Re: Extracting some elements, members of another list

*To*: mathgroup at smc.vnet.net*Subject*: [mg112433] Re: Extracting some elements, members of another list*From*: Valeri Astanoff <astanoff at gmail.com>*Date*: Wed, 15 Sep 2010 04:40:08 -0400 (EDT)*References*: <i6nedc$g8u$1@smc.vnet.net>

On 14 sep, 11:14, Nacho <ncc1701... at gmail.com> wrote: > Hi! > > I would like your advice about this problem: > > I have a list of elements, consisting in phone numbers with some data: > > list1= { { phone1, data1}, {phone2, data2}, {phone3, data3} .... } > > And a list of phones, a subset of the phone numbers in list1 > > list2= {phone2, phone3, phone7... } > > I'd like to extract the elements from list1 whose phone numbers are > present in list 2, that is: > > result= { { phone2, data2}, {phone3, data3, {phone7, data7} .... } > > I've used this with small lists and it works fine: > > result = Select[list1, MemberQ[list2, #[[1]]] &]; > > The problem is that now I would like to use it with big lists. list1 > is over 1.000.000 elements long and list2 is about 500.000 elements > long. Ordering is not a problem, I could resort the lists. > > Any hint to extract this list faster? It seems to take a lot of time > (estimation is about 5 hours and I have to do it repeatedly) > > Thanks! Good day, Assuming both lists are sorted, you could match them this way (an example with size n=10000): In[1]:= n = 10000; list1 = Table[{k,RandomInteger[{1,n}]},{k,1,n}] ; list2 = list1[[All,1]][[Floor[n/4];; Floor[3n/4]]] ; In[4]:= (list3 = {#,True}& /@ list2; list4 = {#,False}& /@ Complement[list1[[All,1]],list2]; list5 = Union[list3, list4]; list6 = Transpose[{list1, list5}]; Cases[list6,{{_,_},{_,True}}][[All,1]])//Timing//Short Out[4]//Short= {0.094,{{2500,8261},{2501,8320},<<4997>>,{7499,6791}, {7500,943}}} In[5]:= Select[list1,MemberQ[list2,#[[1]]]&]//Timing//Short Out[5]//Short= {9.219,{{2500,8261},{2501,8320},<<4997>>,{7499,6791}, {7500,943}}} In[7]:= 9.219/0.094 Out[7]= 98.0745 Matching lists seems to be 100 times faster then MemberQ hth -- Valeri Astanoff