Or alternatively: std(reshape(A, 10, div(length(A), 10)), 1)
Simon On Thursday, October 9, 2014 7:10:11 PM UTC-4, Patrick O'Leary wrote: > > "optionally *along dimensions in region*" (emphasis mine). You are > attempting to read along the tenth dimension of the array. > > You're trying to split the array into groups of ten elements, it sounds > like. > > [std(A[10(n-1)+1:10n]) for n in 1:length(A)./10] > > On Thursday, October 9, 2014 5:56:01 PM UTC-5, K leo wrote: >> >> I am hoping to get the std's of every 10 consecutive elements in A. >> >> std(v[, region]) >> Compute the sample standard deviation of a vector or array v, optionally >> along dimensions in region. The algorithm returns an estimator of the >> generative distribution’s standard deviation under the assumption that >> each entry of v is an IID drawn from that generative distribution. This >> computation is equivalent to calculating sqrt(sum((v - mean(v)).^2) / >> (length(v) - 1)). Note: Julia does not ignore NaN values in the >> computation. For applications requiring the handling of missing data, >> the DataArray package is recommended. >> >> On 2014年10月10日 06:49, Patrick O'Leary wrote: >> > On Thursday, October 9, 2014 5:42:40 PM UTC-5, K leo wrote: >> > >> > julia> std(A, 10) >> > >> > >> > A only has elements along the first dimension. What behavior do you >> > expect here? >> >>
