would appreciate feedback on how to reshape general ND matrix from a file in octave. To set the problem,
if I read data like for 2D case x y z
0 10 13
0 11 -1
0 12 12
1 10 5
1 11 4
1 12 3
2 10 1
2 11 6
2 12 2
reading Z values
matrix = dlmread('data.dat');
z = matrix(:,3);
z = reshape(z, 3, 3).';
x = [0 1 2];
y = [10 11 12];
xi = linspace (min (x), max (x), 200);
yi = linspace (min (y), max (y), 300);
fcn = interpn (x, y, A, xi, yi, "spline");
Now, transposing a matrix doesn't work in ND which here needs to be done. How would one do a ND case pulling from data so to format this for interpn,
thanks, Damir
Related
I have an NxM matrix in MATLAB that I would like to reorder in similar fashion to the way JPEG reorders its subblock pixels:
(image from Wikipedia)
I would like the algorithm to be generic such that I can pass in a 2D matrix with any dimensions. I am a C++ programmer by trade and am very tempted to write an old school loop to accomplish this, but I suspect there is a better way to do it in MATLAB.
I'd be rather want an algorithm that worked on an NxN matrix and go from there.
Example:
1 2 3
4 5 6 --> 1 2 4 7 5 3 6 8 9
7 8 9
Consider the code:
M = randi(100, [3 4]); %# input matrix
ind = reshape(1:numel(M), size(M)); %# indices of elements
ind = fliplr( spdiags( fliplr(ind) ) ); %# get the anti-diagonals
ind(:,1:2:end) = flipud( ind(:,1:2:end) ); %# reverse order of odd columns
ind(ind==0) = []; %# keep non-zero indices
M(ind) %# get elements in zigzag order
An example with a 4x4 matrix:
» M
M =
17 35 26 96
12 59 51 55
50 23 70 14
96 76 90 15
» M(ind)
ans =
17 35 12 50 59 26 96 51 23 96 76 70 55 14 90 15
and an example with a non-square matrix:
M =
69 9 16 100
75 23 83 8
46 92 54 45
ans =
69 9 75 46 23 16 100 83 92 54 8 45
This approach is pretty fast:
X = randn(500,2000); %// example input matrix
[r, c] = size(X);
M = bsxfun(#plus, (1:r).', 0:c-1);
M = M + bsxfun(#times, (1:r).'/(r+c), (-1).^M);
[~, ind] = sort(M(:));
y = X(ind).'; %'// output row vector
Benchmarking
The following code compares running time with that of Amro's excellent answer, using timeit. It tests different combinations of matrix size (number of entries) and matrix shape (number of rows to number of columns ratio).
%// Amro's approach
function y = zigzag_Amro(M)
ind = reshape(1:numel(M), size(M));
ind = fliplr( spdiags( fliplr(ind) ) );
ind(:,1:2:end) = flipud( ind(:,1:2:end) );
ind(ind==0) = [];
y = M(ind);
%// Luis' approach
function y = zigzag_Luis(X)
[r, c] = size(X);
M = bsxfun(#plus, (1:r).', 0:c-1);
M = M + bsxfun(#times, (1:r).'/(r+c), (-1).^M);
[~, ind] = sort(M(:));
y = X(ind).';
%// Benchmarking code:
S = [10 30 100 300 1000 3000]; %// reference to generate matrix size
f = [1 1]; %// number of cols is S*f(1); number of rows is S*f(2)
%// f = [0.5 2]; %// plotted with '--'
%// f = [2 0.5]; %// plotted with ':'
t_Amro = NaN(size(S));
t_Luis = NaN(size(S));
for n = 1:numel(S)
X = rand(f(1)*S(n), f(2)*S(n));
f_Amro = #() zigzag_Amro(X);
f_Luis = #() zigzag_Luis(X);
t_Amro(n) = timeit(f_Amro);
t_Luis(n) = timeit(f_Luis);
end
loglog(S.^2*prod(f), t_Amro, '.b-');
hold on
loglog(S.^2*prod(f), t_Luis, '.r-');
xlabel('number of matrix entries')
ylabel('time')
The figure below has been obtained with Matlab R2014b on Windows 7 64 bits. Results in R2010b are very similar. It is seen that the new approach reduces running time by a factor between 2.5 (for small matrices) and 1.4 (for large matrices). Results are seen to be almost insensitive to matrix shape, given a total number of entries.
