Maybe I should just go with a for loop but I want to see if there is a more efficient/faster way to do it.
I have a matrix of numbers, let's say 10x10. I want to multiply 1,1 by 1,2, then 1,3 times 1,4, etc and then sum those results for row 1. Then move to the next row and do the same thing. The end result would be a vector of 10.
It is possible for this matrix to be 1000x1000 so I want it to be as fast as possible. Thanks!
I would use
v = sum(M(:,1:2:end-1).*M(:,2:2:end),2);
Here M(:,1:2:end-1).*M(:,2:2:end) does multiplication: every element of an odd-numbered column of M is multiplied by its neighbor to the right. (This assumes even number of columns, otherwise the process you described is ill-defined.) Then every row is added up by the sum command.
On my computer, doing this for a 1000 by 1000 matrix takes 0.04 seconds.
Related
given the 10x10 matrix m:(10 10)#100?1
Find the average of every row and column of the matrix
Find the average of the matrix
Take the first element from the first row, the second from the second row etc
find the diagonal elements of a matrix
First I'm going to change the input so you can better see the solution. The 100?1 will give a list of 100 0s which I don't think is what you want. Maybe you wanted either 100 random (0;1) or 100 random values between 0-1. I went for the latter.
q)show m:(10 10)#100?1.
0.4655548 0.8455166 0.7281041 0.7403385 0.5199511 0.199172 0.9548708 0.498..
0.86544 0.3112134 0.3520122 0.4485896 0.6742543 0.2357538 0.7589261 0.318..
0.7053699 0.8153197 0.5051956 0.7546554 0.08613905 0.7824787 0.2080171 0.282..
So, now the questions.
Find the average of every row and column of the matrix.
q)meanRows:avg each m
q)meanCols:avg each flip m
UPDATE From Comment. You can get the average of a matrix column without using each or flip but will return null if any element is null. Another note is that if a column is length 10, 5 of which are null. Then the avg will only consider the average of the 5 non-null values.
So, if you believe there are nulls you may want to get rid of them and then get the mean values:
q)m:^[0;m] //Replace null with 0s if necessary
q)meanCols:avg m //Get avg without flipping or using each
Find the average of the matrix
q)avg avg each m
I think that^ is the quickest way to get the overall mean because it doesn't require razing or flipping.
Take the first element from the first row, the second from the second row etc
q)getVector:{[mtx]mtx'[c;c:til count mtx]}
q)getVector m
0.4655548 0.3112134 0.5051956 0.6333324 0.7258795 0.8671843 0.7556175 0.17954..
Let me know if you have any further questions.
I have .xlsx file n rows and m column. I want to calculate average of 1st 5 rows then next 5 rows likewise ( each 5 interval average) continue for my data.
If A is your matrix, what about:
m=[];
for ii=1:5:20
m(end+1)=mean(mean(A(ii:ii+4,:)));
end
does it work for you?
Here is a "hack" that should be pretty fast if A is big:
m = mean(kron(speye(size(A,1)/5),ones(1,5))*A/5,2);
You can use reshape to get blocks of 5 rows in the first dimension. If A is your matrix
m = squeeze(mean(reshape(A,5,[],size(A,2)),1));
The code works as follows
reshape the matrix to get the blocks of 5 rows in the first dimension
Compute the mean over the blocks of 5 rows.
After the mean, the first dimension is a singleton so it is better to squeeze it so the output is a 2D matrix.
I have a matrix 1x5000 with numbers. Now I am interested in getting values from the matrix in different positions, more precisely in six different places of the matrix. The places should be based on the length, these are the numbers I want to get out:
Number in 1/6 of the matrix length
Number in 2/6 of the matrix length
Number in 3/6 of the matrix length
Number in 4/6 of the matrix length
Number in 5/6 of the matrix length
Number in 6/6 of the matrix length
These values could be printed out in another matrix, so assume the matrix is 1x5000, 3/6 would give the number in the middle of the matrix. I am new in Matlab and therefore the help is much appreciated!
Since your question is unclear I can try to give you an example.
First of all you can use numel function to get matrix's size.
It's easy to get necessary element in Matlab: you can address directly to any element if you know its number (index). So:
x(100) returns 100th element.
Now you got size and know what to do. Last moment - what to do if numel(x) / 6 return non integer?
You can use rounding functions: ceil, floor or round.
index = ceil(numel(x)/6) %if you want NEXT element always
result = x(index)
Next step: there are a lot of ways to divide data. For example now you have just 6 numbers (1/6, 2/6 and so on) but what if there are 1000 of them? You can't do it manually. So you can use for loop, or you can use matrix of indexes or perfect comment Stewie Griffin.
My example:
divider = [6 5 4 3 2 1] % lets take 1/6 1/5 1/4 1/3 1/2 and 1/1
ind = ceil( numel(x)./divider)
res = x(ind)
The colon notation in MATLAB provides an easy way to extract a range of elements from v:
v(3:7) %Extract the third through the seventh elements
You could either manually input range or use a function to convert fractions into suitable ranges
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 have two matrices
r=10,000x2
q=10,000x2
i have to find out those rows of q which are one value or both values(as it is a two column matrix) different then r and allocate them in another matrix, right now i am trying this.i cannot use isequal because i want to know those rows
which are not equal this code gives me the individual elements not the complete rows different
can anyone help please
if r(:,:)~=q(:,:)
IN= find(registeredPts(:,:)~=q(:,:))
end
You can probably do this using ismember. Is this what you want? Here you get the values from q in rows that are different from r.
q=[1,2;3,4;5,6]
r=[1,2;3,5;5,6]
x = q(sum(ismember(q,r),2) < 2,:)
x =
3 4
What this do:
ismember creates an array with 1's in the positions where q == r, and 0 in the remaining positions. sum(.., 2) takes the column sum of each of these rows. If the sum is less than 2, that row is included in the new array.
Update
If the values might differ some due to floating point arithmetic, check out ismemberf from the file exchange. I haven't tested it myself, but it looks good.