I am trying to avoid the for loops and I have been reading through all the old posts there are about it but I am not able to solve my problem. I am new in MATLAB, so apologies for my ignorance.
The thing is that I have a 300x2 cell and in each one I have a 128x128x256 matrix. Each one is an image with 128x128 pixels and 256 channels per pixel. In the first column of the 300x2 cell I have my parallel intensity values and in the second one my perpendicular intensity values.
What I want to do is to take every pixel of every image (for each component) and sum the intensity values channel by channel.
The code I have is the following:
Image_par_channels=zeros(128,128,256);
Image_per_channels=zeros(128,128,256);
Image_tot_channels=zeros(128,128,256);
for a=1:128
for b=1:128
for j=1:256
for i=1:numfiles
Image_par_channels(a,b,j)=Image_par_channels(a,b,j)+Image_cell_par_per{i,1}(a,b,j);
Image_per_channels(a,b,j)=Image_per_channels(a,b,j)+Image_cell_par_per{i,2}(a,b,j);
end
Image_tot_channels(a,b,j)=Image_par_channels(a,b,j)+2*G*Image_per_channels(a,b,j);
end
end
end
I think I could speed it up introducing (:,:,j) instead of specifying a and b. But still a for loop. I am trying to use cellfun without any success due to my lack of expertise. Could you please give me a hand?
I would really appreciate it.
Many thanks and have a nice day!
Y
I believe you could do something like
Image_par_channels=zeros(128,128,256);
Image_per_channels=zeros(128,128,256);
Image_tot_channels=zeros(128,128,256);
for i=1:numfiles
Image_par_channels = Image_par_channels + Image_cell_par_per{i,1};
Image_per_channels = Image_per_channels + Image_cell_par_per{i,2};
end
Image_tot_channels = Image_par_channels + 2*G*Image_per_channels;
I haven't work with matlab in a long time, but I seem to recall you can do something like this. g is a constant.
EDIT:
Removed the +=. Incremental assignment is not an operator available in matlab. You should also note that Image_tot_channels can be build directly in the loop, if you don't need the other two variables later.
Related
I have a number of objects that each have three matrices of distance between own points (x1-x1,x1-x2,x1-x3...;x2-x1,x2-x2,x3-x2...) also with y and z.
I want to find as many nearby points as possible, assuming rotation is not an issue.
I tried something. Since Matlab is supposed to work easy with matrices I am sure something is cumbersome but I don't know how to fix it.
For each object and it's mirror, and for each translation on each axes there is an xyz scenario:
(x1,y1,z1;x2,y2,z2;...)
So I am translating and mirroring one object a million times.
for m=1:object1
for n=1:object2
for i=1:NumRows
for j=1:NumRows2
d_x(m,n,i,j)=obj(m).xyz(i,1)-obj(n).xyz(j,1);
d_y(m,n,i,j)=obj(m).xyz(i,2)-obj(n).xyz(j,2);
d_z(m,n,i,j)=obj(m).xyz(i,3)-obj(n).xyz(j,3);
d_r(m,n,i,j)=sqrt(d_x(m,n,i,j)*d_x(m,n,i,j)+d_y(m,n,i,j)*d_y(m,n,i,j)+d_z(m,n,i,j)*d_z(m,n,i,j));
if d_r(m,n,i,j)>=0 & d_r(m,n,i,j)<1.2
d_r(m,n,i,j)=1.2-d_r(m,n,i,j);
else
d_r(m,n,i,j)=0;
end
sy(m,n)=sy(m,n)+d_r(m,n,i,j);
end
end
end
end
Whenever you start putting indices in variable names, think twice if they maybe should be a single variable. Here we have d_x d_y d_z. My recommendation would be to replace them by a single variable:
d_xyz(m,n,i,j,:)=obj(m).xyz(i,:)-obj(n).xyz(j,:);
And now to your next line, what you are calculating there is actually called a 2-norm. If you know the name, it's simple to shorten:
d_r(m,n,i,j) = norm(squeeze(d_xyz(m,n,i,j,:)),2);
I would like to optimize this piece of Matlab code but so far I have failed. I have tried different combinations of repmat and sums and cumsums, but all my attempts seem to not give the correct result. I would appreciate some expert guidance on this tough problem.
S=1000; T=10;
X=rand(T,S),
X=sort(X,1,'ascend');
Result=zeros(S,1);
for c=1:T-1
for cc=c+1:T
d=(X(cc,:)-X(c,:))-(cc-c)/T;
Result=Result+abs(d');
end
end
Basically I create 1000 vectors of 10 random numbers, and for each vector I calculate for each pair of values (say the mth and the nth) the difference between them, minus the difference (n-m). I sum over of possible pairs and I return the result for every vector.
