Sorting by item_in_another_list

[Cameron Walsh]

Paul McGuire wrote:
> "Cameron Walsh" <cameron.walsh at gmail.com> wrote in message 
> news:ehk3p8$j3q$1 at enyo.uwa.edu.au...
>> Hi,
>>
>> I have two lists, A and B, such that B is a subset of A.
>>
>> I wish to sort A such that the elements in B are at the beginning of A, 
>> and keep the existing order otherwise, i.e. stable sort.  The order of 
>> elements in B will always be correct.
>>
>> for example:
>>
>> A = [0,1,2,3,4,5,6,7,8,9,10]
>> B = [2,3,7,8]
>>
>> desired_result = [2,3,7,8,0,1,4,5,6,9,10]
>>
>>
>> At the moment I have defined a comparator function:
>>
>> def sort_by_in_list(x,y):
>> ret = 0
>> if x in B:
>> ret -= 1
>> if y in B:
>> ret += 1
>> return ret
>>
>> and am using:
>>
>> A.sort(sort_by_in_list)
>>
>> which does produce the desired results.
>>
>> I do now have a few questions:
>>
>> 1.)  Is this the most efficient method for up to around 500 elements? If 
>> not, what would be better?
>> 2.)  This current version does not allow me to choose a different list for 
>> B.  Is there a bind_third function of some description that I could use to 
>> define a new function with 3 parameters, feed it the third (the list to 
>> sort by), and have the A.sort(sort_by_in_list) provide the other 2 
>> variables?
>>
> 
> Think in Python.  Define a function to take the list, and have that function 
> return the proper comparison function.  This gives me the chance to also 
> convert the input list to a set, which will help in scaling up my list to 
> hundreds of elements.  See below.
> 
> -- Paul
> 
> 
> def sort_by_in_list(reflist):
>     reflist = set(reflist)
>     def sort_by_in_list_(x,y):
>         ret = 0
>         if x in reflist: ret -= 1
>         if y in reflist: ret += 1
>         return ret
>     return sort_by_in_list_
> 
> A = [0,1,2,3,4,5,6,7,8,9,10]
> B = [2,3,7,8]
> A.sort( sort_by_in_list(B) )
> print A
> 
> Gives:
> [2, 3, 7, 8, 0, 1, 4, 5, 6, 9, 10]
> 
> 

Looks like our answers crossed-over.  Must learn about sets in Python...

Thanks very much,

Cameron.