Removing trailing zeros from a list

Question

If I have a list like

list1={2334.,342.,0.,0.,2322.,0.,0.,0.,0.,,,,,,};

it's easy to remove to trailing blanks with Select[list1,NumberQ[#]&], and it's easy to remove all the zeroes, but how do I remove only the trailing zeros? I want to leave the mid-list zeroes in place.

Answer 1

Those trailing blanks are actually Nulls (or "" if you're using strings). The trailing nulls and zeros can be removed with patterns:

list1 /. {h__, (0 | 0. | Null) ...} :> {h}
(* {2334., 342., 0., 0., 2322.} *)

Answer 2

If lists are large, something like

list1[[;; (Position[list1, Except[0 |0.| Null], 1][[-1, 1]])]]

should be quite a bit faster, e.g., a quick test on a 50K length list it is over 350X faster than the rule-based solution, and 4X faster than the Take solution.

If the list is "front-loaded" with values (i.e., more wanted than not), using

list1[[;; -Position[Reverse@list1, Except[List | 0 | 0. | Null], 1, 1][[1, 1]]]]

is thousands and hundreds of times faster than rule-based/take posted respectively.

If the lists are precisely the format in your example (desired, chunk of zeroes, chunk of nulls)

Flatten@Split[list1][[;; -3]]

is short, sweet, and very fast.

Answer 3

Block[{Null = 0}, Internal`DeleteTrailingZeros @ Rationalize @ list1]

{2334, 342, 0, 0, 2322}

1. %

{2334., 342., 0., 0., 2322.}