I have a 5D array A and want to vertically stack the blocks
A(x,y,z,:,:) for all x, y and z. I have done the following:
A = rand(128, 128, 64, 20, 3);
a = 1;
for z = 1:Z
for y = 1:Y
for x = 1:X
res(a:a+T-1,:) = squeeze(A(x,y,z,:,:));
a = a + T;
end
end
end
While correct, it appears that this computation is quite slow for my purpose.
Can anyone figure out how to achieve the same goal more efficiently?
You can do this via reshape and permute. First, understand how reshape works.
c = randi(9,[2 2 3])
c(:,:,1) =
5 8
2 6
c(:,:,2) =
4 4
5 1
c(:,:,3) =
3 2
2 3
Let us reshape c into a 2 x 6 matrix.
ans =
5 8 4 4 3 2
2 6 5 1 2 3
Now a 6 x 2 matrix.
ans =
5 4
2 1
8 3
6 2
4 2
5 3
See that while creating the reshaped matrix, MATLAB starts traversing the first dimension of c and puts it along the first dimension (i.e. along rows in the first column) in the reshaped matrix. Specifically,
c(1,1,1) = 5
c(2,1,1) = 2
c(1,2,1) = 8
c(2,2,1) = 6
c(1,1,2) = 4
c(2,1,2) = 5 % stop since you would have filled 6 rows
which forms the first column of the reshaped matrix.
Coming back to your example,
the first column of the reshaped matrix will be along the rows of your 20x3 submatrix i.e. the fourth dimension. The first row will be along your columns of your 20x3 submatrix i.e. along the fifth dimension. However, when you traverse from (1,1) to (1,2) in your reshaped matrix, you will have to accomodate all the elements from your first three dimensions (i.e. 128, 128, 64, why?, because you first traverse your x dimension, then y and then finally z dimension). So your answer is, you want to move along dimensions: 4,1,2,3,5 in that order.
Answer:
sz_A = size(A);
p = sz_A(4)*prod(sz_A(1:3))
reshaped_A = reshape(permute(A,[4, 1, 2, 3, 5]), [p size(A,5)])
Related
I have two (or more but if solved for two, it's solved for any number) 2-by-N matrices which represent points with an x (the first row) and y (the second row) coordinates. The points are always sorted in the increasing x coordinate. What I want to do is I want to merge these two matrices into one 3-by-N matrix so that if two points (one from each matrix) have the same x coordinate, they would form one column in the new matrix, the first row being the x coordinate and the second and third row being the two y coordinates. However, if there is a point in one matrix that has x coordinate different than all other points in the second matrix, I still want to have full 3-element column that is placed such that the x coordinates are still sorted and the missing value from the other matrix is replaced by the nearest value with lower x coordinate (or NaN if there is none).
Better to explain by example.
First matrix:
1 3 5 7 % x coordinate
1 2 3 4 % y coordinate
Second matrix:
2 3 4 7 8 % x coordinate
5 6 7 8 9 % y coordinate
Desired result:
1 2 3 4 5 7 8 % x coordinate
1 1 2 2 3 4 4 % y coordinate from first matrix
NaN 5 6 7 7 8 9 % y coordinate from second matrix
My question is, how can I do it effectively in matlab/octave and numpy? (Effectively because I can always do it "manually" with loops but that doesn't seem right.)
You can do it with interp1 and the keyword 'previous' for strategy (you can also choose 'nearest' if you do not care if it is larger or smaller) and 'extrap' for allowing extrapolation.
Define the matrices
a=[...
1 3 5 7;...
1 2 3 4];
b=[...
2 3 4 7 8;...
5 6 7 8 9];
Then find the interpolation points
x = unique([a(1,:),b(1,:)]);
And interpolate
[x ; interp1(a(1,:),a(2,:),x,'previous','extrap') ; interp1(b(1,:),b(2,:),x,'previous','extrap') ]
Timeit results:
I tested the algorithms on
n = 1e6;
a = cumsum(randi(3,2,n),2);
b = cumsum(randi(2,2,n),2);
and got:
Wolfie: 1.7473 s
Flawr: 0.4927 s
Mine: 0.2757 s
This verions uses set operations:
a=[...
1 3 5 7;...
1 2 3 4];
b=[...
2 3 4 7 8;...
5 6 7 8 9];
% compute union of x coordinates
c = union(a(1,:),b(1,:));
% find indices of x of a and b coordinates in c
[~,~,ia] = intersect(a(1,:),c);
[~,~,ib] = intersect(b(1,:),c);
% create output matrix
d = NaN(3,numel(c));
d(1,:) = c;
d(2,ia) = a(2,:);
d(3,ib) = b(2,:);
% fill NaNs
m = isnan(d);
m(:,1) = false;
i = find(m(:,[2:end,1])); %if you have multiple consecutive nans you have to repeat these two steps
d(m) = d(i);
disp(d);
Try it online!
