python - 满足多个条件的 Pandas ；连接结果以创建新列

Question

我下面有一个代码，您可以看到我正在使用 np.select 来识别列中的字符串是否包含任何代码，并根据逻辑创建带有描述的引用列.

# Creating Score column
col         = 'codes_desc'
conditions  = [(df_merged[col].str.contains('R27', case=False)),
               (df_merged[col].str.contains('R38', case=False)),
               (df_merged[col].str.contains('R52', case=False)),
               (df_merged[col].str.contains('R62', case=False)),
               (df_merged[col].str.contains('R21', case=False)),
               (df_merged[col].str.contains('R22', case=False)),
               (df_merged[col].str.contains('R23', case=False)),
               (df_merged[col].str.contains('R57', case=False)),
               (df_merged[col].str.contains('R82', case=False)),
               (df_merged[col].str.contains('R86', case=False)),
               (df_merged[col].str.contains('R20', case=False)), 
               (df_merged[col].str.contains('R98', case=False)) 
              ]

choices     = [ 
 'The person is a Ninja',
 'The person is a Pirate',
 'The person is a Doctor',
 'The person is a Samurai',
 'The person is a Admiral',
 'The person is a Police',
 'The person is a Teacher',
 'The person is a Singer',
 'The person is a Guitarist',
 'The person is a Chef',
 'The person is a Runner',
 'The person is a Wizard'
]
df_merged["reference"] = np.select(conditions, choices, default= 'Reason Unknown')

但我在数据框中发现“codes_desc”列包含两个代码的情况，例如:

codes_desc

The selected codes are R27, R22.

在这种情况下，我希望我的输出类似于“引用”列中的内容:

1. 'The person is a Ninja'
2. 'The person is a Police'

但是由于 np.select 的工作方式类似于 case 语句；它获取最后的代码描述，那么如何做到这一点？

Answer 1

设置

print(df)
   codes_desc
0  fo bar R20
1   R98 grok 
2    R98, R21
3         R82

解决方案

让我们提取所有匹配的代码，然后使用映射字典将代码映射到相应的值，然后groupby 并使用 join 进行聚合

d = {'R27': 'The person is a Ninja',
     'R38': 'The person is a Pirate',
     'R52': 'The person is a Doctor',
     'R62': 'The person is a Samurai',
     'R21': 'The person is a Admiral',
     'R22': 'The person is a Police',
     'R23': 'The person is a Teacher',
     'R57': 'The person is a Singer',
     'R82': 'The person is a Guitarist',
     'R86': 'The person is a Chef',
     'R20': 'The person is a Runner',
     'R98': 'The person is a Wizard'}


pat = r'\b(%s)\b' % '|'.join(d)
codes = df['codes_desc'].str.extractall(pat)[0]
df['reference'] = codes.map(d).groupby(level=0).agg(', '.join)

结果

   codes_desc                                        reference
0  fo bar R20                           The person is a Runner
1   R98 grok                            The person is a Wizard
2    R98, R21  The person is a Wizard, The person is a Admiral
3         R82                        The person is a Guitarist