In something which made me spent several hours, I have found an inconsistency in how Matlab deals with dimensions. I somebody can explain to me OR indicate me how to report this to Matlab, please enlight me.
For size, ones,zeros,mean, std, and most every other old and common commands existing inside Matlab, the dimension arrangement is like the classical one and like the intended standard (as per the size of every dimension), the first dimension is along the column vector, the second dimension is along the row vector, and the following are the non graphical following indexes.
>x(:,:,1)=[1 2 3 4;5 6 7 8];
>x(:,:,2)=[9 10 11 12;13 14 15 16];
>m=mean(x,1)
m(:,:,1) = 3 4 5 6
m(:,:,2) = 11 12 13 14
m=mean(x,2)
m(:,:,1) =
2.5000
6.5000
m(:,:,2) =
10.5000
14.5000
m=mean(x,3)
m = 5 6 7 8
9 10 11 12
d=size(m)
d = 2 4 2
However, for graphical commands like stream3,streamline and others relying on the meshgrid output format, the dimensions 1 and 2 are swapped!: the first dimension is the row vector and the second dimension is the column vector, and the following (third) is the non graphical index.
> [x,y]=meshgrid(1:2,1:3)
x = 1 2
1 2
1 2
y = 1 1
2 2
3 3
Then, for stream3 to operate with classically arranged matrices, we should use permute(XXX,[2 1 3]) at every 3D argument:
xyz=stream3(permute(x,[2 1 3]),permute(y,[2 1 3]),permute(z,[2 1 3])...
,permute(u,[2 1 3]),permute(v,[2 1 3]),permute(w,[2 1 3])...
,xs,ys,zs);
If anybody can explain why this happens, and can indicate to me why this is not a bug, welcome.
This behavior is not a bug because it is clearly documented as the intended behavior: https://www.mathworks.com/help/matlab/ref/meshgrid.html. Specifically:
[X,Y,Z]= meshgrid(x,y,z) returns 3-D grid coordinates defined by the vectors x, y, and z. The grid represented by X, Y, and Z has size length(y)-by-length(x)-by-length(z).
Without speaking to the original authors, the exact motivation may be a bit obscure, but I suspect it has to do with the fact that the y-axis is generally associated with the rows of an image, while x is the columns.
Columns are either "j" or "x" in the documentation, rows are either "i" or "y".
Some functions deal with spatial coordinates. The documentation will refer to "x, y, z". These functions will thus take column values before row values as input arguments.
Some functions deal with array indices. The documentation will refer to "i, j" (or sometimes "i1, i2, i3, ..., in", or using specific names instead of "i" before the dimension number). These functions will thus take row values before column values as input arguments.
Yes, this can be confusing. But if you pay attention to the names of the variables in the documentation, you will quickly figure out the right order.
With meshgrid in particular, if the "x, y, ..." order of arguments is confusing, use ndgrid instead, which takes arguments in array indexing order.
I'm sure this is a trivial question for a signals person. I need to find the function in Matlab that outputs averaging of contiguous segments of windowsize= l of a vector, e.g.
origSignal: [1 2 3 4 5 6 7 8 9];
windowSize = 3;
output = [2 5 8]; % i.e. [(1+2+3)/3 (4+5+6)/3 (7+8+9)/3]
EDIT: Neither one of the options presented in How can I (efficiently) compute a moving average of a vector? seems to work because I need that the window of size 3 slides, and doesnt include any of the previous elements... Maybe I'm missing it. Take a look at my example...
Thanks!
If the size of the original data is always a multiple of widowsize:
mean(reshape(origSignal,windowSize,[]));
Else, in one line:
mean(reshape(origSignal(1:end-mod(length(origSignal),windowSize)),windowSize,[]))
This is the same as before, but the signal is only taken to the end minus the extra values less than windowsize.
I have a nonlinear curve (found using bicubic interpolation; see below) that describes the transformation of intensity values from the target image to reference image values. (Apologies if I'm not using the correct terminology here).
What is the best way of applying this curve to the target image?
