Re: Extracting some elements, members of another list

• To: mathgroup at smc.vnet.net
• Subject: [mg112435] Re: Extracting some elements, members of another list
• Date: Wed, 15 Sep 2010 04:40:35 -0400 (EDT)

```Dispatch should be a good candidate:

In[1]:= list1 = {{phone1, data1}, {phone2, data2}, {phone3, data3},
{phone4, data4}, {phone5, data5}, {phone6, data6}, {phone7, data7},
{phone8, data8}};
rules = Dispatch[Apply[#1 -> {##1} & , list1, {1}]]
Out[2]= Dispatch[{phone1 -> {phone1, data1}, phone2 -> {phone2, data2},
phone3 -> {phone3, data3}, phone4 -> {phone4, data4},
phone5 -> {phone5, data5}, phone6 -> {phone6, data6},
phone7 -> {phone7, data7}, phone8 -> {phone8, data8}}]

In[5]:= list2 = {phone2, phone3, phone7};
list2 /. rules
Out[6]= {{phone2, data2}, {phone3, data3}, {phone7, data7}}

2010/9/14 Nacho <ncc1701zzz at gmail.com>

> Hi!
>
>
> 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!
>
>

```

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