I have following problem:
I delete entries when in both matrices they are zero. If I have i pairs of such matrices, how to properly write indexing for the loop in matlab here?
code:
x = [0 0 0 1 1 0 5 0 7 0]
y = [0 2 0 1 1 2 5 2 7 0]
idx = ~(x==0 & y==0);
x2 = x(idx)
y2 = y(idx)
can you help me?
If I understand you correctly, you want to match elements where both x and y are zero, so something like this should work (without the not ~):
idx = (x==0 & y==0);
x2 = x(~idx)
y2 = y(~idx)
Edit
Or more simply, as suggested by mutzmatron:
idx = (x ~= 0 | y ~= 0);
x2 = x(idx)
y2 = y(idx)
Like this perhaps ?
x2 = x(find(or((x~=0),(y~=0))))
y2 = y(find(or((x~=0),(y~=0))))
These, obviously, don't do an in-place replacement, so x and y are not changed. I haven't figured out how to do that on both arrays simultaneously yet.
EDIT Read #mutzmatron's comments below
Related
To better explain what I need, here is my first code:
function allpos = f1(x)
allpos=reshape(permute((dec2base(0:power(2,x*x)-1,2)-'0'),[3 2 1]),x,x,[]);
This code does exactly what I need it to. If the users inputs f1(2), it returns every matrix from [0 0; 0 0] to [1 1; 1 1]. However, it also gives me a lot of useless matrices. I only want matrices that are mirrored across the diagonal, with only zeros on the diagonal.
To put more simply, for f1(3), I only want
[0 0 0; 0 0 0; 0 0 0] to [0 1 1; 1 0 1; 1 1 0]. That means that if I run the new f1(3); it will return 8 matrices, not 512.
How do I rewrite the function to do this? I know that it will require some sort of addition of a triangular matrix and its transposed self, but I cant piece it together. Thanks!
It can probably be made more concise but the following does what you ask.
function allpos = f1(x)
N = (x-1)*x / 2;
z = permute(dec2base(0:power(2,N)-1,2)-'0',[3 2 1]);
allpos = zeros(x,x,power(2,N));
idx = repmat(logical(tril(ones(x),-1)),[1,1,power(2,N)]);
allpos(idx) = z(:);
allpos = permute(allpos,[2 1 3]);
allpos(idx) = z(:);
A similar solution to #jodag's one, but more concise...
function allops=f1(x)
allops=zeros(x,x,2^(sum(1:x-1)));
allops(find(triu(ones(x),1))+x^2*(0:2^(sum(1:x-1))-1))=[dec2base(0:2^(sum(1:x-1))-1,2)-'0'].';
allops=allops+permute(allops,[2 1 3])
end
I'm unsure how to phrase the question, but I think an example will help. Suppose I have a vector y = [3;1;4;1;6]. I want to create the matrix Y =
[0 0 1 0 0 0;
1 0 0 0 0 0;
0 0 0 1 0 0;
1 0 0 0 0 0;
0 0 0 0 0 1]
↑ ↑ ↑ ↑ ↑ ↑
1 2 3 4 5 6
where the element on each column is one or zero corresponding to the value in the vector.
I found that I could do it using
Y = []; for k = 1:max(y); Y = [Y (y==k)]; end
Can I do it without a for loop (and is this method more efficient if y has thousands of elements)?
Thanks!
Your method is not efficient because you're growing the size of Y in the loop which is not a good programming practice. Here is how your code can be fixed:
Ele = numel(y);
Y= zeros(Ele, max(y));
for k = 1:Ele
Y (k,y(k))= 1;
end
And here is an alternative approach without a loop:
Ele = numel(y); %Finding no. of elements in y
Y= zeros(Ele, max(y)); % Initiailizing the matrix of the required size with all zeros
lin_idx = sub2ind(size(Y), 1:Ele, y.'); % Finding linear indexes
Y(lin_idx)=1 % Storing 1 in those indexes
You can use bsxfun:
result = double(bsxfun(#eq, y(:), 1:max(y)));
If you are running the code on Matlab version R2016b or later, you can simplify the syntax to
result = double(y(:)==(1:max(y)));
Another approach, possibly more efficient, is to fill in the values directly using accumarray:
result = accumarray([(1:numel(y)).' y(:)], 1);
I found another solution:
E = eye(max(y));
Y = E(y,:);
Another solution:
Y = repmat(1:max(y), size(y)) == repmat(y, 1, max(y))
I'm trying to vectorize the following MATLAB operation:
Given a column vector with indexes, I want a matrix with the
same number of rows of the column and a fixed number of columns. The
matrix is initialized with zeroes and contains ones in the locations
specified by the indexes.
