Re: Re: Need a nice way to do the "left count"

*To*: mathgroup at smc.vnet.net*Subject*: [mg40309] Re: [mg40303] Re: Need a nice way to do the "left count"*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 31 Mar 2003 04:01:04 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

I agree that Bob's algorithm is the simplest and most natural of all except that his use of FoldList seems to me to be intended to mystify and confuse ;-) In fact drbob3[s_List] := Block[{count, n = 0}, count[any_] := 0; Rest@Map[++n - (++count[#]) &, s] ] does the same thing as drbob2 and is, at least on my machine, slightly faster. As for the algorithm, it is essentially exactly what you would do if you had to solve this problem by hand. You simply count how many times an element in the n-th place has occurred and subtract it from n. The only weakness of this algorithm is that when you get a repetition like {b,a,a...} we know that the number of non a's before the second a is the same as the number of non a's before the first a, so there is no need to count but the algorithm wastes a tiny amount of time on recounting this number again. My algorithm for two distinct values avoids this problem. It exploits the fact that to solve the problem for just two distinct values you only need to know: In[23]:= Length/@Split[{a,b,b,a,a,a,b,a,b,a,a}] Out[23]= {1,2,3,1,1,1,2} The numbers you see give you the number of repeats of the same value you are going to see in the final list. You can also (here is the point where the difference between two value case and the general case becomes significant) work out what these values are going to be. The number you are going to get in the k-th position, where k is even will be the sum of the numbers in all the odd positions preceding k, and the value you get in the k-t place when k is odd the sum of all the values in the even positions preceding k. Since my code depends on this fact it obviously can't be expected to work with more than 2 values. Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ On Monday, March 31, 2003, at 10:14 am, Dr Bob wrote: > I actually think my code is the easiest to understand of them all, but > I > suppose every programmer thinks that! > > I had forgotten to time my other code, so here it is: > > drbob2[s_List] := Block[{count, n = 0}, > count[any_] := 0; > Rest@FoldList[++n - (++count[#2]) &, Null, s] > ] > > (Note that the second argument of FoldList, in both my solutions, is > not > used.) > > I timed Selwyn's new solution, and the four best (for more than 2 > values) > are now: > > n = 10000; > test = Array[Round[99Random[]] &, n]; > Timing[fisher[test];] > Timing[selwyn2[test];] > Timing[drbob[test];] > Timing[drbob2[test];] > > {2.171999999999997*Second, Null} > {0.7180000000000035*Second, Null} > {0.10999999999999943*Second, Null} > {0.17199999999999704*Second, Null} > > Bobby > > On Sun, 30 Mar 2003 11:38:29 -0500, Selwyn Hollis > <selwynh at earthlink.net> > wrote: > >> For what it's worth, I get a 4X speed-up by not using PrependTo, and a >> minor improvement by using Block instead of Module within Scan: >> >> countdiffs[s_List] := Module[{members, totals, g, j}, >> members = Union[s]; >> totals = Count[s, #] & /@ members; >> j = Length[s]; >> Scan[Block[{i = First@First@Position[members, #]}, >> g[j] = j - totals[[i]]; >> totals[[i]]--; j--] &, Reverse[s]]; >> Table[g[i], {i, Length[s]}]] >> >> But it's still much slower than drBob's brilliantly incomprehensible >> code >> :^) >> >> --- >> Selwyn >> >> >> On Saturday, March 29, 2003, at 08:13 PM, Dr Bob wrote: >> >>> Here are timings with a binary-valued list: >>> >>> n = 10000; >>> test = Array[Random[Integer] &, n]; >>> Timing[gray[test];] >>> Timing[hanlon[test];] >>> Timing[andrzej1[test];] >>> Timing[fisher[test];] >>> Timing[selwyn[test];] >>> Timing[drbob[test];] >>> Timing[andrzej2[test];] >>> >>> {36.547*Second, Null} >>> {12.406000000000006*Second, Null} >>> {19.437000000000012*Second, Null} >>> {0.29700000000002547*Second, Null} >>> {3.7819999999999823*Second, Null} >>> {0.07800000000003138*Second, Null} >>> {0.06200000000001182*Second, Null} >>> >>> Neither 'gray' nor 'andrzej2' can handle more than two values in a >>> list, >>> and they give different answers in that case, while