Sorting by item_in_another_list

Cameron Walsh cameron.walsh at gmail.com
Tue Oct 24 06:19:35 CEST 2006

```Cameron Walsh wrote:
> 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?
>
>
> Regards to all,
>
> Cameron.

Well I found an answer to the second question with the following:

>>> A=[0,1,2,3,4,5,6,7,8,9,10]
>>> B=[2,3,7,8]
>>> def sort_by_in_list(in_list):
def ret_function(x,y):
ret = 0
if x in in_list:
ret -= 1
if y in in_list:
ret += 1
return ret
return ret_function

>>> A.sort(sort_by_in_list(B))
>>> A
[2, 3, 7, 8, 0, 1, 4, 5, 6, 9, 10]

Hope it helps someone,

Cameron.

```