如何将DataFrame的prediction列转换为指定结构的新DataFrame

先构造示例测试数据 #

import pandas as pd

data = {
    'prediction': [
        [
            {'answer': 'My name is Andrew.', 'text': 'What is your name ?', 'id': 1},
            {'answer': 'I live at California.', 'text': 'Where do you live ?', 'id': 2},
            {'answer': 'I want to become a doctor.', 'text': 'What do you want to do ?', 'id': 3}
        ],
        [
            {'answer': 'My name is Julie.', 'text': 'What is your name ?', 'id': 1},
            {'answer': 'I live at NY.', 'text': 'Where do you live ?', 'id': 2},
            {'answer': 'I want to be a cook.', 'text': 'What do you want to do ?', 'id': 3}
        ]
    ]
}
df = pd.DataFrame(data)

方法一：使用explode + pivot（适合大数据量） #

# 把每行的字典列表拆分为单独行
exploded_df = df.explode('prediction')
# 将字典展开为独立列
expanded_df = exploded_df['prediction'].apply(pd.Series)
# 透视转换为目标结构，重置索引
result_df = expanded_df.pivot(
    index=expanded_df.index, 
    columns='text', 
    values='answer'
).reset_index(drop=True)

方法二：逐行处理字典列表（代码更简洁） #

# 对每行的字典列表生成单行DataFrame，再拼接所有行
result_df = pd.concat(
    df['prediction'].apply(
        lambda lst: pd.DataFrame(lst).set_index('text')['answer'].to_frame().T
    ),
    ignore_index=True
)

两种方法最终都会得到你需要的结构：

What is your name ?    Where do you live ?    What do you want to do ?
0  My name is Andrew.    I live at California.  I want to become a doctor.
1  My name is Julie.     I live at NY.          I want to be a cook.

内容的提问来源于stack exchange，提问作者d_b