Pandas对象类型列数值替换：DataFrame中cited_ids与d列修改需求

原始DataFrame #

id cited_ids        dummy_paper   d      
2  [4]                  NaN        NaN 
4  [9,18,6]             NaN        NaN
6  []                   9          0
7  [2]                  NaN        NaN
9  [4]                   7        0
14 [18,6]                3        0
18 [7]                   1        0

需完成的两项操作 #

• (i) 将df['cited_ids']列表中，对应id的d值为0的元素替换为0；
• (ii) 若df['cited_ids']的列表中存在至少一个0，且该行原始d值不为0，则将该行d值设为1。

第一步操作后结果 #

id cited_ids        dummy_paper   d      
2  [4]                  NaN       NaN 
4  [0,0,6]             NaN        NaN
6  []                   9         0
7  [2]                  NaN       NaN
9  [4]                   7        0
14 [0,6]                 3        0
18 [0]                   1        0

第二步操作后结果 #

id cited_ids        dummy_paper   d      
2  [4]                  NaN       NaN 
4  [0,0,6]             NaN        1
6  []                   9         0
7  [2]                  NaN       NaN
9  [4]                   7        0
14 [0,6]                 3        0
18 [0]                   1        0

补充信息 #

• df['cited_ids']为object类型；
• df.to_dict()输出如下：

{
    'docdb': {0: 2, 1: 4, 2: 6, 3: 7, 4: 9, 5: 14, 6: 18},
    'cited_docdb': {0: [4],
                    1: [9, 18, 6],
                    2: [],
                    3: [2],
                    4: [4],
                    5: [18, 6],
                    6: [7]},
    'fronteer': {0: nan, 1: nan, 2: 9.0, 3: nan, 4: 7.0, 5: 3.0, 6: 1.0},
    'distance': {0: nan, 1: nan, 2: 0.0, 3: nan, 4: 0.0, 5: 0.0, 6: 0.0}
}

实现代码 #

步骤1：替换cited_ids中的目标元素 #

先建立id到d值的映射，再遍历每个列表替换符合条件的元素：

# 创建id与对应d值的映射字典
id_to_d = df.set_index('id')['d'].to_dict()

# 替换cited_ids中对应d值为0的id
df['cited_ids'] = df['cited_ids'].apply(
    lambda lst: [0 if id_to_d.get(x, None) == 0 else x for x in lst]
)

步骤2：更新d值 #

先保存原始d值，再根据条件筛选并更新：

# 保存原始d值用于判断
original_d = df['d'].copy()

# 构建筛选条件：cited_ids包含0，且原始d值不为0（包括NaN）
mask = df['cited_ids'].apply(lambda lst: 0 in lst) & (original_d.isna() | (original_d != 0))

# 对符合条件的行设置d值为1
df.loc[mask, 'd'] = 1

完整可运行代码 #

import pandas as pd
import numpy as np

# 构建原始DataFrame
data = {
    'id': [2,4,6,7,9,14,18],
    'cited_ids': [[4],[9,18,6],[],[2],[4],[18,6],[7]],
    'dummy_paper': [np.nan, np.nan, 9, np.nan,7,3,1],
    'd': [np.nan, np.nan,0,np.nan,0,0,0]
}
df = pd.DataFrame(data)

# 执行步骤1
id_to_d = df.set_index('id')['d'].to_dict()
df['cited_ids'] = df['cited_ids'].apply(
    lambda lst: [0 if id_to_d.get(x, None) == 0 else x for x in lst]
)

# 执行步骤2
original_d = df['d'].copy()
mask = df['cited_ids'].apply(lambda lst: 0 in lst) & (original_d.isna() | (original_d != 0))
df.loc[mask, 'd'] = 1

print(df)

内容的提问来源于stack exchange，提问作者Lusian