Mr_and_Mrs_D Mr_and_Mrs_D - 2 years ago 77
Python Question

Sort a subset of a python list to have the same relative order as in other list

So having a list say

b = [b1, b2, b3]
I want to be able to sort a list
in such a way that all
's that also exist in
have the same relative order as in
- leaving the rest of
's elements alone. So

a = [ b1, x, b3, y, b2] -> [ b1, x, b2, y, b3]
a = [ b1, x, b2, y, b3] -> no change
a = [ b1, x, y, b2] -> no change
a = [ b3, x, b1, y, b2] -> [ b1, x, b2, y, b3]

b of course may be a tuple or any other ordered structure. What I came up with

bslots = dict((x, a.index(x)) for x in a if x in b)
bslotsSorted = sorted(bslots.keys(), key=lambda y: b.index(y))
indexes = sorted(bslots.values())
for x,y in zip(bslotsSorted, indexes):
a[y] = x

is clumsy and O(n^2)

Answer Source
  • First create a dictionary using items from b where the key is the item and value is its index, we will use this to sort the matched items in a later on.

  • Now filter out item from a that are present in that dict, dict provides O(1) lookup.

  • Now sort this list of filtered items and convert it to an iterator.

  • Now loop over a again and for each item check if is present in dict then fetch its value from iterator otherwise use it as is.

def solve(a, b):
    dct = {x: i for i, x in enumerate(b)}
    items_in_a = [x for x in a if x in dct]
    it = iter(items_in_a)
    return [next(it) if x in dct else x for x in a]
>>> b = ['b1', 'b2', 'b3']
>>> a = [ 'b1', 'x', 'b3', 'y', 'b2']
>>> solve(a, b)
['b1', 'x', 'b2', 'y', 'b3']
>>> a = [ 'b1', 'x', 'b2', 'y', 'b3']
>>> solve(a, b)
['b1', 'x', 'b2', 'y', 'b3']
>>> a = [ 'b1', 'x', 'y', 'b2']
>>> solve(a, b)
['b1', 'x', 'y', 'b2']
>>> a = [ 'b3', 'x', 'b1', 'y', 'b2']
>>> solve(a, b)
['b1', 'x', 'b2', 'y', 'b3']

Overall time complexity is going to be max of (O(len(a)), O(len(b)), O(items_in_a_length log items_in_a_length).

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