I have a problem with find. I don't know what is the problem.
I want to find indices in a matrix by respect to another matrix that have same X , Y and different Z.
My matrix are selectedl and selectedf with similar rows and different rows so.
exP=find(selectedl(:,[1,2])==selectedf(:,[1,2]) & selectedl(:,3)~=selectedf(:,3));
what is the issue in the line?
A more readable version would be:
A = selectedl == selectedf;
exP = find(A(:,1) & A(:,2) & ~A(:,3));
If I understand correctly (first column is x, second is y, third is z, each row represents a different point), you just need to add row-wise all in the first part:
exP=find(all(selectedl(:,[1,2])==selectedf(:,[1,2]),2) & selectedl(:,3)~=selectedf(:,3));
Or, perhaps more readable:
exP=find(selectedl(:,1)==selectedf(:,1) & selectedl(:,2)==selectedf(:,2) & selectedl(:,3)~=selectedf(:,3));
This assumes both matrices have the same size and order matters. If not, see #Dan's answer.
You probably should be using ismember instead of find:
[incl, indices] = ismember(selectedl(:,1:2), selectedf(:,1:2), 'rows'); %// Or possibly ismember(selectedf(:,1:2), selectedl(:,1:2), 'rows') depending on what you're after
excl = ismember(selectedl, selectedf, 'rows');
indices(incl & ~excl)
This way your rows don't have to correspond 1 to 1 across the two matrices and teh matrices don't even need to be of the same length.
Related
Introduction to problem:
I'm modelling a system where i have a matrix X=([0,0,0];[0,1,0],...) where each row represent a point in 3D-space. I then choose a random row, r, and take all following rows and rotate around the point represented by r, and make a new matrix from these rows, X_rot. I now want to check whether any of the rows from X_rot is equal two any of the rows of X (i.e. two vertices on top of each other), and if that is the case refuse the rotation and try again.
Actual question:
Until now i have used the following code:
X_sim=[X;X_rot];
if numel(unique(X_sim,'rows'))==numel(X_sim);
X(r+1:N+1,:,:)=X_rot;
end
Which works, but it takes up over 50% of my running time and i were considering if anybody in here knew a more efficient way to do it, since i don't need all the information that i get from unique.
P.S. if it matters then i typically have between 100 and 1000 rows in X.
Best regards,
Morten
Additional:
My x-matrix contains N+1 rows and i have 12 different rotational operations that i can apply to the sub-matrix x_rot:
step=ceil(rand()*N);
r=ceil(rand()*12);
x_rot=x(step+1:N+1,:);
x_rot=bsxfun(#minus,x_rot,x(step,:));
x_rot=x_rot*Rot(:,:,:,r);
x_rot=bsxfun(#plus,x_rot,x(step,:));
Two possible approaches (I don't know if they are faster than using unique):
Use pdist2:
d = pdist2(X, X_rot, 'hamming'); %// 0 if rows are equal, 1 if different.
%// Any distance function will do, so try those available and choose fastest
result = any(d(:)==0);
Use bsxfun:
d = squeeze(any(bsxfun(#ne, X, permute(X_rot, [3 2 1])), 2));
result = any(d(:)==0);
result is 1 if there is a row of X equal to some row of X_rot, and 0 otherwise.
How about ismember(X_rot, X, 'rows')?
i have two matrices
r=10,000x2
q=10,000x2
i have to find out those rows of q which are one value or both values(as it is a two column matrix) different then r and allocate them in another matrix, right now i am trying this.i cannot use isequal because i want to know those rows
which are not equal this code gives me the individual elements not the complete rows different
can anyone help please
if r(:,:)~=q(:,:)
IN= find(registeredPts(:,:)~=q(:,:))
end
You can probably do this using ismember. Is this what you want? Here you get the values from q in rows that are different from r.
q=[1,2;3,4;5,6]
r=[1,2;3,5;5,6]
x = q(sum(ismember(q,r),2) < 2,:)
x =
3 4
What this do:
ismember creates an array with 1's in the positions where q == r, and 0 in the remaining positions. sum(.., 2) takes the column sum of each of these rows. If the sum is less than 2, that row is included in the new array.
Update
If the values might differ some due to floating point arithmetic, check out ismemberf from the file exchange. I haven't tested it myself, but it looks good.
Lets say I have two vectors A and B with different lengths Length(A) is not equal to Length(B) and the Values in Vector A, are not the same as in Vector B. I want to compare each value of B with Values of A (Compare means if Value B(i) is almost the same value of A(1:end) for example B(i)-Tolerance<A(i)<B(i)+Tolerance.
How Can I do this without using for loop since the data is huge?
I know ismember(F), intersect,repmat,find but non of those function can really help me
You may try a solution along these lines:
tol = 0.1;
N = 1000000;
a = randn(1, N)*1000; % create a randomly
b = a + tol*rand(1, N); % b is "tol" away from a
a_bin = floor(a/tol);
b_bin = floor(b/tol);
result = ismember(b_bin, a_bin) | ...
ismember(b_bin, a_bin-1) | ...
ismember(b_bin, a_bin+1);
find(result==0) % should be empty matrix.
