Thanks to both for explanations. "along dimensions in region" sounds
pretty confusing to me. Can that be stated more clearly? Pardon my
English.
I guess this is what I wanted.
julia> [std(A[i:i+9]) for i=1:length(A)-9]
91-element Array{Any,1}:
0.395761
0.391694
0.392545
0.363307
⋮
0.322292
0.325662
0.345799
On 2014年10月10日 07:17, Simon Kornblith wrote:
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?
