average over a certain range of index ?

I have vector A = [1 7 12 20 23 31]; and B = [6 11 19 22 30 35].

now i want to average another matrix C with 1x35 elements over the index range "1:6, 7:11, 12:19, 20:22, 23:30, 30:35".

is it possible to do something like mean(C(A:B))?? However this does not work. Can anyone please help in this regard.?

[Jan]

How large are the inputs? Can the intervals overlap? Is A(2:end) and B(1:end-1)+1 equal in every case?

Answer 1

[randerss simil]

clear
clc
A = [1 7 12 20 23 31];
B = [6 11 19 22 30 35];
C = rand(1,35)

for i = 1:numel(A)	
	kk(i) = mean(C(A(i):B(i)));
end 

[Megha]

randerss simil thank you so much for your reply.

Unfortunately, this is my last option... I know this stuff... but i wanted to do it quickly without using FOR loop...

Answer 2

[Jan]

nC = 1e6;            % Number of elements of C
nA = 1e5;            % Number of intervals
C  = rand(1, nC);    % Test data

% Create indices: ----------------------
  % non-overlapping intervals with gaps
  % v = sort(randperm(nC, nA * 2));
  % A = v(1:2:end);
  % B = v(2:2:end);

  % overlapping:
  v  = sort(randi([1, nC], nA, 2), 2);
  A  = v(:, 1);
  B  = min(v(:, 2), A + 100);           % Maximum width: 101

  % non-overlapping without gaps:
  % B = [sort(randperm(1e6-1, 1e5-1)), 1e6];
  % A = [1, B(1:end-1) + 1];

tic;
for i = 1:numel(A)	
   kk1(i) = mean(C(A(i):B(i)));
end
toc

tic;
kk2 = zeros(1, numel(A));
for i = 1:numel(A)	
   kk2(i) = mean(C(A(i):B(i)));
end
toc

tic;
kk3 = zeros(1, numel(A));
for i = 1:numel(A)	
   kk3(i) = sum(C(A(i):B(i))) / (B(i)-A(i)+1);
end
toc

isequal(kk1, kk2, kk3)  % Yes, the outputs are identical

The timings:

% Elapsed time is 0.211296 seconds.   No pre-allocation
% Elapsed time is 0.185444 seconds.   With pre-allocation
% Elapsed time is 0.014207 seconds.   SUM/Len instead of MEAN

This shows, that mean() is the bottleneck of the standard approach.

[EDITED] I'm working on a C-Mex version, which is 10 times faster. I'm going to publish it in the FEX.

Answer 3

[Image Analyst]

Megha, here is a non-for loop way of doing it using splitapply():

A = [1 7 12 20 23 31];
B = [6 11 19 22 30 35];
C = rand(1,35)

% Method 1 : using splitapply() and no for loop:
% Create group labels.
r = repelem(1:length(A), B - A + 1)
% Get the mean of each group.
theMeans = splitapply(@mean, C, r)

% Method 2 : for loop.   
% Double check first method by comparing to the for loop method:
for i = 1:numel(A)	
	kk(i) = mean(C(A(i):B(i)));
end 
kk  % Report to command window.  The values are the same, as expected.

[Jan]

splitapply calls accumarry internally. Then this is a shortcut:

m = accumarray(r.', C, [], @mean);

Calling the function handle @mean is surpringly slower than SUM/Len. Some timings:

% Dense and not overlapping intervals:
C = rand(1, 1e6);
B = [sort(randperm(1e6-1, 1e5-1)), 1e6];
A = [1, B(1:end-1) + 1];

tic;
r  = repelem(1:length(A), B - A + 1);
m1 = splitapply(@mean, C, r);
toc

tic;
r  = repelem(1:length(A), B - A + 1);
m2 = accumarray(r.', C, [], @mean);
toc

tic;  % Implicit @sum and specify the output size:
r  = repelem(1:length(A), B - A + 1);
m3 = accumarray(r.', C, [numel(A), 1]) ./ (B(:) - A(:) + 1);
toc

isequal(m1(:), m2, m3)  % true

% R2018b, Win10, Intel i7
% Elapsed time is 2.463368 seconds.   SPLITAPPLY
% Elapsed time is 0.927217 seconds.   ACCUMARRAY(@mean)
% Elapsed time is 0.009275 seconds.   ACCUMARRAY(sum) ./ Len

Factor 265 faster than splitapply and factor 100 compared to @mean.

Answer 4

[Jan]

Another method assuming, that the intervals are dense and not overlapping:

A = [1 7 12 20 23 31];
B = [6 11 19 22 30 35];  % A(2:end) == B(1:end-1) + 1 !!!
C = rand(1,35)

S  = cumsum(C);
SB = S(B);
kk = [SB(1), diff(SB)] ./ (B - A + 1)

This is less accurate for large C, because cumsum accumulates rounding errors.

Timings:

C = rand(1, 1e6);

% Dense and not overlapping intervals:
B = [sort(randperm(1e6-1, 1e5-1)), 1e6];
A = [1, B(1:end-1) + 1];

tic;
kk3 = zeros(1, numel(A));
for i = 1:numel(A)	
   kk1(i) = sum(C(A(i):B(i))) / (B(i)-A(i)+1);
end
toc

tic
S  = cumsum(C);
SB = S(B);
kk2 = [SB(1), diff(SB)] ./ (B - A + 1);
toc

max(abs(kk1 - kk2))   % 2.91e-11 !!! Is this sufficient?!

Timings:

% Elapsed time is 0.040201 seconds.   Loop SUM/Len
% Elapsed time is 0.004538 seconds.   CUMSUM (rounding limitations!)
% Elapsed time is 0.009275 seconds.   ACCUMARRAY SUM/Len
% Elapsed time is 0.002097 seconds.   C-Mex (published soon)