given a number of signals in a matrix, say m [T x N]
and a grouping variable g [ 1 x N ] with a number of labels L < N
is there an efficient, in-place way to compute mean time signals for every label?
A for loop
ml = zeros (T,L)
for i = 1:T
for j = 1:L
ml(i,j) = mean ( m(i,find(g==j)) )
end
end
is a straightforward way to do it, but could there be a faster/cleaner way, possibly using vectorised code? Just to (a) get rid of for-loops and (b) assign m = labelled_means(m, ...) in one go. I read about statarray, but that looks much less efficient even than this for loop.
Well you could get rid of your outer loop really easily:
for j = 1:L
ml(:,j) = mean(m(:,find(g==j)), 2)
end
You can use arrayfun. It will be cleaner, don't know if faster:
result = arrayfun(#(n) mean(A(:,find(g==n)),2),1:2,'UniformOutput',false);
result = reshape([result{:}],[],L)
For example, with data
A = [1 2 3 4;
5 6 7 8;
9 10 11 12];
L = 2;
g = [1 1 2 2];
this gives
result =
1.5000 3.5000
5.5000 7.5000
9.5000 11.5000
You can use accumarray
[X,Y] = ndgrid(1:size(A,1), g);
accumarray([X(:) Y(:)], A(:), [], #mean);
Related
Let A be of size [n,m], i.e. it has n rows and m columns. Given I of size [n,1] with max(I)<=m, what is the fastest way to return B of size [n,1], such that B(i)=A(i,I(i))?
Example:
A =
8 1 6
3 5 7
4 9 2
and
I =
1
2
2
I want B to look like
B =
8
5
9
There obviously exist several ways to implement this, but in my case n is in the order of 1e6 and m in the order of 1e2, which is why I'm interested in the fastest implementation. I would like to avoid ind2sub or sub2ind since they both appear to be too slow as well. Any idea is greatly appreciated! Thanks!
You can replicate the behavior of sub2ind yourself. This gives me a speedup in my test:
clear
%% small example
A = rand(4,6)
I = [3 2 2 1]
inds = (I-1)*size(A,1) + (1:length(I));
B = A(inds)
%% timing
n = 1e4;
m = 1e2;
A = rand(n, m);
I = ceil(rand(1,n) * m);
% sub2ind
F = #() A(sub2ind(size(A), 1:size(A,1), I));
timeit(F)
% manual
F = #() A((I-1)*size(A,1) + (1:length(I)));
timeit(F)
You can also use something like this:
A(meshgrid(1:size(A,2),1:size(A,1)) == repmat(I,1,size(A,2)))
which will give you the same result, with no loop and no sub2ind.
I have a task to find the maximum product of elements in a matrix. The input arguments are the matrix and the scalar n that tells me how many elements to multiply. If n=2, then I should multiply two by two. The maximum product can lie either in rows, or in columns on in diagonal. For example in this case A's elements were multiplied 2 by 2 along rows (B) and columns (C).
A =
8 1 6
3 5 7
4 9 2
B =
8.0000 6.0000
15.0000 35.0000
36.0000 18.0000
C =
24.0000 5.0000 42.0000
12.0000 45.0000 14.0000
I do it using the loop
c=[]
for ii = 1:(length(A)-n+1)
p = prod(A(:,ii:ii+n-1));
c=[c p];
end
or
for i=1:size(A,2)
B(i,:)=real(exp(conv(log(A(i,:)),ones(1,n),'valid')));
C(:,i)=real(exp(conv(log(A(:,i)),ones(1,n),'valid')));
end
In both cases I receive the product but when it comes to getting the maximum among that products, (in my case that is the product of A(2,2)*A(3,2)=45) I cannot find the indices of original matrix elements that formed that product, i.e. (2,2) and (3,2).
Any help will be appreciated.
