问题
The problem:
>>> a = dict(a=1,b=2 )
>>> b = dict( b=3,c=2)
>>> c = ???
c = {'a': 1, 'b': 5, 'c': 2}
So, the idea is two add to dictionaries by int/float values in the shortest form. Here's one solution that I've devised, but I don't like it, cause it's long:
c = dict([(i,a.get(i,0) + b.get(i,0)) for i in set(a.keys()+b.keys())])
I think there must be a shorter/concise solution (maybe something to do with reduce and operator module? itertools?)... Any ideas?
Update: I'm really hoping to find something more elegant like "reduce(operator.add, key = itemgetter(0), a+b)". (Obviously that isn't real code, but you should get the idea). But it seems that may be a dream.
Update: Still loking for more concise solutions. Maybe groupby can help? The solution I've come up with using "reduce"/"groupby" isn't actually concise:
from itertools import groupby
from operator import itemgetter,add
c = dict( [(i,reduce(add,map(itemgetter(1), v))) \
for i,v in groupby(sorted(a.items()+b.items()), itemgetter(0))] )
回答1:
Easiest to just use a Counter
>>> from collections import Counter
>>> a = dict(a=1,b=2 )
>>> b = dict( b=3,c=2)
>>> Counter(a)+Counter(b)
Counter({'b': 5, 'c': 2, 'a': 1})
>>> dict(Counter({'b': 5, 'c': 2, 'a': 1}))
{'a': 1, 'c': 2, 'b': 5}
回答2:
solving not in terms of "length" but performance, I'd do the following:
>>> from collections import defaultdict
>>> def d_sum(a, b):
d = defaultdict(int, a)
for k, v in b.items():
d[k] += v
return dict(d)
>>> a = {'a': 1, 'b': 2}
>>> b = {'c': 2, 'b': 3}
>>> d_sum(a, b)
{'a': 1, 'c': 2, 'b': 5}
it's also py3k-compatible, unlike your original code.
回答3:
In my first impression, I will write:
>>> c = a.copy()
>>> for k in b: c[k] = c.get(k, 0) + b[k]
回答4:
If you want short code, you're there.
If you want clean code, inherit from Ber's defaultdict
and overload __add__
:
from collections import defaultdict
class summable(defaultdict):
def __add__(self, rhs):
new = summable()
for i in (self.keys() + rhs.keys()):
new[i] = self.get(i, 0) + rhs.get(i, 0)
return new
a = summable(int, a=1, b=2)
b = summable(int, b=3, c=4)
c = a + b
print c
Gives:
>>>
defaultdict(None, {'a': 1, 'c': 4, 'b': 5})
>>>
回答5:
I think one line of code is already pretty short :)
I may become "half a line", it you use defaultdict and remove some unnecessary list and set creations:
from collections import defaultdict
a = defaultdict(int, a=1, b=2)
b = defaultdict(int, b=3, c=4)
c = dict((k, a[k]+b[k]) for k in (a.keys() + b.keys()))
print c
回答6:
The first thing I think of is a bit more efficient and (IMO) a bit more elegant, but still too much typing. Actually, it's about equivalent to kcwu's.
c = reduce(lambda(d, k): [d.update({k: d.get(k, 0) + b[k]}), d][1], b, a.copy())
It's really a shame that dict.update
doesn't return self
. I guess it's not the Python way. If it did, the [..., d][1]
trick would be unnecessary.
Perl: "Easy things are easy, hard things are possible"
%a = (a => 1, b => 2);
%b = (b => 3, c => 2);
%c = (%a, map {$_ => $a{$_} + $b{$_}} keys %b);
Haskell:
"Easy things are hard, hard things are easy""Hard things are easy, the impossible just happened"
import qualified Data.Map as M
a = M.fromList [('a', 1), ('b', 2)]
b = M.fromList [('b', 3), ('c', 2)]
c = M.unionWith (+) a b
回答7:
Comment for @John Pirie's answer:
Here's implementation that doesn't use (self.keys() + rhs.keys())
:
from collections import defaultdict
class sumdict(defaultdict):
def __add__(self, rhs):
d = self.copy()
d += rhs
return d
__radd__ = lambda self, lhs: self + lhs
def __iadd__(self, rhs):
for k, v in rhs.items():
self[k] += v
return self
a = sumdict(int, a=1, b=2)
b = dict(b=3, c=4)
c = b + a
a += b
assert a == c == {'a': 1, 'c': 4, 'b': 5} != b
回答8:
def GenerateSum():
for k in set(a).union(b):
yield k, a.get(k, 0) + b.get(k, 0)
e = dict(GenerateSum())
print e
or, with a one liner:
print dict((k, a.get(k,0) + b.get(k,0)) for k in set(a).union(b))
来源:https://stackoverflow.com/questions/877295/python-dict-add-by-valuedict-2