Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2002
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2002

[Date Index] [Thread Index] [Author Index]

Search the Archive

RE: Pattern Matching in Lists (correct.)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35567] RE: [mg35547] Pattern Matching in Lists (correct.)
  • From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
  • Date: Fri, 19 Jul 2002 06:08:33 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

In writing this post I made a common error, see below

> -----Original Message-----
> From: Wolf, Hartmut 
To: mathgroup at smc.vnet.net
> Sent: Thursday, July 18, 2002 12:41 PM
> Subject: [mg35567] RE: [mg35547] Pattern Matching in Lists
> 
> 
> 
> 
> > -----Original Message-----
> > From: Anthony Mendes [mailto:amendes at zeno.ucsd.edu]
To: mathgroup at smc.vnet.net
> > Sent: Thursday, July 18, 2002 9:07 AM
> > Subject: [mg35567] [mg35547] Pattern Matching in Lists
> > 
> > 
> > Hello,
> > 
> > Suppose w={1,1,1,0,0,1,0,1,0,0,1,0,0}.
> > 
> > How can I count the number of occurrences of a 1 in w immediately
> > followed by a 0 in w?
> > 
> > I have tried every incarnation of Count[] I can think of; 
> for example,
> > 
> > Count[w,{___,1,0,___}]
> > 
> > does not seem to work.  In general, how can I count the number of
> > occurrences of a 1 followed by a 0 in a list of 1's and 0's?  
> > Thank you!
> > 
> > 
> > -- 
> > Tony
> > _____________________
> > amendes at math.ucsd.edu
> > 
> > 
> Tony,
> 
> there are certainly many solutions (and many will appear here 
> in this newsgroup). Just to pass some ideas
> 
> Count[w /. {a___, 1, 0} :> {a, {1, 0}} 
>         //. {a___, 1, 0, b__} :> {a, {1, 0}, b}, {1, 0}]
> 
> Count[w //. {a___, 1, 0, b___} :> {a, \[HappySmiley], b}, 
> \[HappySmiley]]
> 
> Length[ReplaceList[w, {a___, 1, 0, b___} :> {a, {}, b}]]
> 
> specialCount[w_] := 
>   Module[{ww = w //. {a___, x_, x_, b___} :> {a, x, b}}, 
>     If[ww[[1]] === 0, Quotient[Length[ww] - 1, 2], 
> Quotient[Length[ww], 2]]]
> 
> 
> this will lead to another idea:
> 
> specialCount2[w_] := With[{l = Length[Split[w]]},
>     If[w[[1]] === 1, Quotient[l, 2], Quotient[l - 1, 2]]]
> 
> 
> The last two methods wouldn't work if elements other than 0 
> or 1 were present, so best use a pattern test for w
**> 
**>   w_?VectorQ[#, MatchQ[#, 0 | 1] &] &
**> 
**>   w_?VectorQ[#, IntegerQ[#] && 0 <= # <= 1 & ] &
**> 

Should read
 
   w_?(VectorQ[#, MatchQ[#, 0 | 1] &] &)
 
   w_?(VectorQ[#, IntegerQ[#] && 0 <= # <= 1 & ] &)

(Although the intended meaning should be clear, this happens, when you don't
test the trivia.) PatternTest has a very high priority, whereas function a
relatively low one. I once wrote into my personal style guide to _always_
enclose 

    (expr &)

in parentheses, e.g. for better reading. But when the anonymous Function is
an argument to a higher function or appears at the left of an expression (as
is mostly the case) then we don't need them.

Of similar interest are the cases where you need to write

     (expr)&

(or better: ((expr)&) ) In practice this is when expr contains an
assignment, a compound expression or (of course) a // function application .
 

> 
> We may also convert the problem to the numeric domain
> 
> specialCount3[w_] := 
>   Block[{t = 1, c = 0}, 
>     NestWhile[(If[Mod[#, 2] == 0, t = 0, If[t === 0, ++c; t = 1]]; 
>           Quotient[#, 2]) &, FromDigits[w, 2], Positive]; c]
> 
> --
> Hartmut
> 


To reconcile you here some more solutions:

Count[NestList[Rest, w, Length[w]], {1, 0, ___}]


Last[Fold[Switch[{#1[[1]], #2}, 
                 {1, 0}, {#2, #1[[2]] + 1}, 
                 _, {#2, #1[[2]]}] & , {0, 0}, w]]


sc4[{n_, {a_, b_, c___}}] := 
  Switch[{a, b}, {1, 0}, {n + 1, {b, c}}, _, {n, {b, c}}]

sc4[else_] := else

specialCount4[w_] := First[FixedPoint[sc4, {0, w}]]


--
Hartmut



  • Prev by Date: RE: Re: Factoring question
  • Next by Date: Re: Re: Re: Re: Factoring question
  • Previous by thread: Comparison of Mathematica on Various Computers
  • Next by thread: Repost recent messages - Administration