Sorting a list depending of the indexes of another sorted list

[Peter Otten]

Steven D'Aprano wrote:

>> The following relies on undocumented (I hope) behaviour:

>>>>> preferences = [10, 30, 20]
>>>>> hosts = [ "mx1.domain.com", "anotherhost.domain.com", 
>>... "mx2.domain.com"] 
>>>>> hosts.sort(key=lambda x, p=iter(preferences).next: p())
>>>>> preferences.sort()
>>>>> hosts
>> ['mx1.domain.com', 'mx2.domain.com', 'anotherhost.domain.com']

> What bit are you suggesting is undocumented?

The lambda spits out the items in preferences in the same order as they
occur in that list. If hosts.sort(key=...) in its C-implemented decoration
phase would iterate over the items in hosts in, say, reverse order hosts
would not be sorted correctly. Here's an illustration in Python:

>>> def dsu(items, key, reorder=lambda x: x):
...     for i in reorder(range(len(items))):
...             items[i] = key(items[i]), items[i]
...     items.sort()
...     items[:] = [v for k, v in items]
...     return items
... 
>>> dsu([1,2,3], lambda x, n=iter("acb").next: n())
[1, 3, 2]
>>> dsu([1,2,3], lambda x, n=iter("acb").next: n(), reversed)
[3, 1, 2]

Peter