如何从字符串列表中返回一个非类型？

Question

我在使用此功能时遇到问题。当我尝试运行它时，它不起作用。 任何人都可以帮我修好吗？

def string_avg_update(L): 
    '''(list of str) -> NoneType 
    Given a list of strings where each string has the format: 
    'name, grade, grade, grade, ...' update the given 
    list of strs to be a list of floats where each item 
    is the average of the corresponding numbers in the 
    string. Note this function does NOT RETURN the list. 
    >>> L = ['Anna, 50, 92, 80', 'Bill, 60, 70', 'Cal, 98.5, 100, 95.5, 98'] 
    >>> string_avg_update(L) 
    >>> L 
    [74.0, 65.0, 98.0] 
    ''' 
    average = 0 
    for item in L: 
     if item.isdigit(): 
      average = sum(item)/len(item) 

Answer 1

def string_avg_update(L): 
    for x in range(len(L)): 
     split = L[x].split(',') 
     for y in range(1, len(split)): 
      split[y] = float(split[y]) 
     summation = sum(split[z] for z in range(1, len(split))) 
     average = summation/(len(split)-1) 
     L[x] = average 

L = ['Anna, 50, 92, 80', 'Bill, 60, 70', 'Cal, 98.5, 100, 95.5, 98'] 

string_avg_update(L) 

print(L) 

返回：

[74.0, 65.0, 98.0] 

更新后：

遍历L和创建新的列表，拆分，其将在L，其中有逗号的元素。然后在需要的地方将字符串更改为浮动。然后运行花车上的总和和平均值。将L中的元素设置为计算的平均值。

Answer 2

更新

现在我明白你正在试图做的（与@PM 2Ring的帮助下）什么，这里是一个完全不同的答案（这个问题实际上比我原以为简单多）。这将有有益的，如果你回答要么我原来的答复或从自己和他人各种意见，并解释你试图更清晰地做......

def string_avg_update(L): 
    '''(list of str) -> NoneType 

    Given a list of strings where each string has the format: 
    'name, grade, grade, grade, ...' update the given 
    list of strs to be a list of floats where each row 
    is the average of the corresponding numbers in the 
    string. Note this function does NOT RETURN the list. 

    >>> L = ['Anna, 50, 92, 80', 'Bill, 60, 70', 'Cal, 98.5, 100, 95.5, 98'] 
    >>> string_avg_update(L) 
    >>> L 
    [74.0, 65.0, 98.0] 
    ''' 
    # Replace contents of L with average of numeric values in each string elem. 
    for i, record in enumerate(L): 
     grades = [float(grade) for grade in record.split(',')[1:]] 
     L[i] = sum(grades)/len(grades) 

    # Function will implicitly return None since there's no return statement. 

if __name__ == '__main__': 

    L = ['Anna, 50, 92, 80', 'Bill, 60, 70', 'Cal, 98.5, 100, 95.5, 98'] 
    print(L) 
    string_avg_update(L) 
    print(L) # -> [74.0, 65.0, 98.0] 

Answer 3

我们遍历使用enumerate名单。这给了我们每个列表项目（idx）和字符串本身（student）的索引。

然后我们调用拆分每个字符串来提取其内容，保存第一项为name，其余为data。然后我们将data中的字符串转换为浮点数。然后我们计算这些浮点数的平均值。然后我们将平均值保存回适当的列表项。

def string_avg_update(L): 
    '''(list of str) -> NoneType 
    Given a list of strings where each string has the format: 
    'name, grade, grade, grade, ...' update the given 
    list of strs to be a list of floats where each item 
    is the average of the corresponding numbers in the 
    string. Note this function does NOT RETURN the list. 
    >>> L = ['Anna, 50, 92, 80', 'Bill, 60, 70', 'Cal, 98.5, 100, 95.5, 98'] 
    >>> string_avg_update(L) 
    >>> L 
    [74.0, 65.0, 98.0] 
    ''' 
    for idx, student in enumerate(L): 
     name, *data = student.split(',') 
     data = [float(u) for u in data] 
     L[idx] = sum(data)/len(data) 

L = ['Anna, 50, 92, 80', 'Bill, 60, 70', 'Cal, 98.5, 100, 95.5, 98'] 
print(L) 

string_avg_update(L) 
print(L) 

输出

['Anna, 50, 92, 80', 'Bill, 60, 70', 'Cal, 98.5, 100, 95.5, 98'] 
[74.0, 65.0, 98.0] 

Answer 4

试试这个。

def string_avg_update(lst): 

    for i in range(len(lst[:])): 
     grades = lst[i].split(',')[1:] 
     float_grades = [float(grade) for grade in grades]   
     average = sum(float_grades)/len(float_grades)   
     lst[i] = average 



L = ['Anna, 50, 92, 80', 'Bill, 60, 70', 'Cal, 98.5, 100, 95.5, 98'] 
string_avg_update(L)  

>>> print(L) 
[74.0, 65.0, 98.0] 

我知道enumerate()对此更好。