Standard Deviation for X,Y data over 30 Years

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I have 30 years of monthly average temperature for global latitude/longitude in a matrix A (dimensions Y,X,Year; size 361,721,30). I want to compute the standard deviation for each lat/long over the 30-year period so the resulting matrix should be size (361,721).
I've tried std(A,[],3) but the standard deviation isn't correct. Any thoughts on what I'm doing wrong or how to do this using the STD function?
  2 个评论
dpb
dpb 2023-7-2
What makes you think the result of std is in error? I've never had it fail...sometimes there are ~isfinite() values or other bogus data that can fool you, but that's not the fault of std
Attach enough to show us what isn't what you think it should be; the obvious way would be to attach the .mat file of your data...
dpb
dpb 2023-7-2
编辑:dpb 2023-7-2
A=randi(100,5,5,5);
S1=std(A,[],3)
S1 = 5×5
29.5888 36.7056 33.7831 27.3807 29.1153 34.1980 26.2145 24.1661 37.8061 17.2279 26.7526 36.8673 32.9424 41.0122 32.5346 16.1864 25.1734 17.5926 36.8578 15.4272 19.0342 23.5266 16.1245 20.5353 21.6102
S2=std(A,0,3)
S2 = 5×5
29.5888 36.7056 33.7831 27.3807 29.1153 34.1980 26.2145 24.1661 37.8061 17.2279 26.7526 36.8673 32.9424 41.0122 32.5346 16.1864 25.1734 17.5926 36.8578 15.4272 19.0342 23.5266 16.1245 20.5353 21.6102
std(A(1,1,:))
ans = 29.5888
v=squeeze([A(1,1,:)]);v=v-mean(v);
sqrt(dot(v,v)/(numel(v)-1))
ans = 29.5888
sqrt(v.'*v/(numel(v)-1))
ans = 29.5888
Shows it all seems to work as expected. I noticed the doc stated to use the 0 weight argumet explicitly so just checking that [] did as expected (sure it would) as well...

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回答(1 个)

the cyclist
the cyclist 2023-7-2
I agree with @dpb's comment. The only other thing I can think of is that the syntax
std(A,[],3)
which seems to be equivalent to
std(A,0,3)
gives you normalization by N-1, which would typically what you want for sampled data. If instead you wanted normalization by N (typically for population data), then you can do
std(A,1,3)
instead. But I expect that is not the issue you mean.

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