您使用lambda函数是正确的，但我将这样使用它：

# Create the dataframe from your question
df = pd.DataFrame({'ColumnA': ['prefix_1', 'prefix_2'], 'Column B': [['A', 'B'], ['C', 'D']]})

# Create Column C accordingly
df['Column C'] = df.apply(lambda row: [row['ColumnA'] + '-' + elem for elem in row['Column B']], axis=1)

确保使用axis=1，以便lambda函数按行应用。

使用列表解析。
如果B列中有列表：

df['Column C'] = [[f'{p}-{x}' for x in l] for p, l in
                   zip(df['ColumnA'], df['Column B'])]

如果B列中有字符串：

df['Column C'] = [[f'{p}-{x}' for x in l] for p, l in
                   zip(df['ColumnA'],
                       df['Column B'].str[1:-1].str.split(',\s*')
                       )]

如果你想要一个字符串作为输出：

df['Column C'] = ['['+', '.join([f'{p}-{x}' for x in l])+']'
                  for p, l in
                   zip(df['ColumnA'],
                       df['Column B'].str[1:-1].str.split(',\s*')
                       )]

输出：

ColumnA Column B                  Column C
0  prefix_1   [A, B]  [prefix_1-A, prefix_1-B]
1  prefix_2   [C, D]  [prefix_2-C, prefix_2-D]

可重现输入：

# as string
df = pd.DataFrame({'ColumnA': ['prefix_1', 'prefix_2'],
                   'Column B': ['[A, B]', '[C, D]']})

# as lists
df = pd.DataFrame({'ColumnA': ['prefix_1', 'prefix_2'],
                   'Column B': [['A', 'B'], ['C', 'D']]})