Re:Challenge! ....RESULTS..., (Corrected)

*To*: mathgroup at christensen.cybernetics.net*Subject*: [mg1069] Re:[mg950] Challenge! ....RESULTS..., [mg986] (Corrected)*From*: Allan Hayes <hay at haystack.demon.co.uk>*Date*: Fri, 12 May 1995 16:50:58 -0400

I have corrected some typing errors in my previous posting. The corrected version ([2]CORRECTED COPY) follows the first section. I have also, got closer to seeing why codes involving Module and With can be inefficient. I suggest some ways of using them more efficiently. Since this is of more general consequence I will explain it at once. [1] USING MODULE AND WITH ************************* (*6*) Module[{a=A},B] and With[{a=A},B] can be slow when A or B are big. Some Suggestions: Name A, preferaby inside Block,: eg Block[{a = A}, With[{a=a},B]]; Minimize B by placing Module and With as deep inside the expression as possible; consistent with its working as required. Try naming all or parts of B (eg. Block[{b = B}, With[{a=A},b]]) but don't remove symbols that you wish With to replace. In the test below it turns out that we need to deal with both A and B to get a significant improvement. *********************** Examples: For brevity I deal only with With. Please note that thes are not realstic examples: they are chosen to investigate the behaviour of Module and With. The fastest form of he code is simply test[L_] := {5 len, L[[1,1]]}. data =Array[#,{300}]&/@(Range[100]); First the direct application of With testWith[L_] := With[ {len = Length[L]}, {5 len, L[[1,1]]} ] testWith[data]//Timing {1.48333 Second, {500, 1[1]}} Now to try the suggestions individually and together testWith2[L_] := Block[{len = Length[L]}, (*name A *) With[ {len = len}, {5 len, L[[1,1]]} ] ]; testWith2[data]//Timing {1.46667 Second, {500, 1[1]}} testWith3[L_] := {With[{len = Length[L]},5 len], L[[1,1]]};(*With inside*) testWith3[data]//Timing {1.41667 Second, {500, 1[1]}} testWith4[L_] := Block[{len = Length[L]}, (*name A*) { With[ {len = len},5 len ], L[[1,1]] } (*With inside*) ] testWith4[data]//Timing {0. Second, {500, 1[1]}} testWith5[L_] := Block[{len = Length[L], l = L}, (*name A and B*) With[ {len = len}, {5 len, l[[1,1]]} ] ]; testWith5[data]//Timing {0. Second, {500, 1[1]}} Why do we need to deal with both A and B before any significant improvement is seen? There is another way: use replacement directly like this testReplace[L_] := Block[{len}, {5 len, L[[1,1]]} /. len -> Length[L] ]; testReplace[data]//Timing {0. Second, {500, 1[1]}} For repeated use we have the following timings Do[testWith4[data],{1000}]//Timing Do[testReplace[data],{1000}]//Timing {1.46667 Second, Null} {1.61667 Second, Null} So using With seems better. Unfortunately this speed up does not seem to carry through to the interleaving code. (*previous code using replacement*) interleave3Replace[L:{s_,___}] := Block[{len}, Transpose[ L[[(Random[Integer,{1,len}]&/.len-> Length[L])/@s ]], (*4*) {1,1} ] ]; Do[interleave3Replace[data],{30}]//Timing//First 3.55 Second (*new code using With*) interleave3With[L:{s_,___}] := Block[{len = Length[L]}, Transpose[ L[[With[{len=len},Random[Integer,{1,len}]&]/@s ]], (*4*) {1,1} ] ]; Do[interleave3With[data],{30}]//Timing//First 3.51667 Second One final point: we can avoid the need to take With inside as follows interleave3WithA[L_] := Block[{len = Length[L]L}, With[{len=len}, Transpose[ L[[Random[Integer,{1,len}]&/@First[LL] ]], {1,1} ] ]]; Do[interleave3WithA[data],{30}]//Timing//First