Python 在不使用 .values() 等的情况下查找字典中值的均值
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Finding mean of a values in a dictionary without using .values() etc
提问by Rachel
I have a dictionary that looks like:
我有一本字典，看起来像：
G={'E': 18.0, 'D': 17.0, 'C': 19.0, 'B': 15.0, 'A': 0}
I have to find the mean of the values e.g. mean(18,17,19,15,0) using a simple forloop without using built in functions like .values(), .items()and so on. I tried the following but am getting an error:
我必须使用一个简单的for循环而不使用诸如.values(),.items()等内置函数来找到值的平均值，例如 mean(18,17,19,15,0) 。我尝试了以下操作，但出现错误：
d=[float(sum(values)) / len(values) for key, values in G]
return (d)
ValueError: need more than 1 value to unpack
Can someone help me fix this????
有人可以帮我解决这个问题吗？？？？
采纳答案by Aaron Hall
To do this with a "simple for loop", using your constraints against using the dict methods:
要使用“简单的 for 循环”执行此操作，请使用您的约束而不是使用 dict 方法：
G = {'E': 18.0, 'D': 17.0, 'C': 19.0, 'B': 15.0, 'A': 0}
count = 0
_sum = 0
for key in G:
count += 1
_sum += G[key]
print('this is the mean: ', _sum/count)
If you're supposed to avoid dict methods, clearly this is an academic exercise.
如果你应该避免使用 dict 方法，显然这是一个学术练习。
Without that constraint:
没有这个限制：
The statisticsmodule in the standard library has a meanmethod, which would be my first thought (as the standard library does not require third party packages.):
statistics标准库中的模块有一个mean方法，这是我的第一个想法（因为标准库不需要第三方包。）：
>>> G={'E': 18.0, 'D': 17.0, 'C': 19.0, 'B': 15.0, 'A': 0}
>>> from statistics import mean
>>> mean(G[k] for k in G)
13.8
Third party packages like numpy and pandas have objects with a meanmethod:
像 numpy 和 pandas 这样的第三方包有一个带有mean方法的对象：
>>> from numpy import array
>>> array([G[k] for k in G]).mean()
13.8
>>> from pandas import Series
>>> Series([G[k] for k in G]).mean()
13.8
If we allow ourselves to use the values()method, this gets a little simpler with iterable unpacking. For some reason the other answers violate that condition, so I figure I should show the more efficient way of doing it:
如果我们允许自己使用该values()方法，通过可迭代解包，这会变得更简单一些。出于某种原因，其他答案违反了该条件，所以我想我应该展示更有效的方法：
>>> Series([*G.values()]).mean()
13.8
回答by Gian
Iteration over a dictionary iterates over its keys. Try just using for key in G, and then using G[key]appropriately instead of values.
对字典的迭代会对其键进行迭代。尝试只使用for key in G, 然后G[key]适当地使用而不是values。
Alternatively, use the iteritems()method of the dictionary to get key, valuepairs from G, i.e.:
或者，使用iteritems()字典的方法key, value从G中获取对，即：
d=[float(sum(values)) / len(values) for key, values in G.iteritems()]
(For the record, your actual method of computing a mean doesn't look right to me, but you may as well fix the iteration problem first).
（为了记录，您计算均值的实际方法在我看来并不正确，但您最好先解决迭代问题）。
回答by ebarr
You want:
你要：
mean = sum([G[key] for key in G])/float(len(G))
Your original code will also produce a:
您的原始代码还将产生：
TypeError: 'int' object is not iterable
when you try to sum the values.
当您尝试对值求和时。
回答by Gassa
Use G.values()to get all the values from a dictionary.
用于G.values()从字典中获取所有值。
G = {'E': 18.0, 'D': 17.0, 'C': 19.0, 'B': 15.0, 'A': 0}
d = float(sum(G.values())) / len(G)
print (d)
This prints 13.8.
这打印13.8.
Note that there is a difference between Python 2 and Python 3 here. In Python 2, G.values()is a newly constructed list of values. In Python 3, it is a generator, which can be thought of as a “lazy list”. The same thing is called G.itervalues()in Python 2.
请注意，此处 Python 2 和 Python 3 之间存在差异。在 Python 2 中，G.values()是一个新构造的值列表。在 Python 3 中，它是一个生成器，可以被认为是一个“惰性列表”。同样的事情G.itervalues()在 Python 2 中被调用。
回答by barrios
In Python 3.4 upwards there is a very clear way:
在 Python 3.4 以上有一个非常明确的方法：
import statistics
numbers = [G[key] for key in G]
mean_ = statistics.mean(numbers)
回答by Jerome Montino
Another for-loop, this one not needing a counter for items.
另一个 for 循环，这个循环不需要物品的计数器。
G = {'E': 18.0, 'D': 17.0, 'C': 19.0, 'B': 15.0, 'A': 0}
sum = 0
for k in G:
sum += float(G[k])
print "Mean: " + str(sum/len(G))
Results to:
结果：
Mean: 13.8
[Finished in 0.3s]
Actually, given your data, there's no need to use float. On my end, removing floatand just leaving sum += G[k]still outputs the same since the values are not strings anyway.
实际上，鉴于您的数据，没有必要使用float. 最后，删除float和离开sum += G[k]仍然输出相同，因为值无论如何都不是字符串。
回答by Adam Erickson
What I suggest instead of the current answers is adopting a functional programming paradigm that is reusable and flexible. For example, creating a function to calculate any statistic on listvalues contained within a simple dict:
我建议采用一种可重用且灵活的函数式编程范式，而不是当前的答案。例如，创建一个函数来计算listsimple 中包含的值的任何统计信息dict：
def summarize_dict(dictionary, function):
dict_new = {}
for k,v in dictionary.items():
dict_new[k] = function(v)
return dict_new
Testing:
测试：
import numpy as np
keys = ["a","b","c","d","e"]
values = [range(2),range(4),range(6),range(8),range(10)]
dictionary = dict(zip(keys, values))
summarize_dict(dictionary, np.mean)
Yields:
产量：
{'a': 0.5, 'b': 1.5, 'c': 2.5, 'd': 3.5, 'e': 4.5}
回答by Sergio R
If you use numpy:
如果您使用 numpy：
import numpy as np
np.array(list(dict.values())).mean()
回答by insanely_sin
import numpy as np
np.mean(list(dict.values()))