I hope this explanation is clear,
Thanks a lot in advance.
It is at least easy to vectorize your inner loop:
Result=zeros(S,1);
for c=1:T-1
d=(X(c+1:T,:)-X(c,:))-((c+1:T)'-c)./T;
Result=Result+sum(abs(d),1)';
end
Here, I'm using the new automatic singleton expansion. If you have an older version of MATLAB you'll need to use bsxfun for two of the subtraction operations. For example, X(c+1:T,:)-X(c,:) is the same as bsxfun(#minus,X(c+1:T,:),X(c,:)).
What is happening in the bit of code is that instead of looping cc=c+1:T, we take all of those indices at once. So I simply replaced cc for c+1:T. d is then a matrix with multiple rows (9 in the first iteration, and one fewer in each subsequent iteration).
Surprisingly, this is slower than the double loop, and similar in speed to Jodag's answer.
Next, we can try to improve indexing. Note that the code above extracts data row-wise from the matrix. MATLAB stores data column-wise. So it's more efficient to extract a column than a row from a matrix. Let's transpose X:
X=X';
Result=zeros(S,1);
for c=1:T-1
d=(X(:,c+1:T)-X(:,c))-((c+1:T)-c)./T;
Result=Result+sum(abs(d),2);
end
This is more than twice as fast as the code that indexes row-wise.
But of course the same trick can be applied to the code in the question, speeding it up by about 50%:
X=X';
Result=zeros(S,1);
for c=1:T-1
for cc=c+1:T
d=(X(:,cc)-X(:,c))-(cc-c)/T;
Result=Result+abs(d);
end
end
My takeaway message from this exercise is that MATLAB's JIT compiler has improved things a lot. Back in the day any sort of loop would halt code to a grind. Today it's not necessarily the worst approach, especially if all you do is use built-in functions.
The nchoosek(v,k) function generates all combinations of the elements in v taken k at a time. We can use this to generate all possible pairs of indicies then use this to vectorize the loops. It appears that in this case the vectorization doesn't actually improve performance (at least on my machine with 2017a). Maybe someone will come up with a more efficient approach.
idx = nchoosek(1:T,2);
d = bsxfun(#minus,(X(idx(:,2),:) - X(idx(:,1),:)), (idx(:,2)-idx(:,1))/T);
Result = sum(abs(d),1)';
Update: here are the results for the running times for the different proposals (10^5 trials):
So it looks like the transformation of the matrix is the most efficient intervention, and my original double-loop implementation is, amazingly, the best compared to the vectorized versions. However, in my hands (2017a) the improvement is only 16.6% compared to the original using the mean (18.2% using the median).
Maybe there is still room for improvement?
I have a csv file which contains data like below:[1st row is header]
Element,State,Time
Water,Solid,1
Water,Solid,2
Water,Solid,3
Water,Solid,4
Water,Solid,5
Water,Solid,2
Water,Solid,3
Water,Solid,4
Water,Solid,5
Water,Solid,6
Water,Solid,7
Water,Solid,8
Water,Solid,7
Water,Solid,6
Water,Solid,5
Water,Solid,4
Water,Solid,3
The similar pattern is repeated for State: "Solid" replaced with Liquid and Gas.
And moreover the Element "Water" can be replaced by some other element too.
Time as Integer's are in seconds (to simplify) but can be any real number.
Additionally there might by some comment line starting with # in between the file.
Problem Statement: I want to eliminate the first dip in Time values and smooth out using some quadratic or cubic or polynomial interpolation [please notice the first change from 5->2 --->8. I want to replace these numbers to intermediate values giving a gradual/smooth increase from 5--->8].
And I wish this to be done for all the combinations of Elements and States.
Is this possible through some sort of coding in Matlab etc ?
Any Pointers will be helpful !!
Thanks in advance :)
You can use the interp1 function for 1D-interpolation. The syntax is
yi = interp1(x,y,xi,method)
where x are your original coordinates, y are your original values, xi are the coordinates at which you want the values to be interpolated at and yi are the interpolated values. method can be 'spline' (cubic spline interpolation), 'pchip' (piece-wise Hermite), 'cubic' (cubic polynomial) and others (see the documentation for details).
You have alot of options here, it really depends on the nature of your data, but I would start of with a simple moving average (MA) filter (which replaces each data point with the average of the neighboring data points), and see were that takes me. It's easy to implement, and fine-tuning the MA-span a couple of times on some sample data is usually enough.
http://www.mathworks.se/help/curvefit/smoothing-data.html
I would not try to fit a polynomial to the entire data set unless I really needed to compress it, (but to do so you can use the polyfit function).