Do[interleave3WithA[{list1,list2}],{1000}]//Timing//First 3.61667 Second 5.68333 Second [2] CORRECTED COPY (I have change the numbering:[1] to [2a] and [2] to [2b]) ************************* Paul Howland <PEHOWLAND at taz.dra.hmg.gb> asked in [mg889] for the most efficient function for generating a new list by randomly interchanging the elements of two other lists. I give in [2a] below a further speed up to my best code in my response [mg986] and find another speed up rule: (*4*) Simple replacements can be very fast But first a correction: in [mg986] I let my keenness to try out larger tests take me away from the original challenge problem.The statement that interleave3 is faster than Dave Wagner's code swap is incorrect for the two lists in the original problem. However, the new code, is faster for these lists. In [2b] I compare this new code interleave3Replace with the results of using straightforward Block, Module and With on interleave3. Module and With come out badly for large data sets. Hence, the rule: (*5*) Try Block for speed. [2a] My previous best interleave3[L:{s_,___}] := Transpose[ L[[Random[Integer,{1,Length[L]}]&/@s ]], {1,1} ]; The new code interleave3Replace[L:{s_,___}] := Block[{len}, Transpose[ L[[(Random[Integer,{1,len}]&/.len-> Length[L])/@s ]], (*4*) {1,1} ] ]; TIMINGS list1 = {a,b,c,d,e,f,g,h,i,j}; list2 = {A,B,C,D,E,F,G,H,I,J}; Timing[Do[interleave3[{list1,list2}],{1000}]][[1]] 7.33333 Second Timing[Do[interleave3Replace[{list1,list2}],{1000}]][[1]] 5.6 Second Dave Wagner's swap code is swap[list1_List,list2_List] := #[[Random[Integer,{1,2}]]]& /@ Transpose[{list1,list2}] and gives the timing Timing[Do[swap[list1,list2],{1000}]][[1]] 6.28333 Second For bigger data interleave3Replace maintains its advantage over interleave3. test =Array[#,{300}]&/@(Range[100]); Do[interleave3[test],{30}]//Timing//First 4.58333 Second Do[interleave3Replace[test],{30}]//Timing//First 3.6 Second [2b] Is the complication of interleave3Replace necessary? Let's try a straightforward use of Block, Module and With interleave3Block[L:{s_,___}] := Block[{len = Length[L]}, Transpose[ L[[Random[Integer,{1,len}]&/@s ]], {1,1} ] ]; interleave3Module[L:{s_,___}] := Module[{len = Length[L]}, Transpose[ L[[Random[Integer,{1,len}]&/@s ]], {1,1} ] ]; interleave3With[L:{s_,___}] := With[{len = Length[L]}, Transpose[ L[[Random[Integer,{1,len}]&/@s ]], {1,1} ] ]; Timings: Timing[Do[interleave3Replace[{list1,list2}],{1000}]][[1]] 5.7 Second Timing[Do[interleave3Block[{list1,list2}],{1000}]][[1]] 5.93333 Second Timing[Do[interleave3Module[{list1,list2}],{1000}]][[1]] 8.3 Second Timing[Do[interleave3With[{list1,list2}],{1000}]][[1]] 6.08333 Second But for bigger data interleave3Replace has more of an advantage over interleave3Block, but Module and With fall dramatically behind. This may be due to their being scoping constructs. For example Module has to look up $ModuleNumber and if it is n it then replaces len with len$n and sets up the assignment len$n = k (* where k is the length of L*) Block simply sets up the assignment len = k. But this seems quite inadequate to explain such large differneces. Do[interleave3Replace[test],{30}]//Timing//First 3.66667 Second Do[interleave3Block[test],{30}]//Timing//First 4.28333 Second Do[interleave3Module[test],{30}]//Timing//First 48.9667 Second (*!*) Do[interleave3With[test],{30}]//Timing//First 48.5333 Second (*!*) Allan Hayes hay at haystack.demon.co.uk