如何用Pandas根据另一DataFrame列的列表更新DataFrame行

现有两个Pandas DataFrame：

# df1
   0  1  2 ECID
0  0  0  0   00
1  0  0  0   01
2  0  0  0   02

# df2
  ECID FailPattern
0   00      [1, 2]
1   01         [1]
2   02         [0]

需要根据df2的FailPattern列，将df1中对应ECID行的指定列（列号与FailPattern中的元素匹配）数值更新为1，最终得到如下结果：

0  1  2 ECID
0  0  1  1   00
1  0  1  0   01
2  1  0  0   02

方法一：合并后逐行更新（直观易懂） #

先通过ECID合并两个DataFrame，再遍历行根据FailPattern修改对应列的值：

import pandas as pd

# 构造示例数据
df1 = pd.DataFrame({'0': [0,0,0], '1': [0,0,0], '2': [0,0,0], 'ECID': ['00','01','02']})
df2 = pd.DataFrame({'ECID': ['00','01','02'], 'FailPattern': [[1,2], [1], [0]]})

# 按ECID合并
merged_df = df1.merge(df2, on='ECID')

# 遍历更新每一行
for idx, row in merged_df.iterrows():
    for col_idx in row['FailPattern']:
        merged_df.loc[idx, str(col_idx)] = 1

# 移除临时列，得到结果
result_df = merged_df.drop('FailPattern', axis=1)
print(result_df)

方法二：Explode+透视表（高效无循环） #

利用Pandas向量化操作替代循环，适合大数据量场景：

import pandas as pd

df1 = pd.DataFrame({'0': [0,0,0], '1': [0,0,0], '2': [0,0,0], 'ECID': ['00','01','02']})
df2 = pd.DataFrame({'ECID': ['00','01','02'], 'FailPattern': [[1,2], [1], [0]]})

# 将FailPattern展开为单行对应单个列索引的格式
exploded_df = df2.explode('FailPattern')
exploded_df['FailPattern'] = exploded_df['FailPattern'].astype(str)  # 转为字符串匹配df1列名
exploded_df['value'] = 1

# 透视表转为宽表，匹配df1结构
pivot_df = exploded_df.pivot(index='ECID', columns='FailPattern', values='value').fillna(0)

# 合并并覆盖原df1的列值，恢复原列顺序
result_df = df1.set_index('ECID').combine_first(pivot_df).reset_index()
result_df = result_df[df1.columns]
print(result_df)

方法三：Apply函数批量处理 #

通过字典映射快速查找对应故障列，再用apply逐行更新：

import pandas as pd

df1 = pd.DataFrame({'0': [0,0,0], '1': [0,0,0], '2': [0,0,0], 'ECID': ['00','01','02']})
df2 = pd.DataFrame({'ECID': ['00','01','02'], 'FailPattern': [[1,2], [1], [0]]})

# 将df2转为ECID到故障列的字典映射
fail_map = df2.set_index('ECID')['FailPattern'].to_dict()

# 定义行处理函数
def update_row(row):
    fail_cols = fail_map[row['ECID']]
    for col in fail_cols:
        row[str(col)] = 1
    return row

# 应用函数到所有行
result_df = df1.apply(update_row, axis=1)
print(result_df)

内容的提问来源于stack exchange，提问作者Nandeep Devendra