I have a struct like this mesh.m_1_0.Deformation_Entformung;
the second field is a struct from m_1_0 till m_3_5 in 6 steps;
the Deformation_Entformung is a matrix with 6 columns and 325562 rows, whereby the first 3 columns contain coordinates (x,y,z).
Now I'm interested in the coordinates that are the closest to (33.5 -88.7801,-0.4480).
This is my code:
SNames = fieldnames(mesh); % SName = m_1_0,m_1_5...m_3_5
for loopIndex = 1:numel(SNames)
stuff = mesh.(SNames{loopIndex}).Deformation_Entformung;
mesh.(SNames{loopIndex}).('Deformation_Entformung_Koordi')=...
stuff(min(stuff(:,1)-33.5) & min(stuff(:,2)--88.7801) & ...
min(stuff(:,3)-0.4480), :);
end
The code runs, but the problem is that the answer is always the first row of the matrix Deformation_Entformung.
I would be glad, if someone could give me a hint.
Well, first of all you mix up indices with values.
min(stuff) returns the minimal value of stuff. So when you write stuff(min(stuff)) that's certainly not doing what you want it to do.
Secondly, if min(stuff(:,1)-33.5) would actually return an index (which it doesn't), then the index would be the same whether you searched for min(stuff(:,1)+100) or min(stuff(:,1)-500000). So the program would still not be doing what you want it to do.
Additionally, the way you are trying to search for the closest point does not even work from a mathematical point of view (even if your programming had no errors). The closest point is not necessarily the closest in each single coordinate. For example, [1 1 1] is certainly closer to [0 0 0] than [20 0 0], [0 20 0] and [0 0 20]. But it is not the closest one in each single coordinate. In fact it is not the closest one in any coordinate.
There might be even more issues with your code, but for starters you should work on how to determine distances. After you master that you should try to pick points with minimal distance. And only after you master both of these should you try to integrate everything into the rest of your stuff. No point in trying to do everything at once.
Related
I am currently working in matlab to design a way to reconstruct 3D data. For this I have two pictures with black points. The difference in the amount of points per frame is key for the reconstruction, but MATLAB gives an error when matrixes are not equal. This is happening becaus the code is not doing what I want it to do, so can anyone hel me with the following?
I have two columns of Xdata: XLI and XRI
What matlab does when I do XLI-XRI is substracting the pairs i.e XLI(1)-XRI(1) etc, but I want to substract each value of XRI of every value of XLI. i.e
XLI(1)-XRI(1,2,3,4 etc)
XLI(2)-XRI(1 2 3 4 etc)
and so on
Can anyone help?
I think you are looking for a way to deduct all combinations from eachother. Here is an example of how you can do that with bsxfun:
xLI = [1 2 3]
xRI = [1 2]
bsxfun(#minus,xLI ,xRI')
I cannot comment on Dennis's post (not enough points on this website) : his solution should work, but depending on your version of Matlab you might get a "Error using ==> bsxfun" and need to transpose either xLI or xRI for that to work :
bsxfun(#minus,xLI' ,xRI)
Best,
Tepp
Introduction to problem:
I'm modelling a system where i have a matrix X=([0,0,0];[0,1,0],...) where each row represent a point in 3D-space. I then choose a random row, r, and take all following rows and rotate around the point represented by r, and make a new matrix from these rows, X_rot. I now want to check whether any of the rows from X_rot is equal two any of the rows of X (i.e. two vertices on top of each other), and if that is the case refuse the rotation and try again.
Actual question:
Until now i have used the following code:
X_sim=[X;X_rot];
if numel(unique(X_sim,'rows'))==numel(X_sim);
X(r+1:N+1,:,:)=X_rot;
end
Which works, but it takes up over 50% of my running time and i were considering if anybody in here knew a more efficient way to do it, since i don't need all the information that i get from unique.
P.S. if it matters then i typically have between 100 and 1000 rows in X.
Best regards,
Morten
Additional:
My x-matrix contains N+1 rows and i have 12 different rotational operations that i can apply to the sub-matrix x_rot:
step=ceil(rand()*N);
r=ceil(rand()*12);
x_rot=x(step+1:N+1,:);
x_rot=bsxfun(#minus,x_rot,x(step,:));
x_rot=x_rot*Rot(:,:,:,r);
x_rot=bsxfun(#plus,x_rot,x(step,:));
Two possible approaches (I don't know if they are faster than using unique):
Use pdist2:
d = pdist2(X, X_rot, 'hamming'); %// 0 if rows are equal, 1 if different.
%// Any distance function will do, so try those available and choose fastest
result = any(d(:)==0);
Use bsxfun:
d = squeeze(any(bsxfun(#ne, X, permute(X_rot, [3 2 1])), 2));
result = any(d(:)==0);
result is 1 if there is a row of X equal to some row of X_rot, and 0 otherwise.
