"Cameron Walsh" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> 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]
--
http://mail.python.org/mailman/listinfo/python-list