the others agree. >>> (They all agree on binary lists.) >>> >>> n = 1000; >>> test = Array[Round[3*Random[]] &, n]; >>> andrzej1[test] == fisher[test] == selwyn[test] == hanlon[test] == >>> drbob[test] >>> gray[test] == drbob[test] >>> andrzej2[test] == drbob[test] >>> gray[test] == andrzej2[test] >>> >>> True >>> False >>> False >>> False >>> >>> Here are timings on a list with 21 different values: >>> >>> n = 10000; >>> test = Array[Round[20Random[]] &, n]; >>> Union[test] >>> Timing[hanlon[test];] >>> Timing[andrzej1[test];] >>> Timing[fisher[test];] >>> Timing[selwyn[test];] >>> Timing[drbob[test];] >>> >>> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, >>> 19, >>> 20} >>> {13.078000000000001*Second, Null} >>> {15.563000000000002*Second, Null} >>> {0.5779999999999959*Second, Null} >>> {2.9849999999999994*Second, Null} >>> {0.07800000000000296*Second, Null} >>> >>> and here are timings with 100 different values: >>> >>> n = 10000; >>> test = Array[Round[99Random[]] &, n]; >>> Timing[hanlon[test];] >>> Timing[andrzej1[test];] >>> Timing[fisher[test];] >>> Timing[selwyn[test];] >>> Timing[drbob[test];] >>> >>> {12.969000000000001*Second, Null} >>> {15.765*Second, Null} >>> {2.2039999999999935*Second, Null} >>> {3.6400000000000006*Second, Null} >>> {0.09399999999999409*Second, Null} >>> >>> 'fisher' seems badly affected by increasing the number of values. >>> >>> Bobby >>> >>> On Sun, 30 Mar 2003 08:00:57 +0900, Andrzej Kozlowski >>> <akoz at mimuw.edu.pl> wrote: >>> >>>> If any one is interested in doing performance tests I suggest using >>>> my >>>> other function, which I also posted to the Mathgroup. The two >>>> functions >>>> were meant to demonstrate the difference between elegance (which I >>>> identify with shortness of code) and efficiency. Here is the >>>> efficient >>>> one (the algorithm works only for lists containing just two distinct >>>> symbols): >>>> >>>> f2[l_List] := Module[{mult = >>>> Length /@ Split[l], list1, >>>> list2, values}, list1 = Rest[FoldList[Plus, 0, Table[mult[[i]], {i, >>>> 1, >>>> Length[mult], >>>> 2}]]]; list2 = >>>> FoldList[Plus, >>>> 0, Table[mult[[i]], {i, 2, 2( >>>> Floor[(Length[mult] + 1)/2]) - 1, 2}]]; >>>> values = Take[Flatten[Transpose[{list2, >>>> list1}]], Length[mult]]; Flatten[Table[Table[values[[i]], { >>>> mult[[i]]}], {i, 1, Length[mult]}]]] >>>> >>>> >>>> >>>> On Sunday, March 30, 2003, at 07:12 am, Steve Gray wrote: >>>> >>>>> Hello all, >>>>> I am again gratified for all the prompt, more-than-competent >>>>> solutions to my problem. I'm enclosing a .nb file with all replies >>>>> received so far so you can compare them. I did not do timing >>>>> because >>>>> the lists I have are relatively short and I don't need to call this >>>>> function very often. >>>>> Thanks again. >>>>> >>>>> Steve Gray >>>>> >>>>> ----- Original Message ----- >>>>> From: BobHanlon at aol.com To: mathgroup at smc.vnet.net > To: mathgroup at smc.vnet.net >>>>> To: stevebg at adelphia.net >>>>> Sent: Saturday, March 29, 2003 4:53 AM >>>>> Subject: [mg40309] [mg40303] Re: [mg40279] Need a nice way to do this >>>>> >>>>> In a message dated 3/29/03 6:11:39 AM, stevebg at adelphia.net writes: >>>>> >>>>> Given a list consisting of only two distinct values, such as >>>>> s={a,b,b,a,a,a,b,a,b,a,a}, I want to derive a list of equal length >>>>> g={0,1,1,2,2,2,4,3,5,4,4}. The rule is: for each position >>>>> 1<=p<=Length[s], look at list s and set g[[p]] to the number of >>>>> elements in s to the left of p which are not equal to s[[p]]. >>>>> In a more general version, which I do not need now, s would >>>>> not be restricted to only two distinct values. >>>>> >>>>> Thank you for any ideas, including other applications where >>>>> this particular calculation is used. The current application is an >>>>> unusual conjecture in geometry. >>>>> >>>>> >>>>> <Left-Count.nb> >>>> Andrzej Kozlowski >>>> Yokohama, Japan >>>> http://www.mimuw.edu.pl/~akoz/ >>>> http://platon.c.u-tokyo.ac.jp/andrzej/ >>>> >>>> >>> >>> >>> >>> -- majort at cox-internet.com >>> Bobby R. Treat<Left-Count.nb> >> >> > > >

**Re: Need a nice way to do this**

**Re: Need a nice way to do this**

**Re: Need a nice way to do the "left count"**

**Re: Re: Need a nice way to do the "left count"**