I am basically looking for the fastest way to achieve this;
for i = 1:length(curve)
I(I==i) = curve(i);
end
which is fairly slow.
Your curve function as look-up table, the simplest way to execute look up table is:
lookuptable=[ 9 8 7 6 5 4 3 2 1 0 ];
I=[ 1 3 4;
5 3 8];
Itransformed=lookuptable(I)
Notice that the index of the lookuptable is accessed by the value of the pixel.
So if the pixel value range is 0-255, first you should use lookuptable with size of 256 use, and second remember to compensate the fact that matlab index is 1:256 so use:
Itransformed=lookuptable(I-1);
I want to reduce a two dimensional matrix to row vector.
But using reshape with large matrices is really slow. The other alternative is to use colon, but i want matrix's transpose to be colon and not the matrix itself.
e.g.
A=magic(3)
A =
8 1 6
3 5 7
4 9 2
A(:) will stack up all the columns one by one. but i am looking for something like this:
AA=A(2:3,:)';
and then reshape or colon AA instead of A.
The issue is i dont want to define additional variable like AA.
Is there anyway to reduce dimension of two dimensional matrix without reshape?
You can avoid the additional variable by linear indexing. For your example:
A([2 5 8 3 6 9])
which gives
3 5 7 4 9 2
What's happening here is that you treat A as if it was already transformed into a vector, and the elements of this one-dimensional array are accessed through indices 1 through 9. Using the colon is a special case of linear indexing, A(:) is the same as A(1 : end).
Figuring out the right linear indices can be tricky, but sub2ind can help with that.
It is possible that this slightly speeds up the code, mainly because (as #Shai wrote) you avoid writing data to an intermediate variable. I wouldn't expect too much, though.
Try looking at subsref. For your example, you could use it as follows:
subsref(A',struct('type','()','subs',{{2:3,':'}}))
Update: I misunderstood the original question; I thought the OP wanted to select rows from 2:3 from the transposed A matrix keeping the columns as is. I will keep the previous answer in case it could be useful to others.
I think he/she could use the following to slice and flatten a matrix:
subsref(A(2:3,:)', struct('type','()','subs',{{':'}}))
This would give as output:
[3 5 7 4 9 2]'
This is a question about Matlab/Octave.
I am seeing some results of the medfilt1(1D Median filter command in matlab) computation by which I am confused.
EDIT:Sorry forgot to mention:I am using Octave for Windows 3.2.4. This is where i see this behavior.
Please see the questions below, and point if I am missing something.
1] I have a 1D data array b=[ 3 5 -8 6 0];
out=medfilt1(b,3);
I expected the output to be [3 3 5 0 0] but it is showing the output as [4 3 5 0 3]
How come? What is wrong here?
FYI-Help says it pads the data at boundaries by 0(zero).
2] How does medfilt2(2D median filter command in matlab) work.
Help says "Each output pixel contains the median value in the m-by-n neighborhood around the corresponding pixel in the input image".
For m=3,n=3, So does it calculate a 3x3 matrix MAT for each of input pixels placed at its center and do median(median(MAT)) to compute its median value in the m-by-n neighbourhood?
Any pointers will help.
thank you. -AD
I was not able to replicate your error with Matlab 7.11.0, but from the information in your question it seems like your version of medfilt1 does not differentiate between an odd or even n.
When finding the median in a vector of even length, one usually take the mean of the two median values,
median([1 3 4 5]) = (3+4)/2 = 3.5
This seems to be what happens in your case. Instead of treating n as odd, and setting the value to be 3, n is treated as even and your first out value is calculated to be
median([0 3 5]) = (3+5)/2 = 4
and so on.. EDIT: This only seems to happen in the endpoints, which suggest that the padding with zeros is not properly working in your Octave code.
For your second question, you are almost right, it calculates a 3x3 matrix in each center, but it does not do median(median(MAT)), but median(MAT(:)). There is a difference!
A = [1 2 3
14 5 33
11 7 13];
median(median(A)) = 11
median(A(:)) = 7