Given inputs A and n, and assuming you don't care which output you give in case of a tie,
%Setup
[r, c] = size(A)
if r <= n
maxProd = prod(A(1:n,1),1);
elseif c <= n
maxProd = prod(A(1,1:n),2);
else
return
end
indOut = zeros(n, 2);
dirOut = 0;
%Check verticals
for ii = 1:(r-n+1)
for jj = 1:c
testProd = prod(A(ii:ii+n-1,jj),1);
if testProd > maxProd
maxProd = testProd;
indOut(1,:) = [ii jj];
dirOut = 1;
end
end
end
%Check horizontals
for ii = 1:r
for jj = 1:(c-n+1)
testProd = prod(A(ii,jj:jj+n-1),2);
if testProd > maxProd
maxProd = testProd;
indOut(1,:) = [ii jj];
dirOut = 2;
end
end
end
%Set output
for ii = 2:n
if dirOut == 1;
indOut(1,n) = indOut(1,n-1) + 1;
else
indOut(2,n) = indOut(2,n-1) + 1;
end
end
If you do care about ties, you'll have to add additional outputs to the side of the indOut matrix
The information can be worked out from knowing where, in B or C, the maximum value occurs. This is returned by the max function as a second argument, albeit in index form. I've tried out your code for producing B and C, and it doesn't seem to produce what you've got, HOWEVER, assuming that you've got some way of producing B and C as in the first part of your question, the following code will return (r1,c1) and (r2,c2) the reference to the two multiplicands (assuming that there is only one maximum - you really do need to check for this):
[maxB,IB]=max(B(:));
[maxC,IC]=max(C(:));
if (maxB>maxC)
[r1,c1]=ind2sub(size(B),IB);
r2=r1; c2=c1+n-1;
else
[r1,c1]=ind2sub(size(C),IC);
r2=r1+n-1; c2=c1;
end
I'm not quite sure whether n is a distance, in which case:
B=A(:,1:end-n+1).*A(:,n:end);
C=A(1:end-n+1,:).*A(n:end,:);
or whether you mean to take the product of all terms within n of the starting point. Either way, the maximum can be found using the above code.
Here is what I want, a 3-D matrix:
K = 2:2.5:10;
den = zeros(1,4,4);
for i = 1:1:4
den(:,:,i) = [1, 5, K(i)-6, K(i)];
end
Or, a cell array is also acceptable:
K = 2:2.5:10;
for i = 1:1:4
den{i} = [1, 5, K(i)-6, K(i)];
end
But I want to know if there is a more efficient way of doing this using vectorized code like:
K = 2:2.5:10;
den = [1, 5, K-6, K];
I know the last code will not get what I wanted. But, like I can use:
v = [1 2 3];
v2 = v.^2;
instead of:
v = [1 2 3];
for i = 1:length(v)
v(i) = v(i)^2;
end
to get the matrix I want. Is there a similar way of doing this so that I can get the 3-D matrix or cell array I mentioned at the beginning more efficiently?
You need to "broadcast" the scalar values in columns so they are of the same length as your K vector. MATLAB does not do this broadcasting automatically, so you need to repeat the scalars and create vectors of the appropriate size. You can use repmat() for this.
K = 2:2.5:10;
%% // transpose K to a column vector:
K = transpose(K);
%% // helper function that calls repmat:
f = #(v) repmat(v, length(K), 1);
%% // your matrix:
den = [f(1) f(5) K-6 K];
This should be more optimized for speed but requires a bit more intermediary memory than the loop does.
Just use reshape with a 1*3 size:
den = reshape([ones(1,length(K));ones(1,length(K))*5; K-6; K],[1 4 length(K)]);
I think the used extra memory by reshape should be low and constant (dependent only on the length of the vector of new sizes).
You can use the classic line equation y=a*x+b, extended to the matrix form:
k = 2:2.5:10 ;
fa = [0 0 1 1].' ; %' // "a" coefficients
fb = [1 5 -6 0].' ; %' // "b" coefficients
d(1,:,:) = fa*k + fb*ones(1,4) ;
The above is better for clarity, but if you're not bothered you can also pack everything in one line:
d(1,:,:) = [0 0 1 1].' * (2:2.5:10) + [1 5 -6 0].' * ones(1,4) ;
If you need to re-use the principle for many different values of k, then you can use an anonymous function to help:
fden = #(k) [0 0 1 1].' * k + [1 5 -6 0].' * ones(1,4) ; %// define anonymous function
k = 2:2.5:10 ;
d(1,:,:) = fden(k) ; %// use it for any value of "k"
I want to write this matrix in matlab,
s=[0 ..... 0
B 0 .... 0
AB B .... 0
. . .
. . .
. . . 0 ....
A^(n-1)*B ... AB B ]
I have tried this below code but giving error,
N = 50;
A=[2 3;4 1];
B=[3 ;2];
[nx,ny] = size(A);
s(nx,ny,N) = 0;
for n=1:1:N
s(:,:,n)=A.^n;
end
s_x=cat(3, eye(size(A)) ,s);
for ii=1:1:N-1
su(:,:,ii)=(A.^ii).*B ;
end
z= zeros(1,60,1);
su1 = [z;su] ;
s_u=repmat(su1,N);
seems like the concatenation of matrix is not being done.
I am a beginner so having serious troubles,please help.
