Best way to count this?

Question

I'm trying to use the Count function for this purpose, but it's not working how I'd like it to. Does anybody know a simple way to do this counting?

I would like to be counting the number of "p"s in expressions like

p[1,2][1]^2 p[3,4][3]

or

p[3,3][1]

I would like the count to return 2 on the first one (even though one of the p's has a squared term), and 1 on the second.

edit: I realized for the purpose of counting I can remove the "square" so the first expression would just look like

p[1,2][1]p[3,4][3]

if that makes it any easier.

@belisariusisforth: Seems like you're violating your Principle of don't-ask-questions-to-expand-the-scope-of-the-question. :) — march, Oct 9, 2015 at 19:02

bill s · Accepted Answer · 2015-10-09 18:57:43Z

4

One way is to turn it into a string and count the number of occurences of p in the string:

StringCount[ToString[p[1, 2][1]^2 p[3, 4][3]], "p"]

answered Oct 9, 2015 at 18:57

bill s

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$\begingroup$ I had just figured this out and was about to answer my own question with that! Thanks though. $\endgroup$
– Alex Mathers
Oct 9, 2015 at 18:59
1

$\begingroup$ Ahem: p["p"]... :D (To be fair to you, the OP didn't mention that they only wanted to count symbols although that's how I interpreted the question) $\endgroup$
– rm -rf ♦
Oct 10, 2015 at 11:22

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J. M.'s lack of A.I. · Accepted Answer · 2015-10-09 19:02:45Z

12

This works:

Count[p[1, 2][1]^2 p[3, 4][3], _p, ∞, Heads -> True]

answered Oct 9, 2015 at 19:02

J. M.'s lack of A.I.♦

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Bob Hanlon · Accepted Answer · 2015-10-09 20:32:07Z

4

"And Now for Something Completely Different"...

expr = p[1, 2][1]^2 p[3, 4][3];

Module[{n = 0}, expr /. p :> n++; n]

(*  2  *)

No claim to be "best"

answered Oct 9, 2015 at 20:32

Bob Hanlon

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Eric Towers · Accepted Answer · 2015-10-09 22:55:12Z

2

Length[Position[p[1, 2][1]^2 p[3, 4][3], p]]

answered Oct 9, 2015 at 22:55

Eric Towers

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4 Answers 4