使用Python中另一个字典的相同键和平均值创建新字典

我有一个名为StudentGrad的现有词典，其数据类型如下：

('Aditya','male','senior'):[0.83,0.87,0.82,0.83],
('Varun','male','senior'):[0.76,0.86,0.88,0.79],
('Shantanu','male','senior'):[0.79,0.81,0.78,0.78],

我想创建一个名为studentAvg的新字典，它具有相同的键，但返回值的平均等级：

('Aditya','male','senior'):[0.84],
('Varun','male','senior'):[0.82],
('Shantanu','male','senior'):[0.79],

任何帮助都将不胜感激

---

谢谢

您可以使用dict理解和自定义函数进行平均：

d1 = {('Aditya','male','senior'):[0.83,0.87,0.82,0.83],
('Varun','male','senior'):[0.76,0.86,0.88,0.79],
('Shantanu','male','senior'):[0.79,0.81,0.78,0.78],
}
def average(list_values):
  "let's assume you want an arithmetic mean"
  return sum(list_values) / len(list_values)
d2 = {k:average(val) for k, val in d1.items()}

如果你真的想要一份平均值列表（如问题中所述）：

试试这个：

studentAvg = {
   ('Aditya','male','senior'):sum(studendGrad["('Aditya','male','senior')"])/len(studendGrad["('Aditya','male','senior')"]),
   ('Varun','male','senior'):sum(studentGrad["(Varun','male','senior')"])/len(studentGrad["('Varun','male','senior')"]),
   ('Shantanu','male','senior'):sum(studentGrad["('Shantanu','male','senior')"]/len(studentGrad["('Shantanu','male','senior')"])
}

尝试这样做：

x  = {('Aditya','male','senior'):[0.83,0.87,0.82,0.83],
('Varun','male','senior'):[0.76,0.86,0.88,0.79],
('Shantanu','male','senior'):[0.79,0.81,0.78,0.78]}

def average(data):
    sum = 0
    for elem in data:
        sum+=elem
    result = sum/len(data)
    return result

y = {}

for elem in x:
    y[elem] = average(x[elem])

---

为什么你的平均数是单元素列表而不是一个数字呢？欢迎来到这里。这不是一个讨论论坛或教程。请花点时间阅读和阅读该页面上的其他链接。花些时间练习这些例子。它会让你了解Python提供的帮助你解决问题的工具。回答了你的问题吗？为什么不使用？？？@wwii有很多方法。但至少通过这种方法（我认为），您可以更好地理解python的机制。（就个人而言，我是熊猫的粉丝，我会这样做：-）

x  = {('Aditya','male','senior'):[0.83,0.87,0.82,0.83],
('Varun','male','senior'):[0.76,0.86,0.88,0.79],
('Shantanu','male','senior'):[0.79,0.81,0.78,0.78]}

def average(data):
    sum = 0
    for elem in data:
        sum+=elem
    result = sum/len(data)
    return result

y = {}

for elem in x:
    y[elem] = average(x[elem])