I have a piece of code that works in the following way.
There is a matrix of size n x 2. Each element is an integer between 1 and some maximum, say m.
I want to search for rows in this matrix, that is, given [v1, v2], output the index of this.
Right now, I am using:
k = find(ismember(edges, [v1, v2], 'rows'));
However, this is the bottleneck in my code because this is in linear time.
I would like to implement some hashmap type structure for fast lookup. What would be an easy way to do this?
Since you know the number of columns how about this (assuming edges is the matrix to be searched):
idx = find(edges(:,1)==v1 & edges(:,2)==v2);
Note, that depending on how exactly you use the index you might be better off using the logical index that is created on the way:
idx = edges(:,1)==v1 & edges(:,2)==v2;
Hope this helps.
R2016b and beyond:
find(all(edges == [v1 v2], 2))
Prior:
find(all(bsxfun(#eq, edges, [v1 v2]), 2))
Try the following code:
M = [5 2; 10 1; 3 2; 4 4; 5 0]
N = [4 4]
ind=find(all(repmat(N,size(M,1),1)==M,2));
ind is the row where the matrix include the specific numbers in N.
Using accumarray can make an adjacency matrix to speed up your search:
A = accumarray(edges,1,[m m],#any,false)
and you can use indexing to search it
if(A(v1,v2))...
If m is very large you can create a sparse matrix:
A = accumarray(edges,1,[m m],#any,false,true)
Or if you need index of it the adjacency matrix can be cereated this way:
A = accumarray(edges,1:size(edgaes,1),[m m],#any,0,true);
so
index = A(v1,v2)
One option that gives me about a 30-40 times speedup over your code is doing the comparison for the entire first column, capturing a set of indices, then only checking the values of the second column at those indices:
ind = find(edges(:, 1) == v1);
k = ind(edges(ind, 2) == v2);
If you still need it done faster than that, you can use the containers.Map class to precompute a mapping of every possible [v1 v2] to the list of row indices where it occurs. One caveat is that the key type for the map can't be a vector of numbers, but since you are dealing with integers (ideally with m not being prohibitively large) you can convert [v1 v2] to a 2-character ASCII key. Here's how you can build the map:
mapObj = containers.Map('KeyType', 'char', 'ValueType', 'any');
[R, C] = meshgrid(1:m);
keys = [R(:) C(:)];
for keyIndex = 1:(m*m)
v = keys(keyIndex, :);
ind = find(edges(:, 1) == v(1));
mapObj(char(v)) = ind(edges(ind, 2) == v(2));
end
And you can access a value from the map very quickly like so:
k = mapObj(char([v1 v2]));
Related
I have two matrices A and B. A(:,1) corresponds to an x-coordinate, A(:,2) corresponds to a y-coordinate, and A(:,3) corresponds to a certain radius. All three values in a row describe the same circle. Now let's say...
A =
[1,4,3]
[8,8,7]
[3,6,3]
B =
[1,3,3]
[1, 92,3]
[4,57,8]
[5,62,1]
[3,4,6]
[9,8,7]
What I need is to be able to loop through matrix A and determine if there are any rows in matrix B that are "similar" as in the x value is within a range (-2,2) of the x value of A (Likewise with the y-coordinate and radius).If it satisfies all three of these conditions, it will be added to a new matrix with the values that were in A. So for example I would need the above data to return...
ans =
[1,4,3]
[8,8,7]
Please help and thank you in advance to anyone willing to take the time!
You can use ismembertol.
result = A(ismembertol(A,B,2,'ByRows',1,'DataScale',1),:)
Manual method
A = [1,4,3;
8,8,7;
3,6,3];
B = [1,3,3;
1,92,3;
4,57,8;
5,62,1;
3,4,6;
9,8,7]; % example matrices
t = 2; % desired threshold
m = any(all(abs(bsxfun(#minus, A, permute(B, [3 2 1])))<=t, 2), 3);
result = A(m,:);
The key is using permute to move the first dimension of B to the third dimension. Then bsxfun computes the element-wise differences for all pairs of rows in the original matrices. A row of A should be selected if all the absolute differences with respect to any column of B are less than the desired threshold t. The resulting variable m is a logical index which is used for selecting those rows.
Using pdist2 (Statistics and Machine Learning Toolbox)
m = any(pdist2(A, B, 'chebychev')<=t, 2);
result = A(m,:);
Ths pdist2 function with the chebychev option computes the maximum coordinate difference (Chebychev distance, or L∞ metric) between pairs of rows.
