如何從字符串列表中返回一個非類型？

Question

我在使用此功能時遇到問題。當我嘗試運行它時，它不起作用。 任何人都可以幫我修好嗎？

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] 

更新後：

遍歷L和創建新的列表，拆分，其將在L，其中有逗號的元素。然後在需要的地方將字符串更改爲浮動。然後運行花車上的總和和平均值。將L中的元素設置爲計算的平均值。

Answer 2

更新

現在我明白你正在試圖做的（與@PM 2Ring的幫助下）什麼，這裏是一個完全不同的答案（這個問題實際上比我原以爲簡單多）。這將有有益的，如果你回答要麼我原來的答覆或從自己和他人各種意見，並解釋你試圖更清晰地做......

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

我們遍歷使用enumerate名單。這給了我們每個列表項目（idx）和字符串本身（student）的索引。

然後我們調用拆分每個字符串來提取其內容，保存第一項爲name，其餘爲data。然後我們將data中的字符串轉換爲浮點數。然後我們計算這些浮點數的平均值。然後我們將平均值保存回適當的列表項。

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) 

輸出

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

Answer 4

試試這個。

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()對此更好。