Here is an example of the script I've already written:
y = [1; 3; 2; 1; 3];
m = size(y, 1);
% For loop
yvec = zeros(m, 3);
for i=1:m
yvec(i, y(i)) = 1;
end
The desired result is:
yvec =
1 0 0
0 0 1
0 1 0
1 0 0
0 0 1
Is it possible to achieve the same result without the for loop? I tried something like this:
% Vectorization (?)
yvec2 = zeros(m, 3);
yvec2(:, y(:)) = 1;
but it doesn't work.
Two approaches you can use here.
Approach 1:
y = [1; 3; 2; 1; 3];
yvec = zeros(numel(y),3);
yvec(sub2ind(size(yvec),1:numel(y),y'))=1
Approach 2 (One-liner):
yvec = bsxfun(#eq, 1:3,y)
Yet another approach:
yvec = full(sparse(1:numel(y),y,1));
You could do this with accumarray:
yvec = accumarray([(1:numel(y)).' y], 1);
I did it this way:
classes_count = 10;
sample_count = 20;
y = randi([1 classes_count], 1, sample_count);
y_onehot = zeros(classes_count, size(y, 2));
idx = sub2ind(size(y_onehot), y, [1:size(y, 2)]);
y_onehot(idx) = 1
Given some multidimensional matrix A in Octave / Matlab,
What's the easiest way to get a matrix of the same size as A where all elements are replaced by their index along the k'th dimension
ie for the matrix
A =
ans(:,:,1) =
0.095287 0.191905
0.226278 0.749100
ans(:,:,2) =
0.076826 0.131639
0.862747 0.699016
I want a function f such that
f(A,1) =
ans(:,:,1) =
1 1
2 2
ans(:,:,2) =
1 1
2 2
f(A,2) =
ans(:,:,1) =
1 2
1 2
ans(:,:,2) =
1 2
1 2
and
f(A, 3) =
ans(:,:,1) =
1 1
1 1
ans(:,:,2) =
2 2
2 2
Also, given a sparse matrix B
What's the easiest way to get another sparse matrix of the same size where the nonzero elements are replaced by their index along the k'th dimension? (so same problem as above, but for only the nonzero elements)
Ideally I'm looking for a way which is well-vectorized for octave (meaning it doesn't explicitly loop over anything)
CLARIFICATION: For the sparse matrix one, I'm looking for a solution which does not involve creating a full size(B) matrix at any point
ndgrid() does what you want, although not in the format you are looking for. If you know the dims of the input A beforehand, you can use the following line to create the N-dimentional mesh grid:
% for matrix a where ndims(a) == 3
[x, y, z] = ndgrid (1:size(a,1), 1:size(a,2), 1:size(a,3));
% x is like f(a, 1)
% y is like f(a, 2)
% z is like f(a, 3)
You may be able to write a custom wrapper around ndgrid() to convert it to the function format you are looking for.
In case anyone's curious, since I didn't know about ndgrid, here's the answer I came up with:
function [y] = indices(a,k)
s = size(a);
n = s(k);
D = length(s);
x = permute(a,[k,1:(k-1),(k+1):D]);
y = reshape(x,n,[]);
y = diag(1:n) * ones(size(y));
y = reshape(y,size(x));
y = permute(y,[(2:k),1,(k+1):D]);
endfunction
function [y] = spindices(a,k)
s = size(a);
n = s(k);
D = length(s);
x = permute(a,[k,1:(k-1),(k+1):D]);
y = reshape(x,n,[]);
y = spdiag(1:n) * spones(y);
y = reshape(y,size(x));
y = permute(y,[(2:k),1,(k+1):D]);
endfunction
Hi everyone this is What I did to carry out an iteration method(gauss seidel) and I want when iteration number greater than 30 it will stop and generate the corresponding result up to 30 iteration. But I wonder why the output result were so weird and I try to check the value on the command window by typing x_ans(:,1) it gives me the correct value. It really made me frustrated why the generate result were not the same. Or any other circumstance or function can be used to set for not converging condition. Sincerely thanks in advance for every single help.
clear;clc
A = [2 8 3 1;0 2 -1 4;7 -2 1 2;-1 0 5 2]
B = [-2;4;3;5]
Es = 1e-5
n = length(B);
x = zeros(n,1);
Ea = ones(n,1);
iter = 0;
while max(Ea) >= Es
if iter <= 30
iter = iter + 1;
x_old = x;
for i = 1:n
j = 1:n;
j(i) = [];
x_cal = x;
x_cal(i) = [];
x(i) = (B(i) - sum(A(i,j) * x_cal)) / A(i,i);
end
else
break
end
x_ans(:,iter) = x;
Ea(:,iter) =abs(( x - x_old) ./ x);
end
result = [1:iter; x_ans; Ea]'
I've gone through the formulas and they are all OK. On a side note, the sum is not necessary. The problem lies with your input data - try reordering! check for example the following, which works
A = [7 -2 1 2;
2 8 3 1;
-1 0 5 2;
0 2 -1 4;]
B = [3;-2;5;4]
see the wiki under convergence.