例如,我有两个字典:
Dict A: {'a': 1, 'b': 2, 'c': 3}
Dict B: {'b': 3, 'c': 4, 'd': 5}
我需要一种python的方式来“组合”两个字典,这样的结果是:
{'a': 1, 'b': 5, 'c': 7, 'd': 5}
也就是说:如果一个键在两个字典中都出现,则将它们的值相加,如果它只在一个字典中出现,则保留其值。
例如,我有两个字典:
Dict A: {'a': 1, 'b': 2, 'c': 3}
Dict B: {'b': 3, 'c': 4, 'd': 5}
我需要一种python的方式来“组合”两个字典,这样的结果是:
{'a': 1, 'b': 5, 'c': 7, 'd': 5}
也就是说:如果一个键在两个字典中都出现,则将它们的值相加,如果它只在一个字典中出现,则保留其值。
当前回答
一个更通用的解决方案,它也适用于非数字值:
a = {'a': 'foo', 'b':'bar', 'c': 'baz'}
b = {'a': 'spam', 'c':'ham', 'x': 'blah'}
r = dict(a.items() + b.items() +
[(k, a[k] + b[k]) for k in set(b) & set(a)])
或者更普通一点:
def combine_dicts(a, b, op=operator.add):
return dict(a.items() + b.items() +
[(k, op(a[k], b[k])) for k in set(b) & set(a)])
例如:
>>> a = {'a': 2, 'b':3, 'c':4}
>>> b = {'a': 5, 'c':6, 'x':7}
>>> import operator
>>> print combine_dicts(a, b, operator.mul)
{'a': 10, 'x': 7, 'c': 24, 'b': 3}
其他回答
另外,请注意a.update(b)比a + b快2倍
from collections import Counter
a = Counter({'menu': 20, 'good': 15, 'happy': 10, 'bar': 5})
b = Counter({'menu': 1, 'good': 1, 'bar': 3})
%timeit a + b;
## 100000 loops, best of 3: 8.62 µs per loop
## The slowest run took 4.04 times longer than the fastest. This could mean that an intermediate result is being cached.
%timeit a.update(b)
## 100000 loops, best of 3: 4.51 µs per loop
>>> A = {'a':1, 'b':2, 'c':3}
>>> B = {'b':3, 'c':4, 'd':5}
>>> c = {x: A.get(x, 0) + B.get(x, 0) for x in set(A).union(B)}
>>> print(c)
{'a': 1, 'c': 7, 'b': 5, 'd': 5}
来自python 3.5:合并和求和
Thanks to @tokeinizer_fsj that told me in a comment that I didn't get completely the meaning of the question (I thought that add meant just adding keys that eventually where different in the two dictinaries and, instead, i meant that the common key values should be summed). So I added that loop before the merging, so that the second dictionary contains the sum of the common keys. The last dictionary will be the one whose values will last in the new dictionary that is the result of the merging of the two, so I thing the problem is solved. The solution is valid from python 3.5 and following versions.
a = {
"a": 1,
"b": 2,
"c": 3
}
b = {
"a": 2,
"b": 3,
"d": 5
}
# Python 3.5
for key in b:
if key in a:
b[key] = b[key] + a[key]
c = {**a, **b}
print(c)
>>> c
{'a': 3, 'b': 5, 'c': 3, 'd': 5}
可重用代码
a = {'a': 1, 'b': 2, 'c': 3}
b = {'b': 3, 'c': 4, 'd': 5}
def mergsum(a, b):
for k in b:
if k in a:
b[k] = b[k] + a[k]
c = {**a, **b}
return c
print(mergsum(a, b))
使用集合。计数器:
>>> from collections import Counter
>>> A = Counter({'a':1, 'b':2, 'c':3})
>>> B = Counter({'b':3, 'c':4, 'd':5})
>>> A + B
Counter({'c': 7, 'b': 5, 'd': 5, 'a': 1})
计数器基本上是dict的一个子类,因此您仍然可以使用它们执行通常使用该类型执行的所有其他操作,例如遍历它们的键和值。
是什么:
def dict_merge_and_sum( d1, d2 ):
ret = d1
ret.update({ k:v + d2[k] for k,v in d1.items() if k in d2 })
ret.update({ k:v for k,v in d2.items() if k not in d1 })
return ret
A = {'a': 1, 'b': 2, 'c': 3}
B = {'b': 3, 'c': 4, 'd': 5}
print( dict_merge_and_sum( A, B ) )
输出:
{'d': 5, 'a': 1, 'c': 7, 'b': 5}