I have calculated 4 standard deviation values for each row of a 4x3 matrix. How to convert these 4 values into a single standard deviation value for the whole matrix?

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a=[2 3 5;12 67 54;3 7 8;14 9 23]
b= std(a,0,2)
  1 个评论
Rik
Rik 2023-6-24
The standard deviation of a large set can only be estimated from the standard deviations of the parts. Are you sure you want to do that instead of calculating the SD for all values?

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回答(3 个)

Torsten
Torsten 2023-6-24
Not possible.
Use
std(a(:))
instead.

VBBV
VBBV 2023-6-24
编辑:VBBV 2023-6-24
a=[2 3 5;12 67 54;3 7 8;14 9 23]
a = 4×3
2 3 5 12 67 54 3 7 8 14 9 23
b = std(a,0,"all")
b = 21.2009
Use the *all* option to get the SD for whole matrix

John D'Errico
John D'Errico 2023-6-24
编辑:John D'Errico 2023-6-24
You should understand why it is not possible to compute the overall standard deviation, merely from the pieces. This is because a standard deviation does not contain information about the mean. That information has essentially been lost to the world. Dumped in the laundry basket of computing, never to be seen again. ;-)
Let me give an example. Consider a case where we have two groups of numbers. I'll call them X and Y.
X = normrnd(0,2,[1,5])
X = 1×5
0.4428 1.5877 -2.3844 -0.6804 0.3617
Y = normrnd(5,4,[1,5])
Y = 1×5
10.7247 5.8261 5.2775 6.5703 0.6086
We can compute the standard deviation of X and Y. That part is easy.
stdX = std(X)
stdX = 1.4921
stdY = std(Y)
stdY = 3.6080
But if you look at the formula for standard deviation, it first subtracts the mean of the set. I'll write it here as a function handle.
S = @(Z) sqrt(sum((Z - mean(Z)).^2/(numel(Z)-1)))
S = function_handle with value:
@(Z)sqrt(sum((Z-mean(Z)).^2/(numel(Z)-1)))
S(X)
ans = 1.4921
Wow! I got it right! But the problem is, unless you know the original means of the subgroups, you cannot infer what the standard deviation of the aggregate group would be. So you can do this:
std([X,Y])
ans = 4.0697
That is the equivalent to what others have suggested, which is to create one large vector of data from the array, and then compute the global standard deviation.

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