这是一个相当直接的方式使用字典理解：

$ b $对于k，v在d.items（））中，b

  sums = {k：sum（i for v in is ifstat（i，int））
 

或在Python 2.6及以下版本：

  sums = dict（（k，sum（i for v in if ifinstance（i，int）））for k，v in d.items（））
 / pre> 

示例：
 >>> {k：sum（i for v in if ifinstance（i，int））for k，v in d.items（）} 
 {'a'：6，'c'：7，'b' 7，'e'：4，'d'：7，'g'：4，'f'：4} 
 
 
I have a dict, 

d = {'a': [4,'Adam', 2], 'b': [3,'John', 4], 'c': [4,'Adam', 3], 'd': [4,'Bill', 3], 'e': [4,'Bob'], 'f': [4, 'Joe'], 'g': [4, 'Bill']}

Is there any quick way to get a sum of the numbers in each of the lists in the dictionary? 

For example, a should return 6, b should return 7, so on.

Currently, I am doing this.
for i in d:
    l2=[]
    for thing in d[i]:
        if type(thing) == int:
            l2.append(thing)
        print sum(l2)
Possible for a quicker fix than having to go through each time and append the numbers to a list?

Thanks!
Here is a fairly straight forward way using a dictionary comprehension:
sums = {k: sum(i for i in v if isinstance(i, int)) for k, v in d.items()}
Or on Python 2.6 and below:
sums = dict((k, sum(i for i in v if isinstance(i, int))) for k, v in d.items())
Example:
>>> {k: sum(i for i in v if isinstance(i, int)) for k, v in d.items()}
{'a': 6, 'c': 7, 'b': 7, 'e': 4, 'd': 7, 'g': 4, 'f': 4}