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Re: newbie list question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg115108] Re: newbie list question
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Thu, 30 Dec 2010 19:07:40 -0500 (EST)

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

l1 = RandomInteger[{1, 10^6}, 5*10^5];
l2 = RandomInteger[{1, 10^4}, 10^3];
r1 = firstPositions2[l1, l2]; // Timing
r2 = firsts[l1, l2]; // Timing
r1 === r2

{1.33669, Null}

{0.089459, Null}

False

Bobby

On Thu, 30 Dec 2010 03:12:17 -0600, Ray Koopman <koopman at sfu.ca> wrote:

> 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]]]
>
>


-- 
DrMajorBob at yahoo.com


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