End Points on movmean don't look right

If you run the simple code (see end of post) you are taking the sliding window filter using matlab "movmean". Per the description:

"The window size is automatically truncated at the endpoints when there are not enough elements to fill the window. When the window is truncated, the average is taken over only the elements that fill the window. M is the same size as A."

If I read this correctly-for element 1 of the dataset, only element 1 "fills the window". Thus the average of A(1) is just A(1)/1 =1. Looking at my plot I do not see this--it appears as thought the first element is wrong--instead of 1, movmean returns 1.5.

What did I miss here?

Thanks--Fritz

========================================

for i=1: 10
    A(i)=i;
end    
for i=11: 20
    A(i)=20-i;
end

plot(A,'b','LineWidth',2);

M=movmean(A,3)

M = 1×20

1.5000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 9.3333 9.0000 8.0000 7.0000 6.0000 5.0000 4.0000 3.0000 2.0000 1.0000 0.5000

hold on; 
plot(M,'k','LineWidth',2);

fprintf("A(1)=%f\n",A(1))  ;

A(1)=1.000000

fprintf("M(1)=%f\n",M(1))  ;

M(1)=1.500000

fclose all;

[Image Analyst]

"A(1)/1 =1" is not correct.

The correct equation is A(1)/1 = A(1).

Answer 1

[Matt J]

If I read this correctly-for element 1 of the dataset, only element 1 "fills the window".

No, the length 3 window is centered on element 1, so element 2 will also be in the window. The value of 1.5 is the average of [1,2]. If you don't want the current element to lie at the right hand edge fo the window, you can do

movmean(A,[2,0])

Answer 2

[Image Analyst]

When the center of your 3-element window is over element 1 of A, which is

A =

1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1 0

the values 1 and 2 are inside the window.

The window does not travel so far that the center of the window is off the left side of A and only the tail overlaps the first element of A. If it did, then the output matrix would be larger than A.

So, the average of 1 and 2 is 1.5, like it is for M.

Answer 3

[Fritz Sonnichsen]

OK--I was placing my window wrong--thanks for the replies

(and sorry for the bad English in my title! FInger error)

cheers

Fritz

[Matt J]

Glad you worked it out, but please Accept-click one of the answers, so that Mathworks has a record that a resolution was reached.