你就快找到答案了。
我所做的是：

data.replace({'Good': '1', 'Average': '2', 'Bad': '3'}, regex=True)

得到你想要的结果
enter image description here

对我来说，第二个解决方案工作，但必要的字符串转换为列表之前：

import ast

df['Feedback'] = df['Feedback'].apply(ast.literal_eval)
#df['Feedback'] = df['Feedback'].str.strip('[]').str.split(',')

使用嵌套字典的第一个解决方案：

df = df.assign(Feedback=[[feedback_dict.get(i,i) for i in x] for x in df['Feedback']])

df['Feedback'] = df['Feedback'].apply(lambda x : [feedback_dict.get(i,i)  for i in list(x)])
print (df)
     ID    Feedback
0  T223  [1, 3, 3]
1  T334  [2, 1, 1]

编辑：如果列表缺少值，则使用if-else语句-将非列表值替换为空列表：

print (df)
     ID             Feedback
0  T223       [Good,Bad,Bad]
1  T334  [Average,Good,Good]
2   NaN                  NaN

feedback_dict = {'Good':1, 'Average':2, 'Bad':3}
df = df.assign(Feedback=[[feedback_dict.get(i,i) for i in x] if isinstance(x, list) else [] 
                            for x in df['Feedback']])

print (df)
     ID   Feedback
0  T223  [1, 3, 3]
1  T334  [2, 1, 1]
2   NaN         []

如果你的用例和这个例子一样简单，我不会推荐这个方法，但是，这里有另一个选项，它可以让你项目的其他部分更容易。

1. df.explode()您的列（假设它是一个列表而不是文本;否则先将其转换为列表）
   1.使用df.replace()执行替换
   1.使用df.groupby()和df.agg()将行重新组合在一起
   对于这个例子，它看起来像这样（假设变量已经像你的问题中那样声明）：

df = df.explode('Feedback')
df['Feedback'] = df['Feedback'].replace(feedback_dict)
df = df.groupby('ID').agg(list)

l , L = [] , []  # two list for adding new values into them

for lst in df.Feeback: # call the lists in the Feeback column
    for i in last: #calling each element in each lists
        if i == 'Good': #if the element is Good then:
            l.append(feedback_dict['Good'])   #append the value 1 to the first created list
        if i == 'Average':  #if the element is Average then:
            l.append(feedback_dict['Average'])  #append the value 2 to the first created list
        if i == 'Bad':   #if the element is Bad then:
            l.append(feedback_dict['Bad']) #append the value 3 to the first created list
L.append(l[:3])  # we need to split half of the first list to add as a list to the second list and the other half as another list to the second list we created
L.append(l[3:])
df['Feeback'] = L  #at the end just put the values of the second created list as feedback column