I have one defaultdict(list) and other normal dictionary
A = {1:["blah", "nire"], 2:["fooblah"], 3:["blahblah"]}
B = {1: "something" ,2:"somethingsomething"}
now lets say that i have something like this
missing_value = "fill_this"
Now, first I want to find what are the keys in B missing from A (like 3 is missing)
and then set those keys to the values missing_value?
What is the pythonic way to do this?
Thanks
You can use setdefault:
for k in A:
B.setdefault(k, "fill_this")
This is essentially the same as the longer:
for k in A:
if k not in B:
B[k] = "fill_this"
However, since setdefault only needs to lookup each k once, setdefault is faster than this "test&set" solution.
Alternatively (and probably slower), determine the set difference and set (no pun intended) those values:
for k in set(A).difference(B):
B[k] = "fill_this"
defaultdict implements its own setdefault.
if k not in B.B.setdefault could even be put out of the loop in some name e.x. B_set_def to avoid methods lookups on B.B.setdefault is probably premature optimization. It's explicitly mentioned in Dubious Python.The solution is to go through A and update B where necessary. It would have O(len(A)) complexity:
for key in A:
if key not in B:
B[key] = missing_value
set in memory. I've already updated the answer to reflect that. set(A)-set(b) is O(len(A)) complexity because the algorithm goes through set(A) and looks for each element of set(A) if it's in set(B) which is O(1) for each element. But eventually you are right :) Because before the sets are subtracted, set(B) is created which is O(len(B)). I'll update the answer.Here's one way:
def test():
A = {1:"blah", 2:"fooblah", 3:"blahblah"}
B = {1: "something" ,2:"somethingsomething"}
keys=set(A.keys()).difference(set(B.keys()))
for k in keys:
B[k]="missing"
print (B)
When I've seen a need for this, there were a standard set of keys to be checked in a dict of dicts. If that's the case and if performance is not a significant factor, I think this is the cleanest syntax.
template = {k: default for k in domain}
for k, d in dicts.items():
dicts[k] = template.copy().update(d)