Here's a non-loop solution zig_zag.m. It looks ugly but it works!:
function [M,index] = zig_zag(M)
[r,c] = size(M);
checker = rem(hankel(1:r,r-1+(1:c)),2);
[rEven,cEven] = find(checker);
[cOdd,rOdd] = find(~checker.'); %'#
rTotal = [rEven; rOdd];
cTotal = [cEven; cOdd];
[junk,sortIndex] = sort(rTotal+cTotal);
rSort = rTotal(sortIndex);
cSort = cTotal(sortIndex);
index = sub2ind([r c],rSort,cSort);
M = M(index);
end
And a test matrix:
>> M = [magic(4) zeros(4,1)];
M =
16 2 3 13 0
5 11 10 8 0
9 7 6 12 0
4 14 15 1 0
>> newM = zig_zag(M) %# Zig-zag sampled elements
newM =
16
2
5
9
11
3
13
10
7
4
14
6
8
0
0
12
15
1
0
0
Here's a way how to do this. Basically, your array is a hankel matrix plus vectors of 1:m, where m is the number of elements in each diagonal. Maybe someone else has a neat idea on how to create the diagonal arrays that have to be added to the flipped hankel array without a loop.
I think this should be generalizeable to a non-square array.
% for a 3x3 array
n=3;
numElementsPerDiagonal = [1:n,n-1:-1:1];
hadaRC = cumsum([0,numElementsPerDiagonal(1:end-1)]);
array2add = fliplr(hankel(hadaRC(1:n),hadaRC(end-n+1:n)));
% loop through the hankel array and add numbers counting either up or down
% if they are even or odd
for d = 1:(2*n-1)
if floor(d/2)==d/2
% even, count down
array2add = array2add + diag(1:numElementsPerDiagonal(d),d-n);
else
% odd, count up
array2add = array2add + diag(numElementsPerDiagonal(d):-1:1,d-n);
end
end
% now flip to get the result
indexMatrix = fliplr(array2add)
result =
1 2 6
3 5 7
4 8 9
Afterward, you just call reshape(image(indexMatrix),[],1) to get the vector of reordered elements.
EDIT
Ok, from your comment it looks like you need to use sort like Marc suggested.
indexMatrixT = indexMatrix'; % ' SO formatting
[dummy,sortedIdx] = sort(indexMatrixT(:));
sortedIdx =
1 2 4 7 5 3 6 8 9
Note that you'd need to transpose your input matrix first before you index, because Matlab counts first down, then right.
Assuming X to be the input 2D matrix and that is square or landscape-shaped, this seems to be pretty efficient -
[m,n] = size(X);
nlim = m*n;
n = n+mod(n-m,2);
mask = bsxfun(#le,[1:m]',[n:-1:1]);
start_vec = m:m-1:m*(m-1)+1;
a = bsxfun(#plus,start_vec',[0:n-1]*m);
offset_startcol = 2- mod(m+1,2);
[~,idx] = min(mask,[],1);
idx = idx - 1;
idx(idx==0) = m;
end_ind = a([0:n-1]*m + idx);
offsets = a(1,offset_startcol:2:end) + end_ind(offset_startcol:2:end);
a(:,offset_startcol:2:end) = bsxfun(#minus,offsets,a(:,offset_startcol:2:end));
out = a(mask);
out2 = m*n+1 - out(end:-1:1+m*(n-m+1));
result = X([out2 ; out(out<=nlim)]);
Quick runtime tests against Luis's approach -
Datasize: 500 x 2000
------------------------------------- With Proposed Approach
Elapsed time is 0.037145 seconds.
------------------------------------- With Luis Approach
Elapsed time is 0.045900 seconds.
Datasize: 5000 x 20000
------------------------------------- With Proposed Approach
Elapsed time is 3.947325 seconds.
------------------------------------- With Luis Approach
Elapsed time is 6.370463 seconds.
Let's assume for a moment that you have a 2-D matrix that's the same size as your image specifying the correct index. Call this array idx; then the matlab commands to reorder your image would be
[~,I] = sort (idx(:)); %sort the 1D indices of the image into ascending order according to idx
reorderedim = im(I);
I don't see an obvious solution to generate idx without using for loops or recursion, but I'll think some more.
Suppose I have an array X of size n by p by q. I would like to reshape it as a matrix with p rows, and in each row put the concatenation of the n rows of size q, resulting in a matrix of size p by nq.
I managed to do it with a loop but it takes a while say if n=1000, p=300, q=300.
F0=[];
for k=1:size(F,1)
F0=[F0,squeeze(X(k,:,:))];
end
Is there a faster way?
