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);
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(:),:);
I have a vector T of length n and m other vectors of the same length with 0 or 1 used as condition to select elements of T. The condition vectors are combined into a matrix I of size n x m.
Is there a one liner to extract a matrix M of values from Tsuch that the i-th column of M are those elements in T that are selected by the condition elements of the i-th column in I?
Example:
T = (1:10)'
I = mod(T,2) == 0
T(I)'
yields
2 4 6 8 10
However
I = mod(T,2:4) == 0
T(I)'
yields an error in the last statement. I see that the columns might select a different number of elements which results in vectors of different lengths (as in the example). However, even this example doesn't work:
I = zeros(10,2)
I(:,1) = mod(T,2)==0
I(:,2) = mod(T,2)==1
Is there any way to achieve the solution in a one liner?
The easiest way I can think of to do something like this is to take advantage of the element-wise multiplication operator .* with your matrix I. Take this as an example:
% these lines are just setup of your problem
m = 10;
n = 10;
T = [1:m]';
I = randi([0 1], m, n);
% 1 liner to create M
M = repmat(T, 1, n) .* I;
What this does is expand T to be the same size as I using repmat and then multiplies all the elements together using .*.
Here is a one linear solution
mat2cell(T(nonzeros(bsxfun(#times,I,(1:numel(T)).'))),sum(I))
First logical index should be converted to numeric index for it we multiply T by each column of I
idx = bsxfun(#times,I,(1:numel(T)).');
But that index contain zeros we should extract those values that correspond to 1s in matrix I:
idx = nonzeros(idx);
Then we extract repeated elements of T :
T2 = T(idx);
so we need to split T2 to 3 parts size of each part is equal to sum of elements of corresponding column of I and mat2cell is very helpful
result = mat2cell(T2,sum(I));
result
ans =
{
[1,1] =
2
4
6
8
10
[2,1] =
3
6
9
[3,1] =
4
8
}
One line solution using cellfun and mat2cell
nColumns = size(I,2); nRows = size(T,1); % Take the liberty of a line to write cleaner code
cellfun(#(i)T(i),mat2cell(I,nRows,ones(nColumns,1)),'uni',0)
What is going on:
#(i)T(i) % defines a function handle that takes a logical index and returns elements from T for those indexes
mat2cell(I,nRows,ones(nColumns,1)) % Split I such that every column is a cell
'uni',0 % Tell cellfun that the function returns non uniform output
I have a vector that has 1's in some places, and I want to create a diagonal using the vector. The vector is called one_vec_two:
n = 4;
one_vec_two = zeros(n*n, 1);
one_vec_two(1,1) = 1;
for k=0:(n-1)
one_vec_two(k*n+1, 1) = 1;
end
non_zero_vecs = [one_vec_two];
placement = [n-1];
A = spdiags(non_zero_vecs, placement, n*n, n*n);
fullA = full(A);
disp(A)
The first element of the vector one_vec_two is a 1:
>> one_vec_two(1)
ans =
1
And, I placed the vector starting at diagonal n-1, which is 3. But, when I go to column 4, I don't see it:
>> fullA(1,4)
ans =
0
Why isn't MATLAB putting my vector in the correct spot?
As per the doc for spdiag,
Note In this syntax, if a column of B is longer than the diagonal it is replacing, and m >= n, spdiags takes elements of super-diagonals from the lower part of the column of B, and elements of sub-diagonals from the upper part of the column of B.
it is placing the lower part of your vector into the location specified. Hence the result is as expected.
It looks like you want something like
A = spdiags(non_zero_vecs([end-placement+1:end 1:end-placement]), placement, n*n, n*n)
or
A = spdiags(non_zero_vecs, -placement, n*n, n*n)'
which both do the same thing, just in slightly different ways.