Use cell arrays and the answer to your previous question
A = [2 3; 4 1];
B = [3 ;2 ];
N = 60;
[cs{1:(N+1),1:N}] = deal( zeros(size(B)) ); %// allocate space, setting top triangle to zero
%// work on diagonals
x = B;
for ii=2:(N+1)
[cs{ii:(N+2):((N+1)*(N+2-ii))}] = deal(x); %//deal to diagonal
x = A*x;
end
s = cell2mat(cs); %// convert cells to a single matrix
For more information you can read about deal and cell2mat.
Important note about the difference between matrix operations and element-wise operations
In your question (and in your previous one) you confuse between matrix power: A^2 and element-wise operation A.^2:
matrix power A^2 = [16 9;12 13] is the matrix product of A*A
element-wise power A.^2 takes each element separately and computes its square: A.^2 = [4 9; 16 1]
In yor question you ask about matrix product A*b, but the code you write is A.*b which is an element-by-element product. This gives you an error since the size of A and the size of b are not the same.
I often find that Matlab gives itself to a coding approach of "write what it says in the equation". That also leads to code that is easy to read...
A = [2 3; 4 1];
B = [3; 2];
Q = 4;
%// don't need to...
s = [];
%// ...but better to pre-allocate s for performance
s = zeros((Q+1)*2, Q);
X = B;
for m = 2:Q+1
for n = m:Q+1
s(n*2+(-1:0), n-m+1) = X;
end
X = A * X;
end
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How to Add a row vector to a column vector like matrix multiplication
I have a nx1 vector and a 1xn vector. I want to add them in a special manner like matrix multiplication in an efficient manner (vectorized):
Example:
A=[1 2 3]'
B=[4 5 6]
A \odd_add B =
[1+4 1+5 1+6
2+4 2+5 2+6
3+4 3+5 3+6
]
I have used bsxfun in MATLAB, but I think it is slow. Please help me...
As mentioned by #b3. this would be an appropriate place to use repmat. However in general, and especially if you are dealing with very large matrices, bsxfun normally makes a better substitute. In this case:
>> bsxfun(#plus, [1,2,3]', [4,5,6])
returns the same result, using about a third the memory in the large-matrix limit.
bsxfun basically applies the function in the first argument to every combination of items in the second and third arguments, placing the results in a matrix according to the shape of the input vectors.
I present a comparison of the different methods mentioned here. I am using the TIMEIT function to get robust estimates (takes care of warming up the code, average timing on multiple runs, ..):
function testBSXFUN(N)
%# data
if nargin < 1
N = 500; %# N = 10, 100, 1000, 10000
end
A = (1:N)';
B = (1:N);
%# functions
f1 = #() funcRepmat(A,B);
f2 = #() funcTonyTrick(A,B);
f3 = #() funcBsxfun(A,B);
%# timeit
t(1) = timeit( f1 );
t(2) = timeit( f2 );
t(3) = timeit( f3 );
%# time results
fprintf('N = %d\n', N);
fprintf('REPMAT: %f, TONY_TRICK: %f, BSXFUN: %f\n', t);
%# validation
v{1} = f1();
v{2} = f2();
v{3} = f3();
assert( isequal(v{:}) )
end
where
function C = funcRepmat(A,B)
N = numel(A);
C = repmat(A,1,N) + repmat(B,N,1);
end
function C = funcTonyTrick(A,B)
N = numel(A);
C = A(:,ones(N,1)) + B(ones(N,1),:);
end
function C = funcBsxfun(A,B)
C = bsxfun(#plus, A, B);
end
The timings:
>> for N=[10 100 1000 5000], testBSXFUN(N); end
N = 10
REPMAT: 0.000065, TONY_TRICK: 0.000013, BSXFUN: 0.000031
N = 100
REPMAT: 0.000120, TONY_TRICK: 0.000065, BSXFUN: 0.000085
N = 1000
REPMAT: 0.032988, TONY_TRICK: 0.032947, BSXFUN: 0.010185
N = 5000
REPMAT: 0.810218, TONY_TRICK: 0.824297, BSXFUN: 0.258774
BSXFUN is a clear winner.
In matlab vectorization, there is no substitute for Tony's Trick in terms of speed in comparison to repmat or any other built in Matlab function for that matter. I am sure that the following code must be fastest for your purpose.
>> A = [1 2 3]';
>> B = [4 5 6];
>> AB_sum = A(:,ones(3,1)) + B(ones(3,1),:);
The speed differential will be much more apparent (at LEAST an order of magnitude) for larger size of A and B. See this test I conducted some time ago to ascertain the superiority of Tony's Trick over repmatin terms of time consumption.
REPMAT is your friend:
>> A = [1 2 3]';
>> B = [4 5 6];
>> AplusB = repmat(A, 1, 3) + repmat(B, 3, 1)
AplusB =
5 6 7
6 7 8
7 8 9