Your example:
a = [1 3 5 7; 1 2 3 4];
b = [2 3 4 7 8; 5 6 7 8 9];
% Get the combined (unique, sorted) `x` coordinates
output(1,:) = unique([a(1,:), b(1,:)]);
% Initialise y values to NaN
output(2:3, :) = NaN;
% Add x coords from `a` and `b`
output(2, ismember(output(1,:),a(1,:))) = a(2,:);
output(3, ismember(output(1,:),b(1,:))) = b(2,:);
% Replace NaNs in columns `2:end` with the previous value.
% A simple loop has the advantage of capturing multiple consecutive NaNs.
for ii = 2:size(output,2)
colNaN = isnan(output(:, ii));
output(colNaN, ii) = output(colNaN, ii-1);
end
If you have more than 2 matrices (as suggested in your question) then I'd advise
Store them in a cell array, and loop over them to do the calls to ismember, instead of having one code line per matrix hardcoded.
The NaN replacement loop is already vectorised for any number of rows.
This is the generic solution for any number of matrices, demonstrated with a and b:
mats = {a, b};
cmats = horzcat(mats);
output(1, :) = unique(cmats(1,:));
output(2:numel(mats)+1, :) = NaN;
for ii = 1:size(mats)
output(ii+1, ismember(output(1,:), mats{ii}(1,:))) = mats{ii}(2,:);
end
for ii = 2:size(output,2)
colNaN = isnan(output(:,ii));
output(colNaN, ii) = output(colNaN, ii-1);
end
I am producing a three dimensional matrix with two rows(e.g. a 2x1000x10 matrix). These contain x- and y-coordinates for dot motion paths, where the first line contains the x-coordinates and the second line the y-coordinates. The number of columns (here: 1000) depends on the length of the motion paths. The number of levels in the third dimension depends on the number of dots (here: 10).
I want to know if some dots are going to overlap, i.e. if any x- and y-coordinates across the third dimension at the same time point t, i.e. any combination of (:, t, :), are identical. So the comparison will be within the matrix.
i = [1:size(matrix, 3)];
j = [1:size(matrix, 3)];
t = [1:size(matrix, 2)];
crash = any(coordinates(:, t, i)==coordinates(:, t, j))
would be trivial because i and j can be the same and thus say that the point is equal to itself.
Do you know how I can detect equal value combinations across matrix dimensions? How do I index them if they exist?
I think this does what you want. To find out if there's a crash:
t = bsxfun(#eq, coordinates, permute(coordinates, [1 2 4 3]));
crash = any(sum(any(all(t,1),2),4)>1,3);
We first use bsxfun to compute a 4D array (t) that tests for equality in the same coordinate (1st dim) and the same position along path (2nd dim), where the 3rd and 4th dims run over all pairs of dots. A crash occurs if both coordinates (all(,...,1)) are equal for any column (any(...,2)) in more than one "other" dot (sum(...,4)>1) for any "reference" dot (any(...,3)). We need to specify "more than one" because any dot is at least equal to itself (for a given position and coordinate).
Example:
>> coordinates = randi(9,2,5,4)
coordinates(:,:,1) =
7 4 3 3 8
7 1 4 2 4
coordinates(:,:,2) =
8 7 8 4 8
4 4 7 2 9
coordinates(:,:,3) =
3 4 7 8 5
7 8 2 9 8
coordinates(:,:,4) =
6 2 7 8 5
2 4 8 3 1
>> t = bsxfun(#eq, coordinates, permute(coordinates, [1 2 4 3]));
>> crash = any(sum(any(all(t,1),2),4)>1,3)
crash =
0
>> coordinates(:,2,4) = coordinates(:,2,1) %// 4th dot equal to 1st dot in column 2
coordinates(:,:,1) =
7 4 3 3 8
7 1 4 2 4
coordinates(:,:,2) =
8 7 8 4 8
4 4 7 2 9
coordinates(:,:,3) =
3 4 7 8 5
7 8 2 9 8
coordinates(:,:,4) =
6 4 7 8 5
2 1 8 3 1
>> t = bsxfun(#eq, coordinates, permute(coordinates, [1 2 4 3]));
>> crash = any(sum(any(all(t,1),2),4)>1,3)
crash =
1
To find the coordinates of the crash:
t = bsxfun(#eq, coordinates, permute(coordinates, [1 2 4 3]));
[m,n,p] = size(coordinates);
ii = find(squeeze(all(bsxfun(#times, t, ...
bsxfun(#ne, reshape(1:p, 1,1,[]), reshape(1:p, 1,1,1,[]) )),1)));
[col, dot_ref, dot_oth] = ind2sub([n p p], ii);
In the latter example, this gives
col =
2
2
dot_ref =
4
1
dot_oth =
1
4
which tells you that the dot 4 equals dot 1 in column 2; and then of course that dot 1 equals dot 4 in column 2.
