合并/追加共享共同项目的列表

Question

The title may be misleading, so feel free to change the wording when the terminology for the real problem arises. =)

In this case, I am aware that the lists can probably be interchanged with tuples, for the most part. The end result can be any iterable as far as I'm concerned.

I have two lists-of-lists. Suppose they are:

list_a = [[1, 'f00d'], [2, 'dead'], [3, 'beef']]
list_b = [[1, 'frankenbeans'], [2, 'chickensoup'], [3, 'spaceballs']]

Neither list is necessarily the same length, nor is guaranteed that they contain a common first element.

What I'm trying to do is create a new list-of-lists/list-of-tuples/list-of-dicts/whatever, as such:

list_c = [[1, 'f00d', 'frankenbeans'], [2, 'dead', 'chickensoup'], [3, 'beef', 'spaceballs']

Updated : Basically, I know the position of the common "ID" in these lists, though it is not necessarily sequential, nor are the lists-of-lists in the same order (but is an integer). I'm looking for an efficient way to create a new set of the sub-lists, based on that common ID.

The naive way:

new_list = []
for list_a_list in list_a:
  for list_b_list in list_b:
    if list_a_list[0] = list_b_list[0]:
      new_list.append([list_a_list[0], list_a_list[1], list_b_list[1]])

... or some such. Giving me the feeling that there's a much "smarter" way to do this, but I kinda suck at that.

Update:
Does it add any bearing if I mention that the list-of-lists each carry thousands to a million items at a time?

Answer 1

from collections import defaultdict
from itertools import chain

final = defaultdict(list)

for idx, value in chain(l1, l2):
  final[idx].append(value)

# and if you have to have a list of lists at the end
finalList = [[k] + v for k, v in final.iteritems()]

Answer 2

Your input lists should be dictionaries in the first place:

dict_a = dict(list_a)
dict_b = dict(list_b)
dict_c = dict((k, [v, dict_b[k]]) for k,v in dict_a.items())

If keys are not guaranteed to occur in both lists, you'll have to be a little more careful:

all_keys = set(dict_a.keys()) | set(dict_b.keys())
dict_c = dict((k, (dict_a.get(k), dict_b.get(k))) for k in all_keys)

For example, for list_a = [(1, 'a')] and list_b = [(1, 'b'), (2, 'c')] , the above would set dict_c to {1: ('a', 'b'), 2: (None, 'c')} .

Answer 3

itertools.groupby() is helpful for this kind of task:

from itertools import groupby, chain
from operator import itemgetter

list_a = [[1, 'f00d'], [2, 'dead'], [3, 'beef']]
list_b = [[1, 'frankenbeans'], [2, 'chickensoup'], [3, 'spaceballs']]

combined = [(k, [v[1] for v in g]) for k, g in
            groupby(sorted(list_a+list_b), key=itemgetter(0))]

print combined

Note that it was necessary to create a new sorted list combining list_a and list_b before we can use groupby, since groupby assumes that the list will already be sorted by the key.