Create a 4D array from 3D array with 3 columns, and 1D (row only) array with 3 columns
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When I load a test data file, Matlab tells me it has 4 dimensions, and when I check size, I see:
>> size(myVar)
ans =
3 241 161 161
'3' because I think the variable contains a 3D array.
'241 161 161' because the variable also contain a 1D array that has one row with three columns.
(Or '241 161 161' could mean there are three, 1D arrays with one column in each? I don't know how to check.)
Either way, I know that 241, 161, 161 is the dimension of a space in a 3D grid-like structure.
I want to create a .mat file with similar structure from the following two arrays:
1st array:
I have a 3D array, with size {576 3 4}, it looks something like,
(:,:,1) =
-0.0336 0.6440 -0.1464
-0.0336 0.6441 -0.1467
...... ....... .......
576 rows.
Also similar structure for (:,:,2), (:,:,3) and (:,:,4).
2nd array:
1D, row-only array with three columns
L = [ 100 100 100];
How do I create a 4D variable that has data from arr1 and arr2, and looks like what I wrote in the beginning?
My guess is its size would look like '3 100 100 100'.
The .mat creating part is the easy part:
save('myFile.mat','4D_var');
But I don't know how to create that 4D_var.
Thank you.
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'3' because I think the variable contains a 3D array.
'241 161 161' because the variable also contain a 1D array that has one row with three columns.
No, neither are true. If the size(myVar) reports [3 241 161 161]. It is because it contains one array only, which is a 4D array of dimension 3x241x161x161. Accordingly, the number of elements myVar holds contains is,
numElements=3*241*161*161
Nx = length(myVar(1,:,1,1)); gives 241 ... So, could it be that myVar has 4 arrays inside it?
No.
What would you say about myVar(2,:,3,5)? You will see that length(myVar(2,:,3,5)) is also 241. Wouldn't that imply, by your reasoning, that myVar also contains a 5th array which is not in your list above and which also has 241 "rows"?
More generally, you will see that any myVar(i,:,j,k) gives a different length-241 vector for any combination of (i,j,k). There are many more than 4 such combinations.
Jartok
2024년 1월 15일
채택된 답변
How do I create a 4D variable that has data from arr1 and arr2, and looks like what I wrote in the beginning? My guess is its size would look like '3 100 100 100'.
I don't see what arr1 has to do with anything, but the following might be the "grid space" you're trying to construct,
L=[100,100,100];
[X,Y,Z]=ndgrid( 1:L(1) , 1:L(2) , 1:L(3) );
grid_space=permute(cat(4, X,Y,Z) ,[4,1,2,3] );
size(grid_space)
댓글 수: 24
After reading the data file that has xyz variable (I named it 'myVar' in my example, but it's the same thing) , I extracted a few lines. They are x, y, z coordinates.
Since this is test data, I need to pass my own x, y, z data which is in the form of a 3D array, with size [576 3 4].
Matt J
2024년 1월 16일
You say your xyz data is 576x3x4 but in your posted image, it is 3x5x2.
Stephen23
2024년 1월 16일
Rather than creating (potentially) large intermediate arrays just to measure the size of an array dimension:
length(xyz(1,:,1,1))
you should just call SIZE with its optional DIM input:
size(xyz,2)
You could even get all three in a vector like this:
size(xyz,2:4)
Using SIZE will be simpler and more efficient than creating intermediate arrays that simply get discarded.
(1) One 4D variable will have a 3D array (that has x, y, z coordinate values), and total grid dimension (100 x 100 x 100).
@Jar If your original xyz array is 3x241x161x161, that means it will contain 1870883 elements
3*241*161*161
Now, you say you want the array to be re-organized or transformed somehow into a 100x100x100 array which would contain only 1 million elements. You haven't explained precisely what transformation of xyz is supposed to occur that leads to a reduction in the total number of elements in the final array. Are you trying to do some sort of interpolation?
Matt J
2024년 1월 16일
It might help if you post an image of the scatter plot,
scatter3(xyz(1,:), xyz(2,:), xyz(3,:))
so we can see how the xyz triplets are distributed in space.
It's impossible to see any details. Do they form an equi-spaced lattice of points?
Additionally, you said that you had Bx field values associated with each point, which would mean your Bx variable is currently a 241x161x161 array. You say that this is supposed to turn into a 100x100x100 array. But how? Do you simply want to resize it via interpolation? If so, you can use imresize3, e.g.,
Bx=rand(241,161,161);
Bx_new=imresize3(Bx,[100,100,100]);
whos Bx Bx_new
I want to create, say a 'myXYZ' variable that has x, y, z coordinate values of size 576x3x4 in its 1st dimension, and 100, 100, 100 in its 2nd, 3rd and the 4th dimension.
That part is totally unclear. What will be size(myXYZ) of the final result myXYZ? You should be able to use the rand() function to generate a fake version of myXYZ to demonstrate the shape you want. So show us, with rand(), how you would create a variable of the desired size.
Will it be 576x3x4x100x100x100? Like in,
myVar = rand(576,3,4,100,100,100)
Note that such an array would be 6D, not 4D and would consume 25 GB of memory in single floats.
