I have a very large collection of (p, q) tuples that I would like to convert into a dictionary of lists where the first item in each tuple is a key that indexes a list that
Here is the iterator style of doing it
>>> mylist=[(1, 2), (1, 3), (2, 3)] >>> from itertools import groupby >>> from operator import itemgetter >>> mylist=[(1, 2), (1, 3), (2, 3)] >>> groupby(mylist,itemgetter(0)) >>> list(_) [(1, <itertools._grouper object at 0xb7d402ec>), (2, <itertools._grouper object at 0xb7c716ec>)]
collections.defaultdict works like this:
from collections import defaultdict
dic = defaultdict(list)
for i, j in tuples:
dic[i].append(j)
similar for the dicts:
a, b = {1:[2, 3], 2:[3]}, {1:[4], 3:[1]}
de = defaultdict(list, a)
for i, j in b.items():
de[i].extend(j)
I wanted these done in one line just for fun:
>>> from itertools import groupby
>>> t=(1, 2), (1, 3), (2, 3)
>>> [(i,[x for _,x in list(f)]) for i,f in groupby(sorted(t),lambda t: t[0])]
[(1, [2, 3]), (2, [3])]
>>> b={1:[2, 3], 2:[3]}, {1:[4], 3:[1]}
>>> dict([(key,sum([i[1::][0] for i in elements],[])) for key,elements in groupby(sorted(b[0].items()+b[1].items()),lambda t: t[0])])
{1: [2, 3, 4], 2: [3], 3: [1]}
If the list of tuples is sorted, itertools.groupby
, as suggested by @gnibbler, is not a bad alternative to defaultdict
, but it needs to be used differently than he suggested:
import itertools
import operator
def lot_to_dict(lot):
key = operator.itemgetter(0)
# if lot's not sorted, you also need...:
# lot = sorted(lot, key=key)
# NOT in-place lot.sort to avoid changing it!
grob = itertools.groupby(lot, key)
return dict((k, [v[1] for v in itr]) for k, itr in grob)
For "merging" dicts of lists into a new d.o.l...:
def merge_dols(dol1, dol2):
keys = set(dol1).union(dol2)
no = []
return dict((k, dol1.get(k, no) + dol2.get(k, no)) for k in keys)
I'm giving []
a nickname no
to avoid uselessly constructing a lot of empty lists, given that performance is important. If the sets of the dols' keys overlap only modestly, faster would be:
def merge_dols(dol1, dol2):
result = dict(dol1, **dol2)
result.update((k, dol1[k] + dol2[k])
for k in set(dol1).intersection(dol2))
return result
since this uses list-catenation only for overlapping keys -- so, if those are few, it will be faster.
defaltdict to the rescue (as usual)
from collections import defaultdict
my_dict = defaultdict(list)
for key,value in original_list:
my_dict[key].append(value)
Combining the two dicts can be done like this (note that duplicates will be preserved):
for key,value in orig_dict:
new_dict[key].extend(value)