Re:Challenge! ....RESULTS...,
- To: mathgroup at christensen.cybernetics.net
- Subject: [mg1059] Re:[mg950] Challenge! ....RESULTS..., [mg986]
- From: Allan Hayes <hay%haystack at christensen.cybernetics.net>
- Date: Fri, 12 May 1995 15:10:48 -0400
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 [1] 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 [2] 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.
[1]
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 interleave3A maintains its advantage over interleave3.
test =Array[#,{300}]&/@(Range[100]);
Do[interleave3[test2],{30}]//Timing//First
4.58333 Second
Do[interleave3Replace[test2],{30}]//Timing//First
3.6 Second
[2]
Is the compications 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}
]
];
Module[{len = Length[L]},
Transpose[
L[[Random[Integer,{1,len}]&/@s ]],
{1,1}
]
];
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[test2],{30}]//Timing//First
3.66667 Second
Do[interleave3Block[test2],{30}]//Timing//First
4.28333 Second
Do[interleave3Module[test2],{30}]//Timing//First
48.9667 Second (*!*)
Do[interleave3With[test2],{30}]//Timing//First
48.5333 Second (*!*)
Allan Hayes
hay at haystack.demon.co.uk