Paul McGuire wrote: > "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] > >
Looks like our answers crossed-over. Must learn about sets in Python... Thanks very much, Cameron. -- http://mail.python.org/mailman/listinfo/python-list