How about ismember(X_rot, X, 'rows')?
I'm trying to create a random "path" on a coordinate system on Matlab. I am doing this by creating a for loop where for each iteration it fills in a new value on a matrix that has initial values of zeros.
For example, I have 5 points so I have an initial matrix a=[0 0 0 0 0; 0 0 0 0 0] (row1 = x values, row2 = y values).
The path can move right/left or up/down (no diagonals). In my for loop, I call randi(4) and say something like "if randi(4)=1, then move 1 point to the left (x-1). if randi(4)=2, then move to the right (x+1), etc."
The problem is that you cannot visit a specific point more than once. For example, the path can start at (0,0), then go to (0,1), then (1,1), then (1,0), and then it CANNOT go back to (0,0).. in my current code I don't have this restriction so I was hoping I could get some suggestions..
Since in this example the matrix would look something like a=[0 0 1 1 0; 0 1 1 0 0].
I was thinking of maybe subtracting each new coordinate (here (0,0)) from each column on the matrix a and if any of the columns give me values of zero for both rows (since it's the same coordinate subtracted from itself), then go back one step and let randi(4) run again.. but
How could I tell it to "go back one step" (or two or three)?
How do you compare one column against each column of the already established matrix?
This was just an idea.. are there any functions in Matlab that would let me do this? or maybe compare if two columns are the same within a matrix?
To your questions.
to go back - I suppose this means just throwing away the rightmost columns in your matrix.
to find if it is present you could use ismember
unfortunately it only takes rows so you will need to transpose. Snippet:
a = [1:10; repmat(1:2,1,5)]'
test = ismember(a,[3,2],'rows')
any(test) % not found
test = ismember(a,[3,1],'rows')
any(test) % found
Of course your idea would also work.
I can answer this:
How do you compare one column against each column of the already
established matrix?
Use two different matrices. Compare them using the setdiff() function: http://www.mathworks.com/help/matlab/ref/setdiff.html
I have a matrix of zeros and ones. Each cell with a one value represents a non empty cell.
The cells with the ones is kept in a vector with the coordinates.
I'm iterating through the list and move each element randomly to one of its free neighboring cells.
Is there a way to do this with vector operations?
Thanks
Here's an attempt, not the most elegant or efficient one, but still it should work.
first I assume you have the x,y coordinates of your non empty cells, something like
c=[x y];
The relative positions of the 8 nearest neighbors (n.n.) in a 2D array are given by:
nn=[1 1;1 -1;-1 1;0 -1;-1 0;0 1;1 0;-1 -1];
Let's take all possible permutation of each x,y coordinate in c around its n.n.
permc=bsxfun(#plus,repmat(c,[1 8]),nn(:)');
Now set a vector of random n.n. out of the 8 options for each x,y coordinates:
ri=randi(8,numel(x), 1);
and use it to select random new coordinates
new_c= [ permc(sub2ind(size(permc), (1:numel(x))', 2*ri-1 )) , ...
permc(sub2ind(size(permc), (1:numel(x))', 2*ri))];
Issues:
The code doesn't check if there is a n.n. that is not free, not hard to solve having info of c and permc at hand.
The code doesn't care that a point is on the edge of the array, so it can select a coordinate that is either 0 or one larger than the size of the array. Again, permc has the info to tackle this case and it can be treated separately in several ways.
That's the jist.
Probably, but I think it would be quite tricky to write. You'd have to be careful to prevent two "dots" from moving into the same empty location at the same time.
Looping over the vector and dealing with one dot at a time seems far easier. What's your reason for wanting to do this with vector operations? If you're looking to make your code faster, post your code and ask for suggestions.
I'm a newbie to Matlab and just stumped how to do a simple task that can be easily performed in excel. I'm simply trying to get the percent change between cells in a matrix. I would like to create a for loop for this task. The data is setup in the following format:
DAY1 DAY2 DAY3...DAY 100
SUBJECT RESULTS
I could only perform getting the percent change between two data points. How would I conduct it if across multiple days and multiple subjects? And please provide explanation
Thanks a bunch
FOR EXAMPLE, FOR DAY 1 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5), DAY 2 SUBJECT1(RESULT=2), SUBJECT2(RESULT=8), SUBJECT3(RESULT=10), DAY 3 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5).
