如何计算python字典每个键中不同值的平均值？

Question

I have a dictionary. I want to calculate average of values for each key and print result so that the result shows key and associated average. The following code calculates mean but I don't know how to associate key with the average. My desired answer is Mean = {22:average1, 23:average2, 24:average3}.

          mydict = {22: [1, 0, 0, 1], 23: [0, 1, 2, 1, 0], 24: [3, 3, 2, 1, 0]}

          Mean =[float(sum(values)) / len(values) for key, values in 
          mydict.iteritems()]

Answer 1

You were on the right track. You just needed a dictionary comprehension instead of a list comp.

_dict = {22: [1, 0, 0, 1], 23: [0, 1, 2, 1, 0], 24: [3, 3, 2, 1, 0]}

mean = {key : float(sum(values)) / len(values) for key, values in _dict.iteritems()}
print(mean) 
{22: 0.5, 23: 0.8, 24: 1.8}

---

Notes:

1. .iteritems is replaced with .items in python3
2. (Statutory Warning) Do not use dict as a variable name, it shadows the builtin class with the same name.

Answer 2

Don't use a list comprehension. Use a dictionary comprehension to calculate the average of each list. You can also from __future__ import division to avoid having to use float :

>>> from __future__ import division
>>> d = {22: [1, 0, 0, 1], 23: [0, 1, 2, 1, 0], 24: [3, 3, 2, 1, 0]}
>>> mean = {k: sum(v) / len(v) for k, v in d.iteritems()}
>>> mean
{22: 0.5, 23: 0.8, 24: 1.8}

Answer 3

Update for Python 3.4+

>>> from statistics import mean    # Python 3.4+
>>> d = {22: [1, 0, 0, 1], 23: [0, 1, 2, 1, 0], 24: [3, 3, 2, 1, 0]}
>>> {k:mean(v) for k,v in d.items()}
{22: 0.5, 23: 0.8, 24: 1.8}
>>>