I think this is what you want:
Y = reshape(permute(X, [2 1 3]), size(X,2), []);
Example with n=2, p=3, q=4:
>> X
X(:,:,1) =
0 6 9
8 3 0
X(:,:,2) =
4 7 1
3 7 4
X(:,:,3) =
4 7 2
6 7 6
X(:,:,4) =
6 1 9
1 4 3
>> Y = reshape(permute(X, [2 1 3]), size(X,2), [])
Y =
0 8 4 3 4 6 6 1
6 3 7 7 7 7 1 4
9 0 1 4 2 6 9 3
Try this -
reshape(permute(X,[2 3 1]),p,[])
Thus, for code verification, one can look into a sample case run -
n = 2;
p = 3;
q = 4;
X = rand(n,p,q)
F0=[];
for k=1:n
F0=[F0,squeeze(X(k,:,:))];
end
F0
F0_noloop = reshape(permute(X,[2 3 1]),p,[])
Output is -
F0 =
0.4134 0.6938 0.3782 0.4775 0.2177 0.0098 0.7043 0.6237
0.1257 0.8432 0.7295 0.2364 0.3089 0.9223 0.2243 0.1771
0.7261 0.7710 0.2691 0.8296 0.7829 0.0427 0.6730 0.7669
F0_noloop =
0.4134 0.6938 0.3782 0.4775 0.2177 0.0098 0.7043 0.6237
0.1257 0.8432 0.7295 0.2364 0.3089 0.9223 0.2243 0.1771
0.7261 0.7710 0.2691 0.8296 0.7829 0.0427 0.6730 0.7669
Rather than using vectorization to solve the problem, you could look at the code to try and figure out what may improve performance. In this case, since you know the size of your output matrix F0 should be px(n*q), you could pre-allocate memory to F0 and avoid the constant resizing of the matrix at each iteration of the for loop
n=1000;
p=300;
q=300;
F0=zeros(p,n*q);
for k=1:size(F,1)
F0(:,(k-1)*q+1:k*q) = squeeze(F(k,:,:));
end
While probably not as efficient as the other two solutions, it is an alternative. Try the above and see what happens!
I have a large file (over 2 million cells) and I need the value of each individual cell to be read as the z coordinate of a 3D plot and the original x,y coordinates of the cell to be the x and y coordinates in the graph. For example:
M =
1 2 3
4 5 6
7 8 9
I need it to be read as (1,1,1) (2,1,2) (3,1,3)
(1,2,4) (2,2,5) (3,2,6)
(1,3,7) (2,3,8) (3,3,9)
And from there I need to create a 3D plot. Please help!
Pseudocode:
size = 3 # set this to the maximum value for x
x = 1
y = 1
z = readvalue()
while read succeeded
plot(x, y, z)
x = x + 1
if x > size
x = 1
y = y + 1
z = readvalue()
I'm trying to increase the efficiency of my MATLAB code. What it does is, it replaces the nonzero elements of a matrix with the multiplication of the rest of the nonzero elements in the same row. For instance,
X = [2 3 6 0; 0 3 4 2]
transforms into
X = [18 12 6 0; 0 8 6 12]
It's an easy task to implement within a for loop. It checks every row, finds nonzero values and do its replacements. I want to get rid of the for loop though. Is there a way to implement this without a loop?
Code
X = [2 3 6 0; 0 3 4 2];
X1 = X;
X1(~X) = 1;
out = bsxfun(#rdivide,prod(X1,2),X1).*(X~=0)
Output
out =
18 12 6 0
0 8 6 12
Probably getting the row product first once and then divide by the element you don't want is the simplest way:
X = [2 3 6 0; 0 3 4 2]
Y=X
%get the product of all elements in a row
Y(Y==0)=1
Y=prod(Y,2)
%repeat Y to match the size of X
Y=repmat(Y,1,size(X,2))
%For all but the zero elements, divide Y by X, which is the product of all other elements.
X(X~=0)=Y(X~=0)./X(X~=0)
I here by post the code why I came across while exploring one technique.
Y = repmat((1:m)', [1 n]);
X = repmat(1:n, [m 1]) - labels_left;
X(X<1) = 1;
indices = sub2ind([m,n],Y,X);
final_labels = labels_left;
final_labels(abs(labels_left - labels_right(indices))>=1) = -1;
In above code labels left is single channel image.[m n] is the size of that image. I want to know how this sub2ind works in above code.And Iam also facing problem in the last statement which contains
labels_right(indices)
what the above expression evaluates to.Here labels right is also an image
Maybe a smaller example could help understand:
%# image matrix
M = rand(4,3)
[m n] = size(M)
%# meshgrid, and convert to linear indices
[X,Y] = meshgrid(1:n,1:m)
indices = sub2ind([m,n],Y,X)
%# extract those elements
M(indices)
The matrix M:
>> M
M =
0.95717 0.42176 0.65574
0.48538 0.91574 0.035712
0.80028 0.79221 0.84913
0.14189 0.95949 0.93399
the grid of (x,y) coordinates of all points:
>> X,Y
X =
1 2 3
1 2 3
1 2 3
1 2 3
Y =
1 1 1
2 2 2
3 3 3
4 4 4
converted to linear indices:
>> indices
indices =
1 5 9
2 6 10
3 7 11
4 8 12
then we index into the matrix using those indices.
>> M(indices)
ans =
0.95717 0.42176 0.65574
0.48538 0.91574 0.035712
0.80028 0.79221 0.84913
0.14189 0.95949 0.93399
Note that: M(indices(i,j)) = M(Y(i,j)),X(i,j)).