This computes the same t as above. It then multiplies by a logical mask along the third and fourth dimensions (bsxfun(#ne, reshape(1:p, 1,1,[]), reshape(1:p, 1,1,1,[]) )) to avoid detecting each dot as similar to itself. Finally, it requires that all coordinates (all(...,1)) be the same in order to declare two dots as crashing, and applies find to locate those dots. The result (ii) is a linear index into a 3D array of dimensions path length × number of dots × number of dots, which is translated by ind2sub into position along path and identities of crashing dots.
A=[1 2 ;3 4];B=[5 6;3 9];
A==B
ans =
0 0
1 0
This works in 2D, but also in 3D. You can find the location of the equal numbers by using find:
C=permute(A,[3 2 1]);
D=permute(B,[3 2 1]);
find(C==D)
ans =
3
where 3 is a linear index to the location of the equal number.
for ii = 1:size(A,3)-1
for jj = ii+1:size(A,3)
[r{ii,jj},c{ii,jj},v{ii,jj}] = find(squeeze(A(:,:,ii))==squeeze(A(:,:,jj)));
end
end
This will loop over your matrices in the third dimension, i.e. your physical dots. It will squeeze out the 2D matrix per dot and check that against every other dot's 2D matrix. If it finds any, it will store the row, column and value at the intersection in the cells r, c and v, where the indices correspond to the dot numbers, i.e. r{2,4} will give the row numbers of intersections between dot 2 and 4.
I'm currently bringing some GAUSS code over to Matlab and I'm stuck trying to use the GAUSS "rotater" function.
The command reference entry for rotater says:
Purpose Rotates the rows of a matrix
Format y = rotater(x,r)
Input x: N x K matrix to be rotated. r: N x 1 or 1 x 1 matrix specifying the amount of rotation.
Output y: N x K rotated matrix.
Remarks The rotation is performed horizontally within each row of the matrix. A positive rotation value will cause the elements to move
to the right. A negative rotation will cause the elements to move to
the left. In either case, the elements that are pushed off the end of
the row will wrap around to the opposite end of the same row. If the rotation value is greater than or equal to the number of columns in x, then the rotation value will be calculated using (r % cols(x)).
Example 1
(I'm following Matlab's notation here, with straight brackets for matrices and a semicolon for a new ro)
If x = [1 2 3; 4 5 6], and r = [1; -1],then y = [3 1 2; 5 6 4]
Example 1
If x = [1 2 3; 4 5 6; 7 8 9; 10, 11, 12], and r = [0; 1; 2; 3], then y = [1 2 3; 6 4 5; 8 9 7; 10 11 12]
Maybe someone has found a function like that somewhere or can give me advice how to write it?
This can be done using bsxfun twice:
Compute rotated row indices by subtracting r with bsxfun and using mod. As usual, mod needs indices starting at 0, not 1. The rotated row indices are left as 0-based, because that's more convenient for step 2.
Get a linear index from columns and rotated rows, again using bsxfun. This linear index applied to x gives y:
Code:
[s1 s2] = size(x);
rows = mod(bsxfun(#plus, 1:s2, -r(:))-1, s2); % // step 1
y = x(bsxfun(#plus, rows*s1, (1:s1).')); %'// step 2
circshift is pretty close to what you're looking for except that 1) it works on columns rather than rows, and 2) it shifts the entire matrix by the same offset.
The first one is easy to fix, we just transpose. For the second one I haven't been able to find a vectorized approach, but in the meantime, here's a version with a for loop:
x = [1 2 3; 4 5 6; 7 8 9; 10 11 12]
r = [0 1 2 3]
B = x'
C = zeros(size(B));
for ii = 1:size(B,2)
C(:,ii) = circshift(B(:,ii),r(ii));
end
y = C'
The output is:
x =
1 2 3
4 5 6
7 8 9
10 11 12
r =
0 1 2 3
B =
1 4 7 10
2 5 8 11
3 6 9 12
y =
1 2 3
6 4 5
8 9 7
10 11 12
This can be done using a simple for loop to iterate over each row, and a function called 'circshift' from matlab.
