I could not find something similar.
Given, e.g.
lst = Range[10]
(* {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} *)
the following gives the reverse of lst (I know about Reverse).
lst[[-1 ;; 1 ;; -1]]
(* {10, 9, 8, 7, 6, 5, 4, 3, 2, 1} *)
Why the following does not return the same output?
lst[[;; ;; -1]]
Thanks.
If you evaluate
;; ;; -1
you get
1 ;; All ;; -1
This shows that the default starting point is always 1, regardless of the step size. Note that ;; is short for Span. Generally, the default for a given argument of Span is independent of the other arguments. The default for the first one (beginning) is always 1, the second one (end) always All and the third one (step size) 1.
;; ;; -1 will not work because if we start at 1, we cannot count downwards without hitting the boundary of the array.
;;;;k represents a span from the beginning to the end in steps of k." It is the last point before the "Details" section on the Docs page for Span.
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Commented
Apr 10, 2017 at 16:08
;; ;; -1 is parsed as a whole, and not as a combination of two Spans. It can be seen from comparing it with ;; ;; ;; -1 and with ( ;; ) (;; -1), and also with ToExpression[";;;;-1", InputForm, HoldComplete]. Tangentially related: (123359).
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Commented
Apr 10, 2017 at 16:21
Span, right? Otherwise it starts by default from the beginning. Is there any similar concept to Mathematica with Python's slicing? (Thanks for the commenf by the way!) $\endgroup$SpanandPartcompared to Python's slicing? So far you have only shown that Mathematica uses a slightly different syntax. $\endgroup$lst[::-1]gives thelstreversed without prividing any start/end value. $\endgroup$1and end position toAllit cannot use;; ;; -1to produce the list reversed. See;; ;; -1 // InputForm. $\endgroup$