Is there a way to grow a 3D array in the third dimension using the end index in a loop in Matlab?
In 2D it can be done like
a = [];
for x = y
a(end + 1, :) = f(x);
end
But in 3D the same thing will not work as a(1,1,end) will try to index a(1,1,1) the first iteration (not a(1,1,0) as one might expect). So I can't do
im = [];
for x = y
im(:, :, end + 1) = g(x);
end
It seems the end of a in third dimension is handled a bit differently than in the first two:
>> a = [];
>> a(end,end,end) = 1
Attempted to access a(0,0,1); index must be a positive integer or logical.
Am I missing something about how end indexing works here?
What you're asking...
If you know the size of g(x), initialize im to an empty 3d-array:
im = zeros(n, m, 0); %instead of im = [];
I think your code should work now.
A better way...
Another note, resizing arrays each iteration is expensive! This doesn't really matter if the array is small, but for huge matrices, there can be a big performance hit.
I'd initialize to:
im = zeros(n, m, length(y));
And then index appropriately. For example:
i = 1;
for x = y
im(:, :, i) = g(x);
i = i + 1;
end
This way you're not assigning new memory and copying over the whole matrix im each time it gets resized!
Related
How can I delete all-zero pages from a 3D matrix in a loop?
I have come up with the following code, though it is not 'entirely' correct, if at all. I am using MATLAB 2019b.
%pseudo data
x = zeros(3,2,2);
y = ones(3,2,2);
positions = 2:4;
y(positions) = 0;
xy = cat(3,x,y); %this is a 3x2x4 array; (:,:,1) and (:,:,2) are all zeros,
% (:,:,3) is ones and zeros, and (:,:,4) is all ones
%my aim is to delete the arrays that are entirely zeros i.e. xy(:,:,1) and xy(:,:,2),
%and this is what I have come up with; it doesn't delete the arrays but instead,
%all the ones.
for ii = 1:size(xy,3)
for idx = find(xy(:,:,ii) == 0)
xy(:,:,ii) = strcmp(xy, []);
end
end
Use any to find indices of the slices with at least one non-zero value. Use these indices to extract the required result.
idx = any(any(xy)); % idx = any(xy,[1 2]); for >=R2018b
xy = xy(:,:,idx);
I am unsure what you'd expect your code to do, especially given you're comparing strings in all-numerical arrays. Here's a piece of code which does what you desire:
x = zeros(3,2,2);
y = ones(3,2,2);
positions = 2:4;
y(positions) = 0;
xy = cat(3,x,y);
idx = ones(size(xy,3),1,'logical'); % initialise catching array
for ii = 1:size(xy,3)
if sum(nnz(xy(:,:,ii)),'all')==0 % If the third dimension is all zeros
idx(ii)= false; % exclude it
end
end
xy = xy(:,:,idx); % reindex to get rid of all-zero pages
The trick here is that sum(xy(:,:,ii),'all')==0 is zero iff all elements on the given page (third dimension) are zero. In that case, exclude it from idx. Then, in the last row, simply re-index using logical indexing to retain only pages whit at least one non-zero element.
You can do it even faster, without a loop, using sum(a,[1 2]), i.e. the vectorial-dimension sum:
idx = sum(nnz(xy),[1 2])~=0;
xy = xy(:,:,idx);
I need to create an empty array of matrices, and after fill it with matrices of the same size.
I have made a little script to explain:
result = [];
for i = 0: 4;
M = i * ones(5,5); % create matrice
result = [result,M]; % this would have to append M to results
end
Here result is a matrix of size 5*25 and I need an array of matrices 5*5*4.
I have been researched but I only found this line: result = [result(1),M];
The issue is that [] implicitly concatenates values horizontally (the second dimension). In your case, you want to concatenate them along the third dimension so you could use cat.
result = cat(3, result, M);
But a better way to do it would be to actually pre-allocate your result array using zeros
result = zeros(5, 5, 4);
And then within your loop fill each "slice" of the 3D array with the values.
for k = 0:4
M = k * ones(5,5);
result(:,:,k+1) = M;
end
I tried to check if a 3D array is not all zeros using the next code:
notAll_n0_GreaterThan_ni=1;
while notAll_n0_GreaterThan_ni
notAll_n0_GreaterThan_ni=0;
mask=(n0<ni);
numDimensions=ndims(mask);
for dim_ind=1:numDimensions
if any(mask,dim_ind)
notAll_n0_GreaterThan_ni=1;
break;
end
end
if notAll_n0_GreaterThan_ni
n0(mask)=n0(mask)+1;
end
end
It seems I have error in the code because at the end I get for example: n_0(11,3,69)=21 while ni(11,3,69)=21.1556.