The idea is to discretize the a and b variables to bins of size tol. Then, you ask whether b is found in the same bin as any element from a, or in the bin to the left of it, or in the bin to the right of it.
Advantages: I believe ismember is clever inside, first sorting the elements of a and then performing sublinear (log(N)) search per element b. This is unlike approaches which explicitly construct differences of each element in b with elements from a, meaning the complexity is linear in the number of elements in a.
Comparison: for N=100000 this runs 0.04s on my machine, compared to 20s using linear search (timed using Alan's nice and concise tf = arrayfun(#(bi) any(abs(a - bi) < tol), b); solution).
Disadvantages: this leads to that the actual tolerance is anything between tol and 1.5*tol. Depends on your task whether you can live with that (if the only concern is floating point comparison, you can).
Note: whether this is a viable approach depends on the ranges of a and b, and value of tol. If a and b can be very big and tol is very small, the a_bin and b_bin will not be able to resolve individual bins (then you would have to work with integral types, again checking carefully that their ranges suffice). The solution with loops is a safer one, but if you really need speed, you can invest into optimizing the presented idea. Another option, of course, would be to write a mex extension.
It sounds like what you are trying to do is have an ismember function for use on real valued data.
That is, check for each value B(i) in your vector B whether B(i) is within the tolerance threshold T of at least one value in your vector A
This works out something like the following:
tf = false(1, length(b)); %//the result vector, true if that element of b is in a
t = 0.01; %// the tolerance threshold
for i = 1:length(b)
%// is the absolute difference between the
%//element of a and b less that the threshold?
matches = abs(a - b(i)) < t;
%// if b(i) matches any of the elements of a
tf(i) = any(matches);
end
Or, in short:
t = 0.01;
tf = arrayfun(#(bi) any(abs(a - bi) < t), b);
Regarding avoiding the for loop: while this might benefit from vectorization, you may also want to consider looking at parallelisation if your data is that huge. In that case having a for loop as in my first example can be handy since you can easily do a basic version of parallel processing by changing the for to parfor.
Here is a fully vectorized solution. Note that I would actually recommend the solution given by #Alan, as mine is not likely to work for big datasets.
[X Y]=meshgrid(A,B)
M=abs(X-Y)<tolerance
Now the logical index of elements in a that are within the tolerance can be obtained with any(M) and the index for B is found by any(M,2)
bsxfun to the rescue
>> M = abs( bsxfun(#minus, A, B' ) ); %//' difference
>> M < tolerance
Another way to do what you want is with a logical expression.
Since A and B are vectors of different sizes you can't simply subtract and look for values that are smaller than the tolerance, but you can do the following:
Lmat = sparse((abs(repmat(A,[numel(B) 1])-repmat(B',[1 numel(A)])))<tolerance);
and you will get a sparse logical matrix with as many ones in it as equal elements (within tolerance). You could then count how many of those elements you have by writing:
Nequal = sum(sum(Lmat));
You could also get the indexes of the corresponding elements by writing:
[r,c] = find(Lmat);
then the following code will be true (for all j in numel(r)):
B(r(j))==A(c(j))
Finally, you should note that this way you get multiple counts in case there are duplicate entries in A or in B. It may be advisable to use the unique function first. For example:
A_new = unique(A);
I would like to generate all possible combinations for selecting rows in batches of lets say 'k' size. For example, matrix A has 3 rows and I want all combinations for batch size 2, i.e. rows (1,2)(1,3)(2,3). What would be the simplest way to do it? Then I would like use them for some operation like myfunction();
I think nchoosek function does the trick of selecting the combination but then how can I use each row of the output from nchoosek as index for my matrix?
If you want to use each combination one by one you can do something like this:
A = rand(3);
comb = nchoosek(length(A), 2);
for i = 1:size(comb, 1)
myfunction(A(comb(i, :), :));
end
A(comb(i, :)) is a k x n matrix (here 3 x 2) corresponding to the i-th combination of rows.
This is very closely related to this other question, but that question wanted to avoid sub2ind because of performance concerns. I am more concerned about the "unelegance" of using sub2ind.
Let's suppose I want to create another MxN matrix which is all zeros except for one entry in each column that I want to assign from the corresponding entry in a vector, and choice of row in each column is based on another vector. For example:
z = zeros(10,4);
rchoice = [3 1 8 7];
newvals = [123 456 789 10];
% ??? I would like to set z(3,1)=123, z(1,2)=456, z(8,3)=789, z(7,4)=10
I can use sub2ind to accomplish this (which I used in an answer to a closely related question):
z(sub2ind(size(z),rchoice,1:4)) = newvals
but is there another alternative? Seems like logical addressing could be used in some way but I'm stumped, because in order to set the elements of a logical matrix to 1, you're dealing with the same element positions as in the matrix you actually want to address.
There's a much simpler way of doing it.
nCols=size(z,2);
z(rchoice,1:nCols)=diag(newvals);
You can just add the number of rows in previous columns to rchoice to get the linear index directly.
nRows = size(z,1); %# in case you don't know this already
nCols2write = length(newvals);
z(rchoice+[0:nRows:(nRows*(nCols2write-1)]) = newvals;