I am working with a n x 1 matrix, A, that has repeating values inside it:
A = [0;1;2;3;4; 0;1;2;3;4; 0;1;2;3;4; 0;1;2;3;4]
which correspond to an n x 1 matrix of B values:
B = [2;4;6;8;10; 3;5;7;9;11; 4;6;8;10;12; 5;7;9;11;13]
I am attempting to produce a generalised code to place each repetition into a separate column and store it into Aa and Bb, e.g.:
Aa = [0 0 0 0 Bb = [2 3 4 5
1 1 1 1 4 5 6 7
2 2 2 2 6 7 8 9
3 3 3 3 8 9 10 11
4 4 4 4] 10 11 12 13]
Essentially, each repetition from A and B needs to be copied into the next column and then deleted from the first column
So far I have managed to identify how many repetitions there are and copy the entire column over to the next column and then the next for the amount of repetitions there are but my method doesn't shift the matrix rows to columns as such.
clc;clf;close all
A = [0;1;2;3;4;0;1;2;3;4;0;1;2;3;4;0;1;2;3;4];
B = [2;4;6;8;10;3;5;7;9;11;4;6;8;10;12;5;7;9;11;13];
desiredCol = 1; %next column to go to
destinationCol = 0; %column to start on
n = length(A);
for i = 2:1:n-1
if A == 0;
A = [ A(:, 1:destinationCol)...
A(:, desiredCol+1:destinationCol)...
A(:, desiredCol)...
A(:, destinationCol+1:end) ];
end
end
A = [...] retrieved from Move a set of N-rows to another column in MATLAB
Any hints would be much appreciated. If you need further explanation, let me know!
Thanks!
Given our discussion in the comments, all you need is to use reshape which converts a matrix of known dimensions into an output matrix with specified dimensions provided that the number of elements match. You wish to transform a vector which has a set amount of repeating patterns into a matrix where each column has one of these repeating instances. reshape creates a matrix in column-major order where values are sampled column-wise and the matrix is populated this way. This is perfect for your situation.
Assuming that you already know how many "repeats" you're expecting, we call this An, you simply need to reshape your vector so that it has T = n / An rows where n is the length of the vector. Something like this will work.
n = numel(A); T = n / An;
Aa = reshape(A, T, []);
Bb = reshape(B, T, []);
The third parameter has empty braces and this tells MATLAB to infer how many columns there will be given that there are T rows. Technically, this would simply be An columns but it's nice to show you how flexible MATLAB can be.
If you say you already know the repeated subvector, and the number of times it repeats then it is relatively straight forward:
First make your new A matrix with the repmat function.
Then remap your B vector to the same size as you new A matrix
% Given that you already have the repeated subvector Asub, and the number
% of times it repeats; An:
Asub = [0;1;2;3;4];
An = 4;
lengthAsub = length(Asub);
Anew = repmat(Asub, [1,An]);
% If you can assume that the number of elements in B is equal to the number
% of elements in A:
numberColumns = size(Anew, 2);
newB = zeros(size(Anew));
for i = 1:numberColumns
indexStart = (i-1) * lengthAsub + 1;
indexEnd = indexStart + An;
newB(:,i) = B(indexStart:indexEnd);
end
If you don't know what is in your original A vector, but you do know it is repetitive, if you assume that the pattern has no repeats you can use the find function to find when the first element is repeated:
lengthAsub = find(A(2:end) == A(1), 1);
Asub = A(1:lengthAsub);
An = length(A) / lengthAsub
Hopefully this fits in with your data: the only reason it would not is if your subvector within A is a pattern which does not have unique numbers, such as:
A = [0;1;2;3;2;1;0; 0;1;2;3;2;1;0; 0;1;2;3;2;1;0; 0;1;2;3;2;1;0;]
It is worth noting that from the above intuitively you would have lengthAsub = find(A(2:end) == A(1), 1) - 1;, But this is not necessary because you are already effectively taking the one off by only looking in the matrix A(2:end).
I have a vector whose elements identify the indices (per column) that I need to set in a different matrix. Specifically, I have:
A = 7
1
2
and I need to create a matrix B with some number of rows of zeros, except for the elements identified by A. In other words, I want B:
B = zeros(10, 3); % number of rows is known; num columns = size(A)
B(A(1), 1) = 1
B(A(2), 2) = 1
B(A(3), 3) = 1
I would like to do this without having to write a loop.
Any pointers would be appreciated.
Thanks.
Use linear indexing:
B = zeros(10, 3);
B(A(:).'+ (0:numel(A)-1)*size(B,1)) = 1;
The second line can be written equivalently with sub2ind (may be a little slower):
B(sub2ind(size(B), A(:).', 1:numel(A))) = 1;