I created a function the goes through each row and applies the appropriate shift to it. There may be more efficient ways to implement this, but this way works with your examples. I created a function
function rotated_arr = GaussRotate(input_array, rotation_vector)
[N,K] = size(input_array)
%creates array for return values
rotated_arr = zeros(N,K);
%if the rotation vector is a scalar
if (length(rotation_vector) == 1)
%replicate the value once for each row
rotation_vector = repmat(rotation_vector, [1,N]);
end
%if the rotation vector doesn't have as many entries as there are rows
%in the input array
if (length(rotation_vector) ~= N)
disp('ERROR GaussRotate: rotation_vector is the wrong size')
disp('if input_Array is NxK, rotation_vector must be Nx1 or 1x1')
return
end
%for each row
for idx=1:size(input_array,1)
%shift the row by the appropriate number of columns
%we use [0,shift] because we want to shift the columns, the row
%stays where it is (even though this is a 1xN at this point we
%still specify rows vs columns)
rotated_arr(idx,:) = circshift(input_array(idx,:),[0,rotation_vector(idx)]);
end
end
then simply called it with your examples
x = [1 2 3; 4 5 6];
r = [1; -1];
y = GaussRotate(x,r)
%produces [3 1 2; 5 6 4]
%I also made it support the 1x1 case
r = [-1]
%this will shift all elements one column to the left
y = GaussRotate(x,r)
%produces [2 3 1; 5 6 4]
x = [1 2 3; 4 5 6; 7 8 9; 10, 11, 12]
r = [0; 1; 2; 3]
y = GaussRotate(x,r)
%produces [1 2 3; 6 4 5; 8 9 7; 10 11 12]
suppose that we are creating following matrix from given signal
function [ x ]=create_matrix1(b,l)
n = length(b);
m = n-l+1;
x = zeros(m,l);
for i=1:m
x(i,:)=b(i:i+l-1);
end;
end
with some window length,for example
X=[2;1;3;4;5;7]
X =
2
1
3
4
5
7
>> B=create_matrix1(X,3)
B =
2 1 3
1 3 4
3 4 5
4 5 7
if we have given matrix and windows length ,how can i reconstruct original signal?let say i know that windows length is 3,thanks in advance,i think i should sum elements on anti diagonal and divide by number of elements in this anti diagonal ,but how can i do it by code?thanks in advance
Your original vector is located along the top and right edge of your matrix B and can be reconstructed like so:
>> X_reconstructed = [B(1,1:end-1).'; B(:,end)]
X_reconstructed =
2
1
3
4
5
7
In case the matrix B is some noisy matrix and you actually want to do the averages along the diagonals:
>> BB = fliplr(B);
>> X_mean = arrayfun(#(i) mean(diag(BB,i)), size(B,2)-1:-1:-size(B,1)+1).'
X_mean =
2
1
3
4
5
7
just lets make it simple, assume that I have a 10x3 matrix in matlab. The numbers in the first two columns in each row represent the x and y (position) and the number in 3rd columns show the corresponding value. For instance, [1 4 12] shows that the value of function in x=1 and y=4 is equal to 12. I also have same x, and y in different rows, and I want to average the values with same x,y. and replace all of them with averaged one.
For example :
A = [1 4 12
1 4 14
1 4 10
1 5 5
1 5 7];
I want to have
B = [1 4 12
1 5 6]
I really appreciate your help
Thanks
Ali
Like this?
A = [1 4 12;1 4 14;1 4 10; 1 5 5;1 5 7];
[x,y] = consolidator(A(:,1:2),A(:,3),#mean);
B = [x,y]
B =
1 4 12
1 5 6
Consolidator is on the File Exchange.
Using built-in functions:
sparsemean = accumarray(A(:,1:2), A(:,3).', [], #mean, 0, true);
[i,j,v] = find(sparsemean);
B = [i.' j.' v.'];
A = [1 4 12;1 4 14;1 4 10; 1 5 5;1 5 7]; %your example data
B = unique(A(:, 1:2), 'rows'); %find the unique xy pairs
C = nan(length(B), 1);
% calculate means
for ii = 1:length(B)
C(ii) = mean(A(A(:, 1) == B(ii, 1) & A(:, 2) == B(ii, 2), 3));
end
C =
12
6
The step inside the for loop uses logical indexing to find the mean of rows that match the current xy pair in the loop.
Use unique to get the unique rows and use the returned indexing array to find the ones that should be averaged and ask accumarray to do the averaging part:
[C,~,J]=unique(A(:,1:2), 'rows');
B=[C, accumarray(J,A(:,3),[],#mean)];
For your example
>> [C,~,J]=unique(A(:,1:2), 'rows')
C =
1 4
1 5
J =
1
1
1
2
2
C contains the unique rows and J shows which rows in the original matrix correspond to the rows in C then
>> accumarray(J,A(:,3),[],#mean)
ans =
12
6
returns the desired averages and
>> B=[C, accumarray(J,A(:,3),[],#mean)]
B =
1 4 12
1 5 6
is the answer.