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
Related
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 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 3d matrix H(i,j,k) with dimensions (i=1:m,j=1:n,k=1:o). I will use a simple case with m=n=o = 2:
H(:,:,1) =[1 2; 3 4];
H(:,:,2) =[5 6; 7 8];
I want to filter this matrix and project it to an (m,n) matrix by selecting for each j in 1:n a different k in 1:0.
For instance, I would like to retrieve (j,k) = {(1,2), (2,1)}, resulting in matrix G:
G = [5 2; 7 4];
This can be achieved with a for loop:
filter = [2 1]; % meaning filter (j,k) = {(1,2), (2,1)}
for i = 1:length(filter)
G(:,i) = squeeze(H(:,i,filter(i)));
end
But I'm wondering if it is possible to avoid the for loop via some smart indexing.
You can create all the linear indices to get such an output with the expansion needed for the first dimension with bsxfun. The implementation would look like this -
szH = size(H)
offset = (filter-1)*szH(1)*szH(2) + (0:numel(filter)-1)*szH(1)
out = H(bsxfun(#plus,[1:szH(1)].',offset))
How does it work
(filter-1)*szH(1)*szH(2) and (0:numel(filter)-1)*szH(1) gets the linear indices considering only the third and second dimension elements respectively. Adding these two gives us the offset linear indices.
Add the first dimenion linear indices 1:szH(1) to offset array in elementwise fashion with bsxfun to give us the actual linear indices, which when indexed into H would be the output.
Sample run -
H(:,:,1) =
1 2
3 4
H(:,:,2) =
5 6
7 8
filter =
2 1
out =
5 2
7 4
I have a <206x193> matrix A. It contains the values of a parameter at 206 different locations at 193 time steps. I am interested in the maximum value at each location over all times as well as the corresponding indices. I have another matrix B with the same dimensions of A and I'm interested in values for each location at the time that A's value at that location was maximal.
I've tried [max_val pos] = max(A,[],2), which gives the right maximum values, but A(pos) does not equal max_val.
How exactly does this function work?
I tried a smaller example as well. Still I don't understand the meaning of the indices....
>> H
H(:,:,1) =
1 2
3 4
H(:,:,2) =
5 6
7 8
>> [val pos] = max(H,[],2)
val(:,:,1) =
2
4
val(:,:,2) =
6
8
pos(:,:,1) =
2
2
pos(:,:,2) =
2
2
The indices in idx represent the index of the max value in the corresponding row. You can use sub2ind to create a linear index if you want to test if A(pos)=max_val
A=rand(206, 193);
[max_val, idx]=max(A, [], 2);
A_max=A(sub2ind(size(A), (1:size(A,1))', idx));
Similarly, you can access the values of B with:
B_Amax=B(sub2ind(size(A), (1:size(A,1))', idx));
From your example:
H(:,:,2) =
5 6
7 8
[val pos] = max(H,[],2)
val(:,:,2) =
6
8
pos(:,:,2) =
2
2
The reason why pos(:,:,2) is [2; 2] is because the maximum is at position 2 for both rows.
max is a primarily intended for use with vectors. In normal mode, even the multi-dimensional arrays are treated as a series of vectors along which the max function is applied.
So, to get the values in B at each location at the time where A is maximum, you should
// find the maximum values and positions in A
[c,i] = max(A, [], 2);
// iterate along the first dimension, to retrieve the corresponding values in B
C = [];
for k=1:size(A,1)
C(k) = B(k,i(k));
end
You can refer to #Jigg's answer for a more concise way of creating matrix C
Short Version
How can I do concatMap in MATLAB? I'm trying to build a single vector from a series of smaller, differently sized vectors. I know I can do:
result = [];
for i=1:N
result = [result nextPart(i)];
end
but that has a serious speed impact and there must be a smarter way to do concatMap.
Long Version
I'm trying to write a MATLAB function that returns the counterdiagonals of a block. For example, if you have the block:
1 2 4
3 5 7
6 8 9
then counterDiagonals(block) should return [1 2 3 4 5 6 7 8 9].
I have a function that will find a single counter diagonal of a block. i.e. counterDiagonal(x, 3) will return [4 5 6].
Therefore, counterDiagonals should be as simple as concatMap counterDiagonal(x, i) (1:N) where N is (2*length(block)-1). How can I do this in MATLAB in an efficient way?
One problem with the accepted answer: if the matrix A had zeros, they will be incorrectly removed from the result.. Instead you should work on the indices of the elements:
A = [0 2 4; 3 5 7; 6 8 9]; %# Sample matrix (contains zeros)
ind = reshape(1:numel(A), size(A)); %# indices of elements
ind = fliplr( spdiags( fliplr(ind) ) ); %# get the anti-diagonals (or use ROT90)
ind(ind==0) = []; %# keep non-zero indices
result = A(ind); %# get elements in desired order
This is very similar to this answer I gave in a previous question (the difference was that the anti-digaonals were in reverse order).
I believe what you want to do can be accomplished using the functions ROT90 and SPDIAGS:
A = [1 2 4; 3 5 7; 6 8 9]; %# Sample matrix
result = rot90(A); %# Rotate the matrix counter-clockwise
result = spdiags(result); %# Find all the diagonals
result = result(result ~= 0).'; %'# Remove zero padding and format the results
%# into a row vector
And you should end up with result = [1 2 3 4 5 6 7 8 9].
EDIT: As Amro mentions in a comment, the above code assumes that there are no zeroes in the original matrix A. If there are zeroes in the original matrix, one solution is to replace them with a non-zero flag value that you know doesn't appear in the original matrix (like, for example, NaN), run the above code, then replace the flag values in the result:
A = [0 2 4; 3 0 7; 6 8 0]; %# Sample matrix
result = rot90(A); %# Rotate the matrix counter-clockwise
result(result == 0) = nan; %# Replace zeroes with NaN
result = spdiags(result); %# Find all the diagonals
result = result(result ~= 0).'; %'# Remove zero padding and format the results
%# into a row vector
result(isnan(result)) = 0; %# Put the original zeroes back
Short version:
If you preassign your result array, everything will be a lot faster.
result = zeros(1,knownLengthOfResultsArray); %# such as "numel(block)"
ct = 1;
for i=1:N
tmp = nextPart(i);
nTmp = length(tmp);
result(ct:ct+nTmp-1) = tmp;
ct = ct + nTmp;
end
Long version:
However, it may be even more efficient to rewrite your algorithm. See e.g. the answers to this question (use fliplr on your array first), or #gnovice's answer.