Question

我有一个数据框df，

      plan_year                                    name metal_level_name
0        20118            Gold Heritage Plus 1500 - 02             Gold
1         2018                                     NaN         Platinum
2         2018            Gold Heritage Plus 2000 - 01             Gold

我在下面的plan_year和name列上进行了数据验证，

m4 = ((df['plan_year'].notnull()) & (df['plan_year'].astype(str).str.isdigit()) & (df['plan_year'].astype(str).str.len() == 4))

m1 = (df1[['name']].notnull().all(axis=1))

我在下面得到有效的数据框，

df1 = df[m1 & m4]

我可以得到df1中不存在的行（无效的行）

merged = df.merge(df1.drop_duplicates(), how='outer', indicator=True)
merged[merged['_merge'] == 'left_only']

我想跟踪由于验证而导致哪一行失败。

我想获取一个包含所有无效数据数据框的数据框，如下所示-

 plan_year                                    name metal_level_name    Failed message
0        20118            Gold Heritage Plus 1500 - 02             Gold    Failed due to wrong plan_year
1         2018                                     NaN         Platinum     name column cannot be