根据包含的字符串过滤行，然后比较 Python 中的两列

Question

给定一个玩具数据集如下：

   id room_type company_name
0   1    office      ABC ltd
1   2    office       retail
2   3    office      xyz ltd
3   4    retail       retail
4   5   parking    toy store
5   6      hall          NaN

如果room_type 或company 列包含retail, parking or hall，则比较两列，如果它们不相同，则返回一个带有字符串Invalid company name or room type 的新列check。

我想使用如下代码，因为还有很多其他列要检查：

a = np.where(df['room_type'].str.contains('retail|parking|hall', na = False), 'Invalid company name or room type', None)

# b = np.where(df.area.str.contains('^\d+$', na = True), None,
#                                  'area is not a numbers')  
f = (lambda x: ';'.join(y for y in x if pd.notna(y)) 
                if any(pd.notna(np.array(x))) else np.nan )
df['check'] = [f(x) for x in zip(a)]

预期的结果是这样的：

   id room_type company_name                              check
0   1    office      ABC ltd                                NaN
1   2    office       retail  Invalid company name or room type
2   3    office      xyz ltd                                NaN
3   4    retail       retail                                NaN
4   5   parking    toy store  Invalid company name or room type
5   6      hall          NaN  Invalid company name or room type

如何修改条件a 代码？提前感谢您的帮助。

Answer 1

使用Series.str.cat 连接两列，测试子字符串和第二个条件比较Series.ne 不相等，最后一个链条件| 按位与：

m1 = (df['room_type'].str.cat(df['company_name'], sep=' ', na_rep='')
                     .str.contains('retail|parking|hall', na = False))
m2 = df['room_type'].ne(df['company_name'])

df['check'] = np.where(m1 & m2, 'Invalid company name or room type', None)
print(df)
   id room_type company_name                              check
0   1    office      ABC ltd                               None
1   2    office       retail  Invalid company name or room type
2   3    office      xyz ltd                               None
3   4    retail       retail                               None
4   5   parking    toy store  Invalid company name or room type
5   6      hall          NaN  Invalid company name or room type