind);
The trick is to make use of the matrix aux, which contains 0 if the corresponding entry of X is 0 and its column number is greater than 1; or else contains 2 raised to the column number. Thus, applying cumsum row-wise to this matrix, taking log2 and rounding down (matrix col) gives the column index of the rightmost nonzero entry up to the current entry, for each row (so this is a kind of row-wise "cummulative max" function.) It only remains to convert from column number to linear index (with bsxfun; could also be done with sub2ind) and use that to index X.
This is valid for moderate sizes of X only. For large sizes, the powers of 2 used by the code quickly approach realmax and incorrect indices result.
Example:
X =
0 1 2 2 1 0 0
5 6 3 0 0 2 3
1 1 1 1 0 1 1
gives
>> Y
Y =
0 1 2 2 1 1 1
5 6 3 3 3 2 3
1 1 1 1 1 1 1
You can generalize your own solution as follows:
Y = X.'; %'// Make a transposed copy of X
Y(isnan(Y)) = 0;
idx = find([ones(1, size(X, 1)); Y(2:end, :)]);
Y(idx(2:end)) = diff(Y(idx));
Y = reshape(cumsum(Y(:)), [], size(X, 1)).'; %'// Reshape back into a matrix
This works by treating the input data as a long vector, applying the original solution and then reshaping the result back into a matrix. The first column is always treated as non-zero so that the values don't propagate throughout rows. Also note that the original matrix is transposed so that it is converted to a vector in row-major order.
Modified version of Eitan's answer to avoid propagating values across rows:
Y = X'; %'
tf = Y > 0;
tf(1,:) = true;
idx = find(tf);
Y(idx(2:end)) = diff(Y(idx));
Y = reshape(cumsum(Y(:)),fliplr(size(X)))';
x=[0,1,2,2,1,0;
5,6,3,0,1,2;
1,1,1,1,0,1];
%Do it column by column is easier
x=x';
rm=0;
while 1
%fields to replace
l=(x==0);
%do nothing for the first row/column
l(1,:)=0;
rm2=sum(sum(l));
if rm2==rm
%nothing to do
break;
else
rm=rm2;
end
%replace zeros
x(l) = x(find(l)-1);
end
x=x';
I have a function I use for a similar problem for filling NaNs. This can probably be cutdown or sped up further - it's extracted from pre-existing code that has a bunch more functionality (forward/backward filling, maximum distance etc).
X = [
0 1 2 2 1 0
5 6 3 0 0 2
1 1 1 1 0 1
0 0 4 5 3 9
];
X(X == 0) = NaN;
Y = nanfill(X,2);
Y(isnan(Y)) = 0
function y = nanfill(x,dim)
if nargin < 2, dim = 1; end
if dim == 2, y = nanfill(x',1)'; return; end
i = find(~isnan(x(:)));
j = 1:size(x,1):numel(x);
j = j(ones(size(x,1),1),:);
ix = max(rep([1; i],diff([1; i; numel(x) + 1])),j(:));
y = reshape(x(ix),size(x));
function y = rep(x,times)
i = find(times);
if length(i) < length(times), x = x(i); times = times(i); end
i = cumsum([1; times(:)]);
j = zeros(i(end)-1,1);
j(i(1:end-1)) = 1;
y = x(cumsum(j));
I need to generate a random matrix of K columns and N rows containing ones and zeroes, such that:
a) Each row contains exactly k ones.
b) Each row is different from the other (combinatorics imposes that if N > nchoosek(K, k) there will be nchoosek(K,k) rows).
Assume I want N = 10000 (out of all the possible nchoosek(K, k) = 27405 combinations), different 1ΓK vectors (with K = 30) containing k (with k = 4) ones and K - k zeroes.
This code:
clear all; close
N=10000; K=30; k=4;
M=randi([0 1],N,K);
plot(sum(M,2)) % condition a) not satisfied
does not satisfy neither a) nor b).
This code:
clear all; close;
N=10000;
NN=N; K=30; k=4;
tempM=zeros(NN,K);
for ii=1:NN
ttmodel=tempM(ii,:);
ttmodel(randsample(K,k,false))=1; %satisfies condition a)
tempM(ii,:)=ttmodel;
end
Check=bi2de(tempM); %from binary to decimal
[tresh1,ind,tresh2] = unique(Check);%drop the vectors that appear more than once in the matrix
M=tempM(ind,:); %and satisfies condition b)
plot(sum(M,2)) %verify that condition a) is satisfied
%Effective draws, Wanted draws, Number of possible combinations to draw from
[sum(sum(M,2)==k) N nchoosek(K,k) ]
satisfies condition a) and partially condition b). I say partially because unless NN>>N the final matrix will contain less than N rows each different from each other.
Is there a better and faster way (that possible avoids the for cycle and the need of having NN>>N) to solve the problem?
First, generate N unique k-long permutations of the positions of ones:
cols = randperm(K, N);
cols = cols(:, 1:k);
Then generate the matching row indices:
rows = meshgrid(1:N, 1:k)';
and finally create the sparse matrix with:
A = sparse(rows, cols, 1, N, K);
To obtain the full form of the matrix, use full(A).
Example
K = 10;
k = 4;
N = 5;
cols = randperm(K, N);
cols = cols(:, 1:k);
rows = meshgrid(1:N, 1:k)';
A = sparse(rows, cols , 1, N, K);
full(A)
The result I got is:
ans =
1 1 0 0 0 0 0 1 0 1
0 0 1 1 0 1 0 0 0 1
0 0 0 1 1 0 1 0 1 0
0 1 0 0 0 0 1 0 1 1
1 1 1 0 0 1 0 0 0 0
This computation should be pretty fast even for large values of K and N. For K = 30, k = 4, N = 10000 the result was obtained in less than 0.01 seconds.
You could use randperm(n) to generate random sequences of integers from 1 to n, and store the nonrepeated sequences as rows in a matrix M until size(unique(M,'rows'),1)==size(M,1). Then you could use M to index a logical matrix with the appropriate number of true values in each row.
If you have enough memory for nchoosek(K,k) integers, build an array of those, use a partial Fisher-Yates shuffle to get a proper uniformly random subset of N of those. Now, given the array of N integers, interpret each as the rank of the combination representing each row of your final array. If you use colexicographical ordering of combinations, computing the combination from a rank is pretty simple (though it uses lots of binomial combination functions, so it pays to have a fast one).
I'm not a Matlab guy, but I've done things similar to this in C. This code, for example:
for (i = k; i >= 1; --i) {
while ((b = binomial(n, i)) > r) --n;
buf[i-1] = n;
r -= b;
}
will fill the array buf[] with indices from 0 to n-1 for the rth combination of k out of n elements in colex order. You would interpret these as the positions of the 1s in your row.