I'm trying to find an idiomatic way to do this.
Essentially I have an Nx2 matrix of points of the form
A = [3 4; 3 5; 4 5; 4, 6; 7 3]
I'd like my output to be [3 5; 4 6; 7 3]. In other words I would like each unique x value along with the maximum y value associated with that x.
I was hoping there would be some sort of
unique(A, 'rows', 'highestterm', 2)
method for accomplishing this, but couldn't find anything. Can anyone think of a vectorized way to solve this problem? I can do it pretty easily in a for loop, but would like to avoid that if possible.
I don't know of any single call, like you hoped. But, it can be done fairly tightly (and fully vectorized) like the code below.
%sort by first and then second column
A = sortrows(A,[1 2]);
%find each change in the first column of A
inds = find(diff(A(:,1)) > 0);
%add the last point...because find(diff) doesn't get the last point
inds(end+1) = size(A,1);
%get just those rows that meet the desired criteria
A = A(inds,:);
So, this works by sorting the data and looking for the values in the first column that don't repeat. If there are repeated values, this code grabs the last of the repeating values. Finally, because we sorted by both columns via sortrows(A,[1 2]), the last entry for a repeating value will have the biggest corresponding value from the 2nd column. I think that this hits all of your requirements.
Using accumarray and unique:
[r1, ~, u] = unique(A(:,1));
r2 = accumarray(u, A(:,2), [], #max);
result = [r1 r2];
Related
Good evening! I have a 3D vector. It has the first dimension 1. For clarity, I set it exactly the same as used in my program. "с" is like a number of experiments, in this case there are three, so I calculate the correlation function three times and then add them up.
In fact, the number of experiments is 100. I have to calculate 100 correlation functions and add them.
Tell me how you can do it automatically. And if possible, then no cycles. Thank you.
And yes, in the beginning I set the 3D vector using a loop. Is it possible to set it without a loop as well? This is certainly not my main question, but I would also like to know the answer to it.
d = [1 2 3];
c = [4 2 6];
for i = 1: length(c)
D(1,:,i) = d.*c(i);
end
D
X1 = xcorr(D(:,:,1));
X2 = xcorr(D(:,:,2));
X3 = xcorr(D(:,:,3));
X = X1+X2+X3;
With the help of a loop, my solution looks like this:
d = [1 2 3];
c = [4 2 6];
for i = 1: length(c)
D(1,:,i) = d.*c(i);
x(:,:,i) = xcorr(D(:,:,i));
end
X = sum(x,3)
It seems to be correct. Is it possible to do this without a cycle?
You can easily set your first array D without any loop, even though I don't know why you want to keep the first singleton dimension...
D(1, :, :) = d'.*c;
As for the sum of the autocorrelations, I'm not sure you can do it without a loop. The only thing that you can perhaps do is to not use an array to store the correlation for each index (if memory consumption is a problem for you) and just update the sum:
X = zeros(1, 2*length(d)-1); % initialize the sum array
for i = 1:length(c)
X = X + xcorr(D(:, :, i)); % update the sum
end
I have a situation analogous to the following
z = magic(3) % Data matrix
y = [1 2 2]' % Column indices
So,
z =
8 1 6
3 5 7
4 9 2
y represents the column index I want for each row. It's saying I should take row 1 column 1, row 2 column 2, and row 3 column 2. The correct output is therefore 8 5 9.
I worked out I can get the correct output with the following
x = 1:3;
for i = 1:3
result(i) = z(x(i),y(i));
end
However, is it possible to do this without looping?
Two other possible ways I can suggest is to use sub2ind to find the linear indices that you can use to sample the matrix directly:
z = magic(3);
y = [1 2 2];
ind = sub2ind(size(z), 1:size(z,1), y);
result = z(ind);
We get:
>> result
result =
8 5 9
Another way is to use sparse to create a sparse matrix which you can turn into a logical matrix and then sample from the matrix with this logical matrix.
s = sparse(1:size(z,1), y, 1, size(z,1), size(z,2)) == 1; % Turn into logical
result = z(s);
We also get:
>> result
result =
8
5
9
Be advised that this only works provided that each row index linearly increases from 1 up to the end of the rows. This conveniently allows you to read the elements in the right order taking advantage of the column-major readout that MATLAB is based on. Also note that the output is also a column vector as opposed to a row vector.
The link posted by Adriaan is a great read for the next steps in accessing elements in a vectorized way: Linear indexing, logical indexing, and all that.
there are many ways to do this, one interesting way is to directly work out the indexes you want:
v = 0:size(y,2)-1; %generates a number from 0 to the size of your y vector -1
ind = y+v*size(z,2); %generates the indices you are looking for in each row
zinv = z';
zinv(ind)
>> ans =
8 5 9
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);
I have a 3 dimensional (or higher) array that I want to aggregate by another vector. The specific application is to take daily observations of spatial data and average them to get monthly values. So, I have an array with dimensions <Lat, Lon, Day> and I want to create an array with dimensions <Lat, Lon, Month>.
