dict1 = {a: 5, b: 7}
dict2 = {a: 3, c: 1}
result {a:8, b:7, c:1}
我该怎么获得结果?
dict1 = {a: 5, b: 7}
dict2 = {a: 3, c: 1}
result {a:8, b:7, c:1}
我该怎么获得结果?
这是一个能够完成此操作的一行代码:
dict1 = {'a': 5, 'b': 7}
dict2 = {'a': 3, 'c': 1}
result = {key: dict1.get(key, 0) + dict2.get(key, 0)
for key in set(dict1) | set(dict2)}
# {'c': 1, 'b': 7, 'a': 8}
请注意,set(dict1) | set(dict2)
是您两个字典中所有键的集合。而dict1.get(key, 0)
则是如果该键存在,则返回dict1[key]
,否则返回0
。
这仅适用于较新版本的Python:
{k: dict1.get(k, 0) + dict2.get(k, 0) for k in dict1.keys() | dict2.keys()}
+
的 collections.Counter
:>>> from collections import Counter
>>> dict1 = Counter({'a': 5, 'b': 7})
>>> dict2 = Counter({'a': 3, 'c': 1})
>>> dict1 + dict2
Counter({'a': 8, 'b': 7, 'c': 1})
如果你真的想要将结果作为字典返回,随后你可以将其转换回去:
>>> dict(dict1 + dict2)
{'a': 8, 'b': 7, 'c': 1}
认为这很简单。
a={'a':3, 'b':5}
b= {'a':4, 'b':7}
{i:a[i]+b[i] for i in a.keys()}
{'a': 7, 'b': 12}
这里有一个不错的函数:
def merge_dictionaries(dict1, dict2):
merged_dictionary = {}
for key in dict1:
if key in dict2:
new_value = dict1[key] + dict2[key]
else:
new_value = dict1[key]
merged_dictionary[key] = new_value
for key in dict2:
if key not in merged_dictionary:
merged_dictionary[key] = dict2[key]
return merged_dictionary
通过编写:
dict1 = {'a': 5, 'b': 7}
dict2 = {'a': 3, 'c': 1}
result = merge_dictionaries(dict1, dict2)
结果将是:
{'a': 8, 'b': 7, 'c': 1}
Here is another approach but it is quite lengthy!
d1 = {'a': 5, 'b': 7}
d2 = {'a': 3, 'c': 1}
d={}
for i,j in d1.items():
for k,l in d2.items():
if i==k:
c={i:j+l}
d.update(c)
for i,j in d1.items():
if i not in d:
d.update({i:j})
for m,n in d2.items():
if m not in d:
d.update({m:n})
{
**dict1,
**{ k:(dict1[k]+v if k in dict1 else v)
for k,v in dict2.items() }
}