python compare a list of lists efficiently

问题:

I have a long list of long lists so efficiency is an issue for me. I wondered if there was a neater way of comparing a list of lists other than looping over a list within a loop of the same list (easier to see by example)

matchList=[]
myList = [ ('a',[1,2,3]), ('b', [2,3,4]), ('c', [3,4,5]), ('d', [4,5,6]) ]

tup_num=1
for tup in myList:
    for tup2 in myList[tup_num:]:
        id=str(tup[0])+':'+str(tup2[0])
        matches=set(tup[1]) & set(tup2[1])
        matchList.append((id,matches))
    tup_num+=1

print matchList

Output:

[('a:b', set([2, 3])), ('a:c', set([3])), ('a:d', set([])), ('b:c', set([3, 4])), ('b:d', set([4])), ('c:d', set([4, 5]))]

This works and doesn't repeat comparisons but I'm sure there must be a better way of doing it.

Cheers

回答1:

Using itertools.combinations:

>>> import itertools
>>> matchList = []
>>> myList = [('a',[1,2,3]), ('b', [2,3,4]), ('c', [3,4,5]), ('d', [6,7,8])]
>>> matchList = [
...     ('{}:{}'.format(key1, key2), set(lst1) & set(lst2))
...     for (key1, lst1), (key2, lst2) in itertools.combinations(myList, 2)
... ]
>>> matchList
[('a:b', set([2, 3])), ('a:c', set([3])), ('a:d', set([])), ('b:c', set([3, 4])), ('b:d', set([])), ('c:d', set([]))]

回答2:

• Convert those list to sets at first. That's an O(n) operation, which better avoided.
• itertools.combinations is faster and easier.

Like this:

>>> from itertools import combinations
>>> l
[('a', [1, 2, 3]), ('b', [2, 3, 4]), ('c', [3, 4, 5]), ('d', [6, 7, 8])]
>>> l = [(i, set(j)) for i, j in l]
>>> l
[('a', {1, 2, 3}), ('b', {2, 3, 4}), ('c', {3, 4, 5}), ('d', {8, 6, 7})]
>>> [("%s:%s" % (l1[0], l2[0]), l1[1] & l2[1]) for l1, l2 in combinations(l, 2)]
[('a:b', {2, 3}), ('a:c', {3}), ('a:d', set()), ('b:c', {3, 4}), ('b:d', set()), ('c:d', set())]

回答3:

Using composition and generators makes it clear:

from itertools import combinations

matchList = []
myList = [ ('a',[1,2,3]), ('b', [2,3,4]), ('c', [3,4,5]), ('d', [4,5,6]) ]

def sets(items):
  for name, tuple in items:
    yield name, set(tuple)

def matches(sets):
  for a, b in combinations(sets, 2):
    yield ':'.join([a[0], b[0]]), a[1] & b[1]

print list(matches(sets(myList)))


>>> [('a:b', set([2, 3])), ('a:c', set([3])), ('a:d', set([])), ('b:c', set([3, 4])), ('b:d', set([4])), ('c:d', set([4, 5]))]