Here is a mock example of what I want. Currently, I can get the correct output using a loop, but in practice, my data is very large, so I was hoping for a more efficient solution than the second loop:
% Make the mock data
A = [1 2 3; 4 5 6];
X = zeros(2, 3, 9);
for j = 1:9
X(:, :, j) = A;
A = A + 1;
end
% Aggregate the X values in groups of 3 -- This is the part I would like help on
T = [1 1 1 2 2 2 3 3 3];
X_agg = zeros(2, 3, 3);
for i = 1:3
X_agg(:,:,i) = mean(X(:,:,T==i),3);
end
In 2 dimensions, I would use accumarray, but that does not accept higher dimension inputs.
Before getting to your answer let's first rewrite your code in a more general way:
ag = 3; % or agg_size
X_agg = zeros(size(X)./[1 1 ag]);
for i = 1:ag
X_agg(:,:,i) = mean(X(:,:,(i-1)*ag+1:i*ag), 3);
end
To avoid using the for loop one idea is to reshape your X matrix to something that you can use the mean function directly on.
splited_X = reshape(X(:), [size(X_agg), ag]);
So now splited_X(:,:,:,i) is the i-th part
that contains all the matrices that should be aggregated which is X(:,:,(i-1)*ag+1:i*ag)) (like above)
Now you just need to find the mean in the 3rd dimension of splited_X:
temp = mean(splited_X, 3);
However this results in a 4D matrix (where its 3rd dimension size is 1). You can again turn it into 3D matrix using reshape function:
X_agg = reshape(temp, size(X_agg))
I have not tried it to see how much more efficient it is, but it should do better for large matrices since it doesn't use for loops.
I have a 3x3 matrix, A. I also compute a value, g, as the maximum eigen value of A. I am trying to change the element A(3,3) = 0 for all values from zero to one in 0.10 increments and then update g for each of the values. I'd like all of the other matrix elements to remain the same.
I thought a for loop would be the way to do this, but I do not know how to update only one element in a matrix without storing this update as one increasingly larger matrix. If I call the element at A(3,3) = p (thereby creating a new matrix Atry) I am able (below) to get all of the values from 0 to 1 that I desired. I do not know how to update Atry to get all of the values of g that I desire. The state of the code now will give me the same value of g for all iterations, as expected, as I do not know how to to update Atry with the different values of p to then compute the values for g.
Any suggestions on how to do this or suggestions for jargon or phrases for me to web search would be appreciated.
A = [1 1 1; 2 2 2; 3 3 0];
g = max(eig(A));
% This below is what I attempted to achieve my solution
clear all
p(1) = 0;
Atry = [1 1 1; 2 2 2; 3 3 p];
g(1) = max(eig(Atry));
for i=1:100;
p(i+1) = p(i)+ 0.01;
% this makes a one giant matrix, not many
%Atry(:,i+1) = Atry(:,i);
g(i+1) = max(eig(Atry));
end
This will also accomplish what you want to do:
A = #(x) [1 1 1; 2 2 2; 3 3 x];
p = 0:0.01:1;
g = arrayfun(#(x) eigs(A(x),1), p);
Breakdown:
Define A as an anonymous function. This means that the command A(x) will return your matrix A with the (3,3) element equal to x.
Define all steps you want to take in vector p
Then "loop" through all elements in p by using arrayfun instead of an actual loop.
The function looped over by arrayfun is not max(eig(A)) but eigs(A,1), i.e., the 1 largest eigenvalue. The result will be the same, but the algorithm used by eigs is more suited for your type of problem -- instead of computing all eigenvalues and then only using the maximum one, you only compute the maximum one. Needless to say, this is much faster.
First, you say 0.1 increments in the text of your question, but your code suggests you are actually interested in 0.01 increments? I'm going to operate under the assumption you mean 0.01 increments.
Now, with that out of the way, let me state what I believe you are after given my interpretation of your question. You want to iterate over the matrix A, where for each iteration you increase A(3, 3) by 0.01. Given that you want all values from 0 to 1, this implies 101 iterations. For each iteration, you want to calculate the maximum eigenvalue of A, and store all these eigenvalues in some vector (which I will call gVec). If this is correct, then I believe you just want the following:
% Specify the "Current" A
CurA = [1 1 1; 2 2 2; 3 3 0];
% Pre-allocate the values we want to iterate over for element (3, 3)
A33Vec = (0:0.01:1)';
% Pre-allocate a vector to store the maximum eigenvalues
gVec = NaN * ones(length(A33Vec), 1);
% Loop over A33Vec
for i = 1:1:length(A33Vec)
% Obtain the version of A that we want for the current i
CurA(3, 3) = A33Vec(i);
% Obtain the maximum eigen value of the current A, and store in gVec
gVec(i, 1) = max(eig(CurA));
end
EDIT: Probably best to paste this code into your matlab editor. The stack-overflow automatic text highlighting hasn't done it any favors :-)
EDIT: Go with Rody's solution (+1) - it is much better!