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))
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 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 have a matrix (89x42) of 0's and 1's that I'd like to multiply combinations of rows together.
For example, for matrix
input = [1 0 1
0 0 0
1 1 0];
and with 2 combinations, I want an output of
output = [0 0 0; % (row1*row2)
1 0 0; % (row1*row3)
0 0 0] % (row2*row3)
Which rows to multiply is dictated by "n Choose 2" (nCk), or all possible combinations of the rows n taken k at a time. In this case k=2.
Currently I am using a loop and it works fine for the 89C2 combinations of rows, but when I run it with 89C3 it takes far too long too run.
What would be the most efficient way to do this program so I can do more than 2 combinations?
You can do it using nchoosek and element-wise multiplication.
inp = [1 0 1; 0 0 0; 1 1 0]; %Input matrix
C = nchoosek(1:size(inp,1),2); %Number of rows taken 2 at a time
out = inp(C(:,1),:) .* inp(C(:,2),:); %Multiplying those rows to get the desired output
Several things you can do:
Use logical ("binary") arrays (or even sparse logical arrays) instead of double arrays.
Use optimized combinatorical functions.
bitand or and instead of times (where applicable).
Vectorize:
function out = q44417404(I,k)
if nargin == 0
rng(44417404);
I = randi(2,89,42)-1 == 1;
k = 3;
end
out = permute(prod(reshape(I(nchoosek(1:size(I,1),k).',:).',size(I,2),k,[]),2),[3,1,2]);
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 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 coukldn’t 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));