With for loop
It should work:
A = [1,4,3;
8,8,7;
3,6,3]
B = [1,3,3;
1,92,3;
4,57,8;
5,62,1;
3,4,6;
9,8,7]
index = 1;
for i = 1:size(A,1)
C = abs(B - A(i,:));
if any(max(C,[],2)<=2)
out(index,:) = A(i,:);
index = index + 1
end
end
For each row of A, computes the absolute difference between B and that row, then checks if there exists a row in which the maximum is less than 2.
Without for loop
ind = any(max(abs(B - permute(A,[3 2 1])),[],2)<=2);
out = A(ind(:),:);
Sorry about the bad title, I'm struggling to word this question well. Basically what I want to do is extract elements from a 2d matrix, from row by row, taking out a number of elements (N) starting at a particular column (k). In for loops, this would look like.
A = magic(6);
k = [2,2,3,3,4,4]; % for example
N = 3;
for j = 1:length(A)
B(j,:) = A(j,k(j):k(j)+N-1);
end
I figure there must be a neater way to do it than that.
You could use bsxfun to create an array of indices to use. Then combine this with the row numbers and pass it to sub2ind.
inds = sub2ind(size(A), repmat(1:size(A, 1), 3, 1), bsxfun(#plus, k, (0:(N-1))')).';
B = A(inds);
Or alternately without sub2ind (but slightly more cryptic).
B = A(bsxfun(#plus, 1:size(A,1), ((bsxfun(#plus, k, (0:(N-1)).')-1) * size(A,1))).');
Here's one approach using bsxfun's masking capability and thus logical indexing -
C = (1:size(A,2))';
At = A.';
B = reshape(At(bsxfun(#ge,C,k) & bsxfun(#lt,C,k+N)),N,[]).';
I have an array with n dimensions, and I have a sequence along one dimension at a certain location on all other dimensions. How do I find the location of this sequence? Preferably without loops.
I use matlab. I know what dimension it should be in, but the sequence isnt necessarily there. Find and == dont work. I could make an nd find function using crosscorrelation but Im guessing this is already implemented and I just dont know what function to call.
example:
ND = rand(10,10,10,10);
V = ND(randi(10),randi(10),randi(10),:);
[I1, I2, I3] = find(ND==V);
Edit: The sequence to be found spans the entire dimension it is on, I did not mention this in my original formulation of the problem. Knedlsepp`s solution solves exactly the problem I had, but Luis' solution solves a more general problem for when the sequence doesn't necessarily span the entire dimension.
As there are multiple ways to interpret your question, I will clarify: This approach assumes a 1D sequence of size: numel(V) == size(ND, dimToSearch). So, for V = [1,2] and ND = [1,2,1,2] it is not applicable. If you want this functionality go with Luis Mendo's answer, if not this will likely be faster.
This will be a perfect opportunity to use bsxfun:
We start with some example data:
ND = rand(10,10,10,10);
V = ND(3,2,:,3);
If you don't have the vector V given in the correct dimension (in this case [1,1,10,1]) you can reshape it in the following way:
dimToSearch = 3;
Vdims = ones(1, ndims(ND));
Vdims(dimToSearch) = numel(V);
V = reshape(V, Vdims);
Now we generate a cell that will hold the indices of the matches:
I = cell(1, ndims(ND));
At this point we compute the size of ND if it were collapsed along the dimension dimToSearch (we compute dimToSearch according to V, as at this point it will have the correct dimensions):
dimToSearch = find(size(V)>1);
collapsedDims = size(ND);
collapsedDims(dimToSearch) = 1;
Finally the part where we actually look for the pattern:
[I{:}] = ind2sub(collapsedDims, find(all(bsxfun(#eq, ND, V), dimToSearch)));
This is done in the following way: bsxfun(#eq, ND, V) will implicitly repmat the array V so it has the same dimensions as ND and do an equality comparison. After this we do a check with all to see if all the entries in the dimension dimToSearch are equal. The calls to find and ind2sub will then generate the correct indices to your data.
Let d be the dimension along which to search. I'm assuming that the sought sequence V may be shorter than size(ND,d). So the sequence may appear once, more than once, or never along each dimension-d- "thread".
The following code uses num2cell to reshape ND into a cell array such that each dimension-d-thread is in a different cell. Then strfind is applied to each cell to determine matches with V, and the result is a cell array with the same dimensions as ND, but where the dimension d is a singleton. The contents of each cell tell the d-dimension-positions of the matches, if any.