Keeping simple, take a matrix of ones i.e.
U_iso = ones(72,37)
and some parameters
ThDeg = 0:5:180;
dtheta = 5*pi/180;
dphi = 5*pi/180;
Th = ThDeg*pi/180;
Now the code is
omega_iso = 0;
for i = 1:72
for j=1:37
omega_iso = omega_iso + U_iso(i,j)*sin(Th(j))*dphi*dtheta;
end
end
and
D_iso = (4 * pi)/omega_iso
This code is fine. It take a matrix with dimension 72*37. The loop is an approximation of the integral which is further divided by 4pi to get ONE value of directivity of antenna.
Now this code gives one value which will be around 1.002.
My problem is I dont need 1 value. I need a 72*37 matrix as my answer where the above integral approximation is implemented on each cell of the 72 * 37 matrix. and thus the Directviity 'D' also results in a matrix of same size with each cell giving the same value.
So all we have to do is instead of getting 1 value, we need value at each cell.
Can anyone please help.
You talk about creating a result that is a function essentially of the elements of U. However, in no place is that code dependent on the elements of U. Look carefully at what you have written. While you do use the variable U_iso, never is any element of U employed anywhere in that code as you have written it.
So while you talk about defining this for a matrix U, that definition is meaningless. So far, it appears that a call to repmat at the very end would create a matrix of the desired size, and clearly that is not what you are looking for.
Perhaps you tried to make the problem simple for ease of explanation. But what you did was to over-simplify, not leaving us with something that even made any sense. Please explain your problem more clearly and show code that is consistent with your explanation, for a better answer than I can provide so far.
(Note: One option MIGHT be to use arrayfun. Or the answer to this question might be more trivial, using simple vectorized operations. I cannot know at this point.)
EDIT:
Your question is still unanswerable. This loop creates a single scalar result, essentially summing over the entire array. You don't say what you mean for the integral to be computed for each element of U_iso, since you are already summing over the entire array. Please learn to be accurate in your questions, otherwise we are just guessing as to what you mean.
My best guess at the moment is that you might wish to compute a cumulative integral, in two dimensions. cumtrapz can help you there, IF that is your goal. But I'm not sure it is your goal, since your explanation is so incomplete.
You say that you wish to get the same value in each cell of the result. If that is what you wish, then a call to repmat at the end will do what you wish.
I have been happily using MATLAB to solve some project Euler problems. Yesterday, I wrote some code to solve one of these problems (14). When I write code containing long loops I always test the code by running it with short loops. If it runs fine and it does what it's supposed to do I assume this will also be the case when the length of the loop is longer.
This assumption turned out to be wrong. While executing the code below, MATLAB ran out of memory somewhere around the 75000th iteration.
c=1;
e=1000000;
for s=c:e
n=s;
t=1;
while n>1
a(s,t)=n;
if mod(n,2) == 0
n=n/2;
else
n=3*n+1;
end
a(s,t+1)=n;
t=t+1;
end
end
What can I do to prevent this from happening? Do I need to clear variables or free up memory somewhere in the process? Will saving the resulting matrix a to the hard drive help?
Here is the solution, staying as close as possible to your code (which is very close, the main difference is that you only need a 1D matrix):
c=1;
e=1000000;
a=zeros(e,1);
for s=c:e
n=s;
t=1;
while n>1
if mod(n,2) == 0
n=n/2;
else
n=3*n+1;
end
t=t+1;
end
a(s)=t;
end
[f g]=max(a);
This takes a few seconds (note the preallocation), and the result g unlocks the Euler 14 door.
Simply put, there's not enough memory to hold the matrix a.
Why are you making a two-dimensional matrix here anyway? You're storing information that you can compute just as fast as looking it up.
There's a much better thing to memoize here.
EDIT: Looking again, you're not even using the stuff you put in that matrix! Why are you bothering to create it?
The code appears to be storing every sequence in a different row of a matrix. The number of columns of that matrix will be equal to the length of the longest sequence currently found. This means that a sequence of two numbers will be padded with a bunch of right hand zeros.
I am sure you can see how this is incredibly inefficient. That may be the point of the exercise, or it will be for you in this implementation.
Better is to keep a variable like "Seed of longest solution found" which would store the seed for the longest solution. I would also keep a "length of longest solution found" keep the length. As you try every new seed, if it wins the title of longest, then update those variables.
This will keep only what you need in memory.
Short Answer:Use a 2d sparse matrix instead.
Long Answer: http://www.mathworks.com/access/helpdesk/help/techdoc/ref/sparse.html