I can't find the error. I'll appreciate if someone shows me where I'm wrong and also if there is a simpler way to check existence of nonzero elements in a 3D array.
Let x denote an n-dimensional array. To check if it contains at least one non-zero element, just use
any(x(:))
For example:
>> x = zeros(2,3,4);
>> any(x(:))
ans =
0
>> x(1,2,2) = 5;
>> any(x(:))
ans =
1
Other, more exotic possibilities include:
sum(abs(x(:)))>0
and
nnz(x)>0
This is what you looking for
B = any(your_Array_name_here(:) ==0); no need for loops
the (:) turns the elements of your_Array into a single column vector, so you can use this type of statement on an array of any size
I 've tested this and it works
A = rand(3,7,5) * 5;
B = any(A(:) ==0);
I want to slice an 4D-array into n parts along the 5th Dimension in order to use it in parfor:
X(:,:,:,particles)-->X(:,:,:,particles/n,n)
The Problem is that X is so big that I run out of memory if I start writing it into a new variable, so i want to basically do:
X = cat(5,X(:,:,:,1:particles/n),X(:,:,:,particles/n+1:2*particles/n),...)
I am doing this with
sliced = 'cat(5'
for i=1:n
sliced = strcat(2,sliced,sprintf(',X(:,:,:,(1+(%i-1)*%i):%i*%i)',i,particles/n,i,particles/n))
end
sliced = strcat(2,sliced,')');
X = eval(sliced);
I get:
Error: The input character is not valid in MATLAB statements or expressions.
If i print out the contents of sliced and comment everything and paste the printout of sliced manually into eval('...') it works.
Anyone got a solution for my problem or another way of slicing a 4D array without using additional memory?
Thanks
You can use reshape, which must not use any additional memory -
sz_X = size(X) %// get size
X = reshape(X,sz_X(1),sz_X(2),sz_X(3),sz_X(4)/n,[]); %// reshape and save
%// into same variable and as such must be memory efficient
Ok. I just got things mixed up cat and strcat are not the same... oops :o
n = 4;
particles = 200;
X = rand(6,6,6,particles);
sliced = sprintf('X = cat(5');
for i = 1:n
sliced = cat(2,sliced,sprintf(',X(:,:,:,(1+(%i-1)*%i):%i*%i)',i,particles/n,i,particles/n));
end
sliced = cat(2,sliced,sprintf(');'));
eval(sliced);
works just fine. If somebody has got a better way to slice without memory usage - please feel free to post...
I want to store data coming from for-loops in an array. How can I do that?
sample output:
for x=1:100
for y=1:100
Diff(x,y) = B(x,y)-C(x,y);
if (Diff(x,y) ~= 0)
% I want to store these values of coordinates in array
% and find x-max,x-min,y-max,y-min
fprintf('(%d,%d)\n',x,y);
end
end
end
Can anybody please tell me how can i do that. Thanks
Marry
So you want lists of the x and y (or row and column) coordinates at which B and C are different. I assume B and C are matrices. First, you should vectorize your code to get rid of the loops, and second, use the find() function:
Diff = B - C; % vectorized, loops over indices automatically
[list_x, list_y] = find(Diff~=0);
% finds the row and column indices at which Diff~=0 is true
Or, even shorter,
[list_x, list_y] = find(B~=C);
Remember that the first index in matlab is the row of the matrix, and the second index is the column; if you tried to visualize your matrices B or C or Diff by using imagesc, say, what you're calling the X coordinate would actually be displayed in the vertical direction, and what you're calling the Y coordinate would be displayed in the horizontal direction. To be a little more clear, you could say instead
[list_rows, list_cols] = find(B~=C);
To then find the maximum and minimum, use
maxrow = max(list_rows);
minrow = min(list_rows);
and likewise for list_cols.
If B(x,y) and C(x,y) are functions that accept matrix input, then instead of the double-for loop you can do
[x,y] = meshgrid(1:100);
Diff = B(x,y)-C(x,y);
mins = min(Diff);
maxs = max(Diff);
min_x = mins(1); min_y = mins(2);
max_x = maxs(1); max_y = maxs(2);
If B and C are just matrices holding data, then you can do
Diff = B-C;
But really, I need more detail before I can answer this completely.
So: are B and C functions, matrices? You want to find min_x, max_x, but in the example you give that's just 1 and 100, respectively, so...what do you mean?