So I\'ve got a comprehension to the effect of:
dict((x.key, x.value) for x in y)
The problem, of course, is that if there\'s multiple
You can add the elements one by one to a dictionary that contains empty lists by default:
import collections
result_dict = collections.defaultdict(list)
for x in y:
result_dict[x.key].append(x.value)
You can also do something very similar without having to use the collections module:
result_dict = {}
for x in y:
result_dict.setdefault(x.key, []).append(x.value)
but this is arguably slightly less legible.
An equivalent, more legible (no need to "parse" the less common setdefault
) but more pedestrian, base Python approach is:
result_dict = {}
for x in y:
if x.key not in result_dict:
result_dict[x.key] = []
result_dict[x.key].append(x.value)
The first solution is clearly the preferred one, as it is at the same time concise, legible, and fast.
I'm not saying it's the right thing to do, but, just out of sheer intellectual curiosity..
You can use itertools.groupby
and lambda
to do it in one dict comprehension, if that's what you really want to do: (where l
is the list of tuples you want to make a dict
out of:
dict((k, [v[1] for v in vs]) for (k, vs) in itertools.groupby(l, lambda x: x[0]))
Nope. You cannot do this in a comprehension.
But you can use itertools.groupby
.