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Re: newbie list question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg115078] Re: newbie list question
  • From: Ray Koopman <koopman at sfu.ca>
  • Date: Thu, 30 Dec 2010 04:12:17 -0500 (EST)

At 3:00 am on Dec 29, 2010, Carl Woll <carlw at wolfram.com> wrote:
> On 12/28/2010 5:53 AM, Daniel Lichtblau wrote:
>> ----- Original Message -----
>> 
>>> From: "Gareth Edwards"<gareth.edwards at cubicmotion.com>
>>> To: mathgroup at smc.vnet.net
>>> Sent: Sunday, December 26, 2010 3:02:23 AM
>>> Subject:  newbie list question
>>> Hi,
>>> 
>>> Liking Mathematica a lot, but struggling with the early part of
>>> the learning curve, i.e. don't know what I don't know...
>>> 
>>> What would be the neatest syntax for finding the first location
>>> of elements from one list in another? For example:
>>> 
>>> listA = { 4,5,8,2,6,4 }
>>> listB = { 8,4,2 }
>>> 
>>> I would like a function to return {3,1,4} in this case (the first
>>> location in A of each element in B)
>>> 
>>> Many thanks!
>> 
>> Neatest is subject to debate. A straightforward and simple method
>> is to Map Position (of element in first list) over the second list.
>> 
>> firstPositions1[l1_, l2_] := Flatten[Map[Position[l1, #, 1, 1]&, l2]]
>> 
>> This will be fine provided the second list is not too large; it
>> scales as m1*m2 where mj is the length of list j. A method that is
>> slower per element of list2, but scales as O(m1+m2), is shown below.
>> 
>> firstPositions2[l1_, l2_] := Module[
>>    {stor, elem, tab},
>>    tab = ConstantArray[0, Length[l2]];
>>    MapIndexed[If[! IntegerQ[stor[#1]], stor[#1] = #2[[1]]]&, l2];
>>    Do[
>>     elem = stor[l1[[j]]];
>>     If[IntegerQ[elem],
>>      stor[l1[[j]]] = False;
>>      tab[[elem]] = j;
>>      ], {j, Length[l1]}];
>>    tab]
>> 
>> Quick test that these work.
>
>> listA = {4, 5, 8, 2, 6, 4};
>> listB = {8, 4, 2};
>> 
>> In[85]:= firstPositions1[listA, listB] ==
>>   firstPositions2[listA, listB] ==
>>   IntegerDigits[IntegerPart[100*Pi]]
>> 
>> Out[85]= True
>> 
>> If the second list has 10 or so elements, the second method is
>> noticeably slower than the first. But it does in fact scale much
>> better. Here we compare speed when second list is 1000 elements
>> (not necessarily unique).
>> 
>> l1 = RandomInteger[10^6, 5*10^5];
>> l2 = RandomInteger[10^6, 10^3];
>> 
>> In[73]:= Timing[pos1 = firstPositions1[l1, l2];]
>> Out[73]= {68.11000000000001, Null}
>> 
>> In[74]:= Timing[pos2 = firstPositions2[l1, l2];]
>> Out[74]= {1.933999999999969, Null}
>> 
>> Notice there is a difference in behavior beyond speed. If the second
>> list contains values not found in the first, the second result will
>> have zeros in those corresponding slots. I'm guessing this is
>> probably desirable behavior, as compared to omitting them altogether
>> (since you won't otherwise know what element in result corresponds
>> to what element in second list).
>> 
>> In[75]:= pos1 == DeleteCases[pos2, 0]
>> Out[75]= True
>> 
>> In[76]:= Length[pos1]
>> Out[76]= 379
>> 
>> Daniel Lichtblau
>> Wolfram Research
> 
> If we assume that all of the list elements are positive integers,
> then the following is much faster:
> 
> firsts[l1_, l2_] :=  Normal[SparseArray[Automatic, {Max[l1]}, 0,
> {1, {{0, Length[l1]}, List /@ l1}, Range[Length[l1]]}][[l2]]]
> 
> Example:
> 
> l1 = RandomInteger[{1, 10^6}, 5*10^5];
> l2 = RandomInteger[{1, 10^6}, 10^3];
> 
> In[130]:= r1 = firstPositions2[l1, l2]; // Timing
> r2 = firsts[l1, l2]; // Timing
> r1 === r2
> 
> Out[130]= {2.387, Null}
> 
> Out[131]= {0.109, Null}
> 
> Out[132]= True
> 
> Carl Woll
> Wolfram Research

If Max[l2] can be > Max[l1] then we need 

 firsts[l1_, l2_] :=  Normal[SparseArray[Automatic, {Max[l1,l2]}, 
  0, {1, {{0, Length[l1]}, List /@ l1}, Range[Length[l1]]}][[l2]]]



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