Sorting by item_in_another_list
cameron.walsh at gmail.com
Tue Oct 24 06:42:01 CEST 2006
Paul McGuire wrote:
> "Cameron Walsh" <cameron.walsh at gmail.com> wrote in message
> news:ehk3p8$j3q$1 at enyo.uwa.edu.au...
>> 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:
>> 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
> 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
> [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,
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