Question

我有一本字典

alice = {
    "name": "Alice",
    "homework": [100.0, 92.0, 98.0, 100.0],
    "quizzes": [82.0, 83.0, 91.0],
    "tests": [89.0, 97.0]
}

和一个取平均值的函数

def average(numbers):
    return float(sum(numbers))/len(numbers)

我想要python函数get_average（）来读取字典。之后，我希望函数取字典中列表的平均值，并将该列表的平均值乘以一个特定的数字（对于作业，评分权重为.10，测验为.30，测试为.60），并给出加权作业，测验和测试的总和。我正在使用if语句来检查从字典中的for循环中选择的列表是否是作业，测验或测试。

def get_average(alice):
    hw=0
    qz=0
    ts=0

    for assignment in student:
        if assignment== ["homework"]:  
            hw=average(assignment)*.10
        if assignment==["quizzes"]:
            qz=average(assignment)*.30
        if assignment==["tests"]:
            ts=average(assignment)*.60

    return hw+qz+ts

这只返回零。它应该是91.15

Answer 1

其他答案绰绰有余。我只是想尝试一些不同的东西。

使用此方法，您可以轻松扩展标记条件的深度，而不会使用所有额外的if语句扩展函数的大小。它也更容易编辑，IMO，重复性较低，而且很重要。

alice = {
    "name": "Alice",
    "homework": [100.0, 92.0, 98.0, 100.0],
    "quizzes": [82.0, 83.0, 91.0],
    "tests": [89.0, 97.0]
}

def average(numbers):
    return float(sum(numbers))/len(numbers)

def get_average(student):
    criteria = {'homework': 0.10, 'quizzes': 0.30, 'tests': 0.60}
    return sum(y * average(student[x]) for (x,y) in criteria.iteritems())

alice['average'] = get_average(alice)

print alice

希望这有帮助。

Answer 2

通过字典循环可以获得字典中的键。您需要使用平均函数的值。您还可以使用iteritems：

同时遍历键和值

这应该有效：

for assignment, scores in student.iteritems():
    if assignment=="homework":  
        hw=average(scores)*.10
    if assignment=="quizzes":
        qz=average(scores)*.30
    if assignment=="tests":
        ts=average(scores)*.60

return hw+qz+ts

你也可以循环浏览dict而不是得分有学生[赋值]，但每次循环迭代需要两次查找，所以最好使用iteritems。

Answer 3

您的代码中存在一些错误：

assignment在'name'函数中迭代学生的密钥，即'homework'，get_average()等。因此，请将其与字符串'homework'进行比较，而不是与列表['homework']进行比较。
正如我所说assignment包含密钥，因此您需要使用该密钥的值来调用average，即student[assignment]
您正在student函数中迭代get_average()变量，因此它应该是get_average(student)而不是get_average(alice)。

演示：

>>> def get_average(student):  
...     hw=0
...     qz=0
...     ts=0
...     for assignment in student:
...     # assignment iterates over keys of student i.e., 'name', 'homework' etc. So compare it with string 'homework' not with list ['homework']
...         if assignment== "homework":  
...         # as I said assignment contains keys, so you need to call average with value of that key i.e., student[assignment]
...             hw=average(student[assignment])*.10
...         if assignment=="quizzes":
...             qz=average(student[assignment])*.30
...         if assignment=="tests":
...             ts=average(student[assignment])*.60
...     return hw+qz+ts
... 
>>> def average(numbers):
...     return float(sum(numbers))/len(numbers)
... 
>>> alice = {
...     "name": "Alice",
...     "homework": [100.0, 92.0, 98.0, 100.0],
...     "quizzes": [82.0, 83.0, 91.0],
...     "tests": [89.0, 97.0]
... }
>>> get_average(alice)
91.14999999999999

Answer 4

如果你正在使用python 3，请使用items()，因为iteritems()已从此版本中删除：

def get_average(alice):

    tot = 0
    for tests, grades in alice.items():   #iterates over keys and values
        if tests == "homework":
            tot += average(grades)*0.1
        if tests == "quizzes":
            tot += average(grades)*0.3
        if tests == "tests":
            tot += average(grades)*0.6

    return tot

另一种使用dict理解的方法：

def get_average(alice):

    tot = 0
    gen = {test: average(grades) for test, grades in alice.items() if isinstance(grades, list)}
    for test, grades in gen.items():
        if test == "homework":
            tot += grades*0.1
        if test == "quizzes":
            tot += grades*0.3
        if test == "tests":
            tot += grades*0.6

    return tot

如果从字典列表比较

4 个答案: