Well, I might argue that the slices are along the axes rather than them being the same thing.
> On Oct 9, 2014, at 10:05 PM, John Myles White <[email protected]> > wrote: > > I’d suggest slices for consistency with the function mapslices. > > — John > >> On Oct 9, 2014, at 6:15 PM, Tim Holy <[email protected]> wrote: >> >> Would be great to have it clarified in the manual. >> >> I think I've brought this up before, and if there was a consensus I don't >> recall it. In my opinion, the various usages of "dims" and "region" in the >> manual and help are pretty confusing. It would be nice to standardize >> terminology. I confess to being fond of talking about the "axes" of an >> array, >> but I am fine with other choices too. >> >> --Tim >> >>> On Friday, October 10, 2014 07:29:06 AM K Leo wrote: >>> 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 >> 0.392545 >>> ⋮ >>> 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? >
