Re: How does Position work?

*To*: mathgroup at smc.vnet.net*Subject*: [mg85380] Re: [mg85313] How does Position work?*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Thu, 7 Feb 2008 22:31:28 -0500 (EST)*References*: <30638300.1202384648040.JavaMail.root@m08>*Reply-to*: drmajorbob at bigfoot.com

Your expression is a Map: something /@ list It does "something" to each element of "list", one at a time and returns each result. list = Riffle[ConstantArray[0, 10^4], ConstantArray[1, 10^4]]; Timing[Flatten[Position[list, #1]] & /@ list;] {9.31428, Null} There's NO REASON this should go faster, since you're doing an operation 20,000 times and assembling 20,000 results. In your mind, perhaps, you're doing something different, as in: Timing[Flatten[Position[list, #1]] & /@ Union@list;] {0.022799, Null} where Mathematica does "something" TWICE and assembles only TWO results. Bobby On Tue, 05 Feb 2008 18:45:44 -0600, P_ter <peter_van_summeren at yahoo.co.uk> wrote: > Hello, > JÃ¡nos wrote some time ago a solution which involved > lstIn = {2, 3, 4, 4, 2, 1, 1, 5, 4} > Flatten[Position[lstIn, #1]] & /@ lstIn > This last line gives as result: > {{1, 5}, {2}, {3, 4, 9}, {3, 4, 9}, {1, 5}, {6, 7}, {6, 7}, {8}, {3, > 4, 9}} > One can observe that the last 4 in lstIn has the same result as the > first and second 4. Which should be. > But! Did Position make a trie (for example) first and then give the > results? The document about Position only tells that expressions are > searched for in depth-first order. But that does not tell anything about > the data structure. > I used: > lstIn1 = Riffle[ConstantArray[0, 10^4], ConstantArray[1, 10^4]]; > One should expect that this would work very fast, but it takes a long > time. > Does someone know about this? > with friendly greetings, > P_ter > > -- DrMajorBob at bigfoot.com