R语言：基于同维度df2字符串筛选保留df1对应数值

我有两个维度相同的DataFrame（df1与df2）：

• df1为数值型数据：

df1
   V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1   1 11 21 31 41 51 61 71 81  91
2   2 12 22 32 42 52 62 72 82  92
3   3 13 23 33 43 53 63 73 83  93
4   4 14 24 34 44 54 64 74 84  94
5   5 15 25 35 45 55 65 75 85  95
6   6 16 26 36 46 56 66 76 86  96
7   7 17 27 37 47 57 67 77 87  97
8   8 18 28 38 48 58 68 78 88  98
9   9 19 29 39 49 59 69 79 89  99
10 10 20 30 40 50 60 70 80 90 100

• df2为字符串型数据（部分单元格包含多个字符串，例如[1,V10]）：

df2
      V1        V2    V3        V4        V5             V6    V7        V8             V9            V10
1  today     today today     today yesterday          today today     today          today tomorrow,today
2  today     today today     today     today      yesterday today     today          today          today
3  today yesterday today     today     today          today today yesterday          today          today
4  today     today today     today     today          today today     today tomorrow,today          today
5  today     today today yesterday yesterday          today today     today          today       tomorrow
6  today     today today yesterday     today tomorrow,today today     today          today          today
7  today     today today     today     today      yesterday today     today          today tomorrow,today
8  today yesterday today     today     today          today today     today          today          today
9  today     today today     today     today          today today     today tomorrow,today          today
10 today yesterday today yesterday     today          today today yesterday          today          today

df1与df2的单元格一一对应（如df1[1,1]对应df2[1,1]）。需要无需使用for循环的简便方法，仅保留df1中对应df2单元格包含"today"的数值，其余设为NA。期望输出如下：

期望输出
   V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1   1 11 21 31 NA 51 61 71 81  91
2   2 12 22 32 42 NA 62 72 82  92
3   3 NA 23 33 43 53 63 NA 83  93
4   4 14 24 34 44 54 64 74 84  94
5   5 15 25 NA NA 55 65 75 85  NA
6   6 16 26 NA 46 56 66 76 86  96
7   7 17 27 37 47 NA 67 77 87  97
8   8 NA 28 38 48 58 68 78 88  98
9   9 19 29 39 49 59 69 79 89  99
10 10 NA 30 NA 50 60 70 NA 90 100

R语言实现 #

利用向量化操作结合grepl函数直接处理，全程无需循环：

# 生成逻辑矩阵，标记df2中包含"today"的单元格
today_mask <- grepl("today", df2)
# 复制df1，将对应mask为FALSE的位置设为NA
result_df <- df1
result_df[!today_mask] <- NA

也可以用一行代码简化：

result_df <- df1 * NA^!grepl("today", df2)

解释：grepl("today", df2)返回和df2维度一致的逻辑矩阵，!取反后，NA^逻辑值会把TRUE转为NA、FALSE转为1，和df1相乘后，对应位置为NA的会自动将df1的数值转为NA，完全符合需求。

Python（Pandas）实现 #

借助Pandas的向量化字符串方法生成布尔掩码，再通过where方法完成赋值：

import pandas as pd
import numpy as np

# 生成布尔掩码，na=False避免空值干扰判断
today_mask = df2.apply(lambda x: x.str.contains("today", na=False))
# 保留满足条件的数值，其余设为NaN
result_df = df1.where(today_mask)

也可以分步写更直观：

result_df = df1.copy()
result_df[~today_mask] = np.nan

解释：df.where()方法会保留满足条件的原数值，不满足的自动设为NaN；str.contains用于检查每个单元格是否包含"today"，na=False确保空单元格被判定为不包含目标字符串。

内容的提问来源于stack exchange，提问作者user18632888