I want the in-built function Count to count the elements in a list which are greater than a value.
Example:
Count[{1, 1, 2, 3}, (# > 1.5) &]
Why doesn't this work?
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asked Oct 20, 2014 at 18:15
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2$\begingroup$ Inverse of (21206); variant of (18054); beware of (1699). $\endgroup$– Mr.WizardOct 20, 2014 at 19:05
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$\begingroup$ @Mr.Wizard Thanksfor the comment. ;) $\endgroup$– An old man in the sea.Oct 20, 2014 at 21:18
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$\begingroup$ 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.WizardOct 20, 2014 at 21:39
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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. $\endgroup$– bobthechemistOct 20, 2014 at 22:05
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$\begingroup$ @bobthechemist Did you ever cast a reopen vote? $\endgroup$– Mr.WizardOct 30, 2014 at 9:26
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2 Answers
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Your second argument is a function instead of a pattern.
Count[{1, 1, 2, 3}, _?(# > 1.5 &)]
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$\begingroup$ @Fred Simons Dear Fred, could you please kindly comment the following. The function
Selectis quite close toCountby 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? $\endgroup$ Feb 9, 2016 at 9:34 -
$\begingroup$ Check Mr. Wizard's comment to the OP. Specifically the first link. $\endgroup$– EmyFeb 9, 2016 at 9:38
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$\begingroup$ 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. $\endgroup$ Feb 10, 2016 at 9:18
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In V10.0+ you can stick with functions:
CountsBy[{1, 1, 2, 3}, (# > 1.5) &][True]