If I have a list like


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.

  • $\begingroup$ There is ImageCrop/ImageTrim, StringTrim but no ListTrim :/ :) $\endgroup$
    – Kuba
    Mar 19, 2014 at 23:05
  • 1
    $\begingroup$ @Kuba Don't give SW any ideas for "A new kind of list-manipulation" marketing! $\endgroup$
    – rm -rf
    Mar 19, 2014 at 23:19

3 Answers 3


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.} *)
  • $\begingroup$ I suspect the OP just used the successive commas not to indicate Nulls but rather that there were an unknown number of trailing 0s. $\endgroup$
    – murray
    Mar 19, 2014 at 14:48
  • $\begingroup$ @murray I don't think so, because in the next line they say that they remove the "blanks" with Select and NumberQ... you can't remove the zeros in this manner. $\endgroup$
    – rm -rf
    Mar 19, 2014 at 14:49
  • $\begingroup$ You may be right. I thought perhaps he was referring to the similar situation of deleting trailing "blanks" in a string. $\endgroup$
    – murray
    Mar 19, 2014 at 14:54
  • $\begingroup$ Checking closely, it appears that some of the "Null" entries in my actual data are "", not Null. I adjusted your code accordingly. Thanks. $\endgroup$ Mar 19, 2014 at 16:25

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.

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

{2334, 342, 0, 0, 2322}

1. %

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


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