I WANT THE PERCENT CHANGE SO OUTPUT WILL BE DAY 2 SUBJECT1(RESULT=100%), SUBJECT2(RESULT=100%), SUBJECT3(RESULT=100%). DAY3 SUBJECT1(RESULT=50%), SUBJECT2(RESULT=50%), SUBJECT3(RESULT=50%)
updated:
Hi thanks for responding guys. sorry for the confusion. zebediah49 is pretty close to what I'm looking for. My data is for example a 10 x 10 double. I merely wanted to get the percentage change from column to column. For example, if I want the percentage change from rows 1 through 10 on all columns (from columns 2:10). I would like the code to function for any matrix dimension (e.g., 1000 x 1000 double) zebediah49 could you explain the code you posted? thanks
updated2:
zebediah49,
(data(1:end,100)- data(1:end,99))./data(1:end,99)
output=[data(:,2:end)-data(:,1:end-1)]./data(:,1:end-1)*100;
Observing the code above, How would I go about modifying it so that column 100 is used as the index against all of the other columns(1-99)? If I change the code to the following:
(data(1:end,100)- data(1:end,:))./data(1:end,:)
matlab is unable because of exceeding matrix dimensions. How would I go about implementing that?
UPDATE 3
zebediah49,
Worked perfectly!!! Originally I created a new variable for the index and repmat the index to match the matrices which was not a good idea. It took forever to replicate when dealing with large numbers.
Thanks for you contribution once again.
Thanks Chris for your contribution too!!! I was looking more on how to address and manipulate arrays within a matrix.
It's matlab; you don't actually want a loop.
output=input(2:end,:)./input(1:end-1,:)*100;
will probably do roughly what you want. Since you didn't give anything about your matlab structure, you may have to change index order, etc. in order to make it work.
If it's not obvious, that line defines output as a matrix consisting of the input matrix, divided by the input matrix shifted right by one element. The ./ operator is important, because it means that you will divide each element by its corresponding one, as opposed to doing matrix division.
EDIT: further explanation was requested:
I assumed you wanted % change of the form 1->1->2->3->1 to be 100%, 200%, 150%, 33%.
The other form can be obtained by subtracting 100%.
input(2:end,:) will grab a sub-matrix, where the first row is cut off. (I put the time along the first dimension... if you want it the other way it would be input(:,2:end).
Matlab is 1-indexed, and lets you use the special value end to refer to the las element.
Thus, end-1 is the second-last.
The point here is that element (i) of this matrix is element (i+1) of the original.
input(1:end-1,:), like the above, will also grab a sub-matrix, except that that it's missing the last column.
I then divide element (i) by element (i+1). Because of how I picked out the sub-matrices, they now line up.
As a semi-graphical demonstration, using my above numbers:
input: [1 1 2 3 1]
input(2,end): [1 2 3 1]
input(1,end-1): [1 1 2 3]
When I do the division, it's first/first, second/second, etc.
input(2:end,:)./input(1:end-1,:):
[1 2 3 1 ]
./ [1 1 2 3 ]
---------------------
== [1.0 2.0 1.5 0.3]
The extra index set to (:) means that it will do that procedure across all of the other dimension.
EDIT2: Revised question: How do I exclude a row, and keep it as an index.
You say you tried something to the effect of (data(1:end,100)- data(1:end,:))./data(1:end,:). Matlab will not like this, because the element-by-element operators need them to be the same size. If you wanted it to only work on the 100th column, setting the second index to be 100 instead of : would do that.
I would, instead, suggest setting the first to be the index, and the rest to be data.
Thus, the data is processed by cutting off the first:
output=[data(2:end,2:end)-data(2:end,1:end-1)]./data(2:end,1:end-1)*100;
OR, (if you neglect the start, matlab assumes 1; neglect the end and it assumes end, making (:) shorthand for (1:end).
output=[data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100;
However, you will probably still want the indices back, in which case you will need to append that subarray back:
output=[data(1,1:end-1) data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100];
This is probably not how you should be doing it though-- keep data in one matrix, and time or whatever else in a separate array. That makes it much easier to do stuff like this to data, without having to worry about excluding time. It's especially nice when graphing.
Oh, and one more thing:
(data(:,2:end)-data(:,1:end-1))./data(:,1:end-1)*100;
is identically equivalent to
data(:,2:end)./data(:,1:end-1)*100-100;
Assuming zebediah49 guessed right in the comment above and you want
1 4 5
2 8 10
1 4 5
to turn into
1 1 1
-.5 -.5 -.5
then try this:
data = [1,4,5; 2,8,10; 1,4,5];
changes_absolute = diff(data);
changes_absolute./data(1:end-1,:)
ans =
1.0000 1.0000 1.0000
-0.5000 -0.5000 -0.5000
You don't need the intermediate variable, you can directly write diff(data)./data(1:end,:). I just thought the above might be easier to read. Getting from that result to percentage numbers is left as an exercise to the reader. :-)
Oh, and if you really want 50%, not -50%, just use abs around the final line.