检查两个不同字典中的相同密钥，打印密钥和两个值

Question

不确定为什么我不能弄清楚这一点，但我希望程序查看2个字典，如果密钥匹配，我希望它打印密钥随着来自像这样两个字典的值..

[结果]

a 1 2 
b 2 4 
c 3 6 
d 4 8 

[CODE]

from collections import Counter 

a = Counter({'a':1,'b':2,'c':3,'d':4}) 
b = Counter({'a':2,'b':4,'c':6,'d':8}) 

for key in b.most_common(): 
    if set(a.keys()) == set(b.keys()): 
     print(key) 

Answer 1

您可以通过这两个按键的set intersection迭代：

>>> set(a.keys()) & set(b.keys()) 
set(['a', 'c', 'b', 'd']) 

下面是与类型的字典是不完全类似的例子：

from collections import Counter 

a = Counter({'a':1,'b':2,'c':3,'d':4,'e':5}) 
b = Counter({'a':2,'b':4,'c':6,'d':8,'f':7}) 

common_keys = set(a.keys()) & set(b.keys()) 

for key in common_keys: 
    print("%s %d %d" % (key, a[key], b[key])) 
# a 1 2 
# c 3 6 
# b 2 4 
# d 4 8 

请注意，转换到set不preverse的Counter顺序。

你可以转换set为list，并根据这两个值的总和进行排序：

a = {'a':1,'b':2,'c':3,'d':4,'e':5} 
b = {'a':2,'b':4,'c':6,'d':8,'f':7} 

common_keys = list(set(a.keys()) & set(b.keys())) 

common_keys.sort(key=lambda k: a[k] + b[k], reverse=True) 

for key in common_keys: 
    print("%s %d %d" % (key, a[key], b[key])) 
# d 4 8 
# c 3 6 
# b 2 4 
# a 1 2 

最后，如果你不关心根据其值之和排序的关键，这里是你的代码稍加修改的版本：

from collections import Counter 

a = Counter({'a':1,'b':2,'c':3,'d':4}) 
b = Counter({'a':2,'b':4,'c':6,'d':8}) 

for key, a_value in a.most_common(): 
    if key in b: 
     print("%s %d %d" % (key, a_value, b[key])) 

Answer 2

common_keys = sorted(set(a.keys()).intersection(b.keys())) 

for key in common_keys: 
    print(key, a[key], b[key]) 

Answer 3

你几乎在那里，你if条件是关闭，most_common返回元组列表，所以你需要解压的键值对

>>> a = Counter({'a':1,'b':2,'c':3,'d':4}) 
>>> b = Counter({'a':2,'b':4,'c':6,'d':8}) 
>>> for key, bval in b.most_common(): 
...  if key in a: 
...   print(key, a[key], bval) 
... 
d 4 8 
c 3 6 
b 2 4 
a 1 2 
>>> 

如果你不关心顺序，（即你不需要拨打most_common），你可以迭代交叉点。在Python 3，这是很简单的：

>>> a.keys() & b.keys() 
{'d', 'a', 'c', 'b'} 
>>> 

因为.keys实际上返回更像一组比列表的视图。

>>> for common_key in a.keys() & b.keys(): 
...  print(common_key, a[common_key], b[common_key]) 
... 
d 4 8 
a 1 2 
c 3 6 
b 2 4 
>>>