Well, it’s rather simple, but maybe there is already a built-in function to do calculate this? natural_length(s) = s[end]-s[begin] natural_length((1,10)) natural_length((1:10)) It annoying to copy-paste this snippet which is not large enough to bother putting it into a package.

[image] mkitti: This does not work for all iterators. Even last(a) - first(a) seems wrong to me for a non-sorted iterator. A more general solution would be to use the extrema function : diameter(itr) = -(reverse(extrema(itr))...) which is equivalent to the more verbose (but more readable) …

Is there a built-in function for the natural length of a segment represented by an iterator?

New to Julia

stevengj Split this topic April 7, 2024, 7:22pm 10

3 posts were split to a new topic: Transposes vs 1-row matrices

Topic		Replies	Views
Length from iterator? New to Julia iterators	22	7119	August 27, 2019
`iterate` returning iter, next-state and iter's original length General Usage	4	448	September 7, 2018
Iterate with 1:N is faster than 1:length(A) Performance question	12	821	September 16, 2020
Any shortcut to 1:length(myVector)? General Usage	34	2981	January 12, 2021
Why no `last(::Generator)`? Internals & Design	3	672	July 28, 2021

Is there a built-in function for the natural length of a segment represented by an iterator?

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