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	7351	August 27, 2019
Why no `last(::Generator)`? Internals & Design	3	674	July 28, 2021
Getting the last element of an iterator General Usage question	45	7024	November 12, 2021
Implementation of itertools.count New to Julia	15	1698	May 19, 2017
length(Iterators.flatten(...)) not working General Usage question	7	1300	April 7, 2019

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

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