Standard Deviation for X,Y data over 30 Years

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?

[dpb]

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]

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...

Answer 1

[the cyclist]

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.