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Bug with mean() on floating point vector elements

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Victor
Victor on 13 Dec 2017
Commented: Stephen Cobeldick on 13 Dec 2017
Can someone explain this bug?
a = rand(100,1);
mean(a(4), a(6))
Output:
Error using sum
Dimension argument must be a positive integer scalar within indexing range.
Error in mean (line 117)
y = sum(x, dim, flag)/size(x,dim);

  1 Comment

Stephen Cobeldick
Stephen Cobeldick on 13 Dec 2017
"Can someone explain this bug?"
What bug? The fact that you are using mean in a way that is not supported by MATLAB just means that you need to read the documentation.

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Accepted Answer

Jan
Jan on 13 Dec 2017
Edited: Jan on 13 Dec 2017
This is not a bug. Please read the documentation, if you have a problem with a command:
doc mean
The 2nd argument of mean must be the dimension to operate on. Then a floating point number between 0 and 1 will not work. See:
a = magic(3)
mean(a, 1)
mean(a, 2)
You forgot to mention, what you want to calculate. Maybe you mean:
a = rand(100, 1);
mean(a([4,6]))
% or
mean(a(4:6))

  3 Comments

Victor
Victor on 13 Dec 2017
OK, then why this code doesn't throw the same error: a = [3,5,2,5,3]'; mean(a(3), a(4))
Understood - it tries to calculate along singleton dimension.
Steven Lord
Steven Lord on 13 Dec 2017
Looking at your a vector, what is a(4)? It's 5. Can you take the mean of a two-dimensional array in the fifth dimension? Sure. While a(3) doesn't look like it has a fifth dimension, if you ask for the size of a(3) in the fifth dimension what does MATLAB return?
size(a(3), 5) % returns 1
See this Answer for some discussion of trailing singleton dimensions.
The reason your original attempt didn't work is that arrays in MATLAB don't have (for example) a 0.25th dimension.
Stephen Cobeldick
Stephen Cobeldick on 13 Dec 2017
@Viktor: it does not try to calculate along the singleton dimension, it actually does calculate along the singleton dimension.

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