Credit goes to #knedlsepp for his suggestion to use num2cell, which greatly simplified the code.
ND = cat(3, [1 2 1 2; 3 4 5 6],[2 1 0 5; 0 0 1 2] ); %// example. 2x4x2
V = 1:2; %// sought pattern. It doesn't matter if it's a row, or a column, or...
d = 2; %// dimension along which to search for pattern V
result = cellfun(#(x) strfind(x(:).', V(:).'), num2cell(ND,d), 'UniformOutput', 0);
This gives
ND(:,:,1) =
1 2 1 2
3 4 5 6
ND(:,:,2) =
2 1 0 5
0 0 1 2
V =
1 2
result{1,1,1} =
1 3 %// V appears twice (at cols 1 and 3) in 1st row, 1st slice
result{2,1,1} =
[] %// V doesn't appear in 2nd row, 1st slice
result{1,1,2} =
[] %// V appears appear in 1st row, 2nd slice
result{2,1,2} =
3 %// V appears once (at col 3) in 2nd row, 2nd slice
One not very optimal way of doing it:
dims = size(ND);
Vrep = repmat(V, [dims(1), dims(2), dims(3), 1]);
ND_V_dist = sqrt(sum(abs(ND.^2-Vrep.^2), 4));
iI = find(ND_V_dist==0);
[I1, I2, I3] = ind2sub([dims(1), dims(2), dims(3)], iI);
Say I have a nxm matrix and want to treat each row as vectors in a function. So, if I have a function that adds vectors, finds the Cartesian product of vectors or for some reason takes the input of several vectors, I want that function to treat each row in a matrix as a vector.
This sounds like a very operation in Matlab. You can access the ith row of a matrix A using A(i, :). For example, to add rows i and j, you would do A(i, :) + A(j, :).
Given an nxm matrix A:
If you want to edit a single column/row you could use the following syntax: A(:, i) for the ith-column and A(i, :) for ith-row.
If you want to edit from a column/row i to a column/row j, you could use that syntax: A(:, i:j) or A(i:j, :)
If you want to edit (i.e.) from the penultimate column/row to the last one, you could you: A(:, end-1:end) or A(end-1:end, :)
EDIT:
I can't add a comment above because I don't have 50 points, but you should post the function setprod. I think you should be able to do what you want to do, by iterating the matrix you're passing as an argument, with a for-next statement.
I think you're going to have to loop:
Input
M = [1 2;
3 4;
5 6];
Step 1: Generate a list of all possible row pairs (row index numbers)
n = size(M,1);
row_ind = nchoosek(1:n,2)
Step 2: Loop through these indices and generate the product set:
S{n,n} = []; //% Preallocation of cell matrix
for pair = 1:size(row_ind,1)
p1 = row_ind(pair,1);
p2 = row_ind(pair,2);
S{p1,p2} = setprod(M(p1,:), M(p2,:))
end
Transform the matrix into a list of row vectors using these two steps:
Convert the matrix into a cell array of the matrix rows, using mat2cell.
Generate a comma-separated list from the cell array, using linear indexing of the cell contents.
Example: let
v1 = [1 2];
v2 = [10 20];
v3 = [11 12];
M = [v1; v2; v3];
and let fun be a function that accepts an arbitrary number of vectors as its input. Then
C = mat2cell(M, ones(1,size(M,1)));
result = fun(C{:});
is the same as result = fun(v1, v2, v3).
I have two matrices A and B, both of which are Nx3 matrices.
I'm currently getting the maximum value and index for each row of matrix A using:
[maxA, idx] = max(A, [], 2)
idx(j) indicates which column contained the maximum for row j. Now I'd like to select those same positions from matrix B.
I've currently implemented this using a loop:
for j = 1:numel(idx)
maxB(j) = B(j, idx(j))
end
My current implementation is fast enough, although I prefer to avoid unneeded loops so is there a way to express this without a loop?
You can build a vector of linear indices (I expect B to be the same size as A):
vec_indices = sub2ind(size(A), 1:numel(idx), idx);
Then you can use that vector directly for lookup:
maxB = B(vec_indices)
You can construct the single dimension index into the matrix and get them that way. All multidimensional matrices in matlab can be addressed.
You can use
maxB = B(sub2ind([1:length(idx)]',idx(:)));
In one line:
maxB = B(A == max(A, [], 2) * ones(1, 3));
But this is not safe. It assumes unique values in every row of A.