"For my own data, I wanted to create a region of space that is 100x100x100 grid units, and say, I will use the same grid size of 0.25. So you see, the total grid space dimension is not fixed. For simulated data, it happend to be 241x161x161, for my own data, I wanted to use 100x100x100."
myXYZ = myVar(:,1:100,1:100,1:100)
Jartok
2024년 1월 17일
I am sorry, I do not understand.
First, let me use my own variable name (in the recent posts, I am calling it myXYZ) because 'myVar' was meant for test data.
How would I create a 4D myXYZ variable which incorporates a 3D array, and grid space dimension (100x100x100)?
Could you please elaborate on your code? (I can't use myVar, it is for test) I need to create my own variable, with my own 3D data.
On my end, I tried the following:
Error says:
"I need to create my own variable, with my own 3D data."
Based on your screenshot you have
576*3*4
data points. How do you expect to fit those into
100*100*100
array elements?
Jartok
2024년 1월 17일
@Jar What you misunderstand is what happens when you index into a multi-dimensional array. Observe what happens when we apply the size() command to your expressions,
myXYZ=zeros(3,100,100,100);
my3D=ones(576,3,4);
size(myXYZ(1,:,:,:))
size(my3D)
As you can see, the region you reference with the indexing expression myXYZ(1,:,:,:) is a smaller array of size 1x100x100x100 while my3D has a completely different size 576x3x4. There is therefore no way my3D can occupy the region of myXYZ you have indexed. In fact, there is no region within myXYZ where a 576x3x4 array like my3D can be inserted because all dimensions of myXYZ are smaller than 576.
Here is one example of what could work,
myXYZ=zeros(576,3,4,100);
my3D = ones(576,3,4);
size(myXYZ(:,:,:,1)) %The same size as my3D
myXYZ(:,:,:,1)=my3D;
but understand that this is only 1 out of 100 locations where my3D could have been inserted. I could have also inserted my3D as follows,
myXYZ(:,:,:,2)=my3D;
myXYZ(:,:,:,3)=my3D;
...
myXYZ(:,:,:,100)=my3D;
Would it be correct to say that this 3x241x161x161 size means there are {161 pages of 3 rows x 241 columns}, and there are 161 pages of that 3D data (of size 3 x 241 x 161)?
Yes.
But I hope you will agree that this program needs a 3D array at the first dimension.
No. First of all, I don't know what it means to "have a 3D array at the first dimension". That is not intelligble English. Secondly, none of the steps in the code you've shown involve 3D arrays.
You've introduced a new variable called "cube" without telling us anything about it, but I will assume it is an Mx8 array. Similarly, I will assume that myXYZ is a 4D array of size MxNxPxQ. If we look at the first assignment statement,
cube(:,1)=myXYZ(:,i,j,k);
The left hand side cube(:,1) is 1D vector of size Mx1, so it is not 3D. You can verify that with the size() command
The right hand side myXYZ(:,i,j,k) will also be a vector of size Mx1, and so is also not 3D.
The assignment statement works because the left and right hand side both refer to data chunks that are the same size, but none of these chunks are 3D. The same is true in all subsequent lines.
Here is another possible interpretation of what you are trying to do. Suppose you are given ranges of x,y, and z values that span a rectangular 3D region, e.g.,
xrange=[-1.7, +1.7];
yrange=[-3.6, +3.6];
zrange=[-2.0, +2.0];
Now, suppose we want to create a series of x,y,z node locations that evenly sample this region at 10 points in each dimension. You could do so as follows.
[X,Y,Z]=ndgrid(linspace(xrange(1), xrange(2),10),...
linspace(yrange(1), yrange(2),10),...
linspace(zrange(1), zrange(2),10));
myXYZ=permute(cat(4, X,Y,Z),[4,1,2,3]);
Notice that this array has size,
size(myXYZ)
The first dimension is 3, which I think is what you really mean by "assign a 3D array to the first dimension".
Also, we can make a scatter plot of these grid points as follows:
scatter3( myXYZ(1,:) , myXYZ(2,:) , myXYZ(3,:) ,'r' ,'filled','MarkerFaceAlpha',0.4);
xlabel X; ylabel Y; zlabel Z
Jartok
2024년 1월 17일
Jartok
2024년 1월 18일
추가 답변 (2개)
If understood correctly, this is what you are trying to get, e.g.:
rng(13) % for reproducibility
A = rand(5, 3);
size(A)
% B and C have the same size as A
B = rand(5, 3);
C = rand(5, 3);
L(:,:, 1) = A;
L(:,:, 2) = B;
L(:,:, 3) = C;
size(L)
% D array
D = rand(5, 3)
L(:,:,4) = D;
% Final 4-D array
[Row, Col, Layer]=size(L)
L
So, could it be that myVar has 4 arrays inside it
To me, it doesn't sound like you are talking about a 4D array at all, but instead a cell array. A cell array is a variable that can have "4 arrays inside it", for example:
xyz=rand(576,3,4);
Bx=rand(100,100,100);
By=rand(100,100,100);
Bz=rand(100,100,100);
whos xyz Bx By Bz
myVar={ xyz, Bx,By,Bz}
Here the first array is a 3D array of dimension 576x3x4 and the second,third, and fourth arrays are each 100x100x100
size(myVar{1})
size(myVar{2})
size(myVar{3})
size(myVar{4})
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