I want the in-built function Count to count the elements in a list which are greater than a value.


Count[{1, 1, 2, 3}, (# > 1.5) &]

Why doesn't this work?

  • 2
    $\begingroup$ Inverse of (21206); variant of (18054); beware of (1699). $\endgroup$ – Mr.Wizard Oct 20 '14 at 19:05
  • $\begingroup$ @Mr.Wizard Thanksfor the comment. ;) $\endgroup$ – An old man in the sea. Oct 20 '14 at 21:18
  • $\begingroup$ You're welcome. By the way I realize that the close text on this one is a bit dismissive: "simple," "easily," but that is not my intent. I could instead close this as a duplicate of one of the linked questions in the comment above if you would prefer. Also the community may weigh in and decide to reopen it. $\endgroup$ – Mr.Wizard Oct 20 '14 at 21:39
  • 2
    $\begingroup$ @Mr.Wizard I'd vote to open since the function vs. pattern issue is a common mistake for beginner/intermediate MMA users, and I've yet to see a simple Q&A that addresses the issue as this one does. $\endgroup$ – bobthechemist Oct 20 '14 at 22:05
  • $\begingroup$ @bobthechemist Did you ever cast a reopen vote? $\endgroup$ – Mr.Wizard Oct 30 '14 at 9:26

Your second argument is a function instead of a pattern.

Count[{1, 1, 2, 3}, _?(# > 1.5 &)]
  • $\begingroup$ Ups... Many thanks ;) $\endgroup$ – An old man in the sea. Oct 20 '14 at 18:26
  • $\begingroup$ @Fred Simons Dear Fred, could you please kindly comment the following. The function Select is quite close to Count by its functionality. However, it recognizes the pattern argument in the form given in the above question, that is, just #>1.5& as follows: Select[{1, 1, 2, 3}, (# > 1.5 &)] returns {2, 3}. What is the difference between these two cases? $\endgroup$ – Alexei Boulbitch Feb 9 '16 at 9:34
  • $\begingroup$ Check Mr. Wizard's comment to the OP. Specifically the first link. $\endgroup$ – Emy Feb 9 '16 at 9:38
  • $\begingroup$ @AlexeiBoulbitch please see (88220) $\endgroup$ – Mr.Wizard Feb 9 '16 at 11:56
  • $\begingroup$ Dear @Alexei. I was out for some days, so I am a bit late in replying. In the mean time you have already got the link to Mr. Wizard's answer. I can only add that for me, Select and Count are not not so close, still apart from the fact that the first one uses a criterion function and the second one a pattern. Count can be used on any level: Count[{{1,2},3,4}, _Integer, 2]. This cannot be done with Select, that works only on level 1. $\endgroup$ – Fred Simons Feb 10 '16 at 9:18

In V10.0+ you can stick with functions:

CountsBy[{1, 1, 2, 3}, (# > 1.5) &][True]

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