Re: Select from Tuplet using logical expression

*To*: mathgroup at smc.vnet.net*Subject*: [mg117015] Re: Select from Tuplet using logical expression*From*: Ray Koopman <koopman at sfu.ca>*Date*: Mon, 7 Mar 2011 05:50:29 -0500 (EST)*References*: <ikvomg$f8l$1@smc.vnet.net>

On Mar 6, 2:46 am, Heike Gramberg <heike.gramb... at gmail.com> wrote: > On 5 Mar 2011, at 11:08, Peter Pein wrote: >> Am 04.03.2011 09:40, schrieb Ray Koopman: >>> On Mar 1, 2:27 am, Lengyel Tamas<lt... at hszk.bme.hu> wrote: >>>> Hello. >>>> >>>> Skip if needed: >>>> ///I am working on a part combinatorical problem with sets of 3 >>>> differently indexed values (e.g. F_i, F_j, F_k, F denoting >>>> frequency channels) which are subsets of many values (e.g 16 >>>> different frequency channels, denoted F_0, F_1 ... F_15). >>>> >>>> Now, I need to select triplets from these channels, I used Tuplets. >>>> So far so good. From these I need those combinations where indexes >>>> i!=k and/or j!=k, and i=j is allowed (e.g {i,j,k} = {12, 12, 4} is >>>> a valid channel combination, but {3, 12, 3} is not)./// >>>> >>>> So basically I need to generate triplets from a range of integer >>>> numbers, where the first and second elements of these triplets do >>>> not match the third. I thought Select would help, but I don't know >>>> if there exists an option to control elements' values in a condition. >>>> >>>> From then on I must use these triplets' elements in a function. >>>> >>>> But first I am asking your help in generating thos triplets of >>>> numbers. >>>> >>>> Thanks. >>>> >>>> Tam s Lengyel >>> >>> 1 - 6 have been posted previously. 7 is new, a modification of 6. >>> 1 - 5 generate all the triples, then delete unwanted ones. >>> 6 & 7 generate only the triples that are wanted. >>> >>> The time differences seem to be reliable. >>> >>> r = Range[32]; >>> AbsoluteTiming[t1 = Select[Tuples[r,3], >>> #[[1]]!=#[[3]]&& #[[2]]!=#[[3]]&]; "1"] >>> AbsoluteTiming[t2 = Select[Tuples[r,3], >>> FreeQ[Most@#,Last@#]&]; "2"] >>> AbsoluteTiming[t3 = Cases[Tuples[r,3], >>> _?(FreeQ[Most@#,Last@#]&)]; "3"] >>> AbsoluteTiming[t4 = DeleteCases[Tuples[r,3], >>> _?(MemberQ[Most@#,Last@#]&)]; "4"] >>> AbsoluteTiming[t5 = DeleteCases[Tuples[r,3], >>> {k_,_,k_}|{_,k_,k_}]; "5"] >>> AbsoluteTiming[t6 = Flatten[Function[ij,Append[ij,#]& /@ >>> Complement[r,ij]] /@ Tuples[r,2], 1]; "6"] >>> AbsoluteTiming[t7 = Flatten[Outer[Append,{#}, >>> Complement[r,#],1]& /@ Tuples[r,2], 2]; "7"] >>> SameQ[t1,t2,t3,t4,t5,t6,t7] >>> >>> {0.378355 Second, 1} >>> {0.390735 Second, 2} >>> {0.409103 Second, 3} >>> {0.420442 Second, 4} >>> {0.140180 Second, 5} >>> {0.128378 Second, 6} >>> {0.085107 Second, 7} >>> True >> >> Ray, >> >> you might want to add >> >> AbsoluteTiming[t8=Flatten/@Flatten[(Distribute[ >> {{#1},Complement[r,#1]},List]&)/@Tuples[r,{2}],1];"8"] >> >> which needs ~95% of the time needed to calculate t7. >> >> Peter > > You could also do > > AbsoluteTiming[ > t9 = Flatten[ > Map[(Tuples[{Complement[r, {#}], Complement[r, {#}], {#}}]) &, r], > 1]; "9"] > > That seems to be about twice as fast as t8, although t9 is in a > different order from t1-t8 (SameQ[t1, t2, t3, t4, t5, t6, t7, t8, > Sort[t9]] is still True though). > > Heike. 7, 8, & 9 below are as in my previous post. 10 below is your 9. Whatever it's called, it's definitely the fastest so far. m[7] := Flatten[ Outer[ Append,{#},Complement[r,#],1]& /@ Tuples[r,2], 2] m[8] := Flatten /@ Flatten[ Distribute[ {{#},Complement[r,#]},List]& /@ Tuples[r,2], 1] m[9] := Flatten[ Distribute[ Append[#,Complement[r,#]],List]& /@ Tuples[r,2], 1] m[10]:= Flatten[ Tuples[ {Complement[r,{#}],Complement[r,{#}],{#}}]& /@ r, 1] Transpose@Table[r = Range[n = 2^k]; Table[AbsoluteTiming@Do[Null,{1*^7}]; AbsoluteTiming[m[j];{j,n}],{j,7,10}],{k,4,6}]//ColumnForm v5.2 {{0.013456, { 7,16}}, {0.091608, { 7,32}}, {0.685204, { 7,64}}} {{0.013743, { 8,16}}, {0.105262, { 8,32}}, {0.822973, { 8,64}}} {{0.006476, { 9,16}}, {0.041605, { 9,32}}, {0.257631, { 9,64}}} {{0.002176, {10,16}}, {0.019507, {10,32}}, {0.167546, {10,64}}} v6.0 {{0.019666, { 7,16}}, {0.129066, { 7,32}}, {0.947086, { 7,64}}} {{0.018183, { 8,16}}, {0.131886, { 8,32}}, {0.986983, { 8,64}}} {{0.007146, { 9,16}}, {0.042182, { 9,32}}, {0.251833, { 9,64}}} {{0.002474, {10,16}}, {0.017437, {10,32}}, {0.145921, {10,64}}} A tiny speedup in 10 may be had by changing Complement[r,{#}] to DeleteCases[r,#]. (Another may be had by using Delete[r,#], but that works only when r = Range[n]. In the OP's example, r was Range[0,n-1], in which case we would Delete[r,#+1]. However, I dislike making routines data-dependent that way. The extra speed is simply not worth the potential cost that could be incurred if someone changed the channel codes.)

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**Re: Select from Tuplet using logical expression**

**Re: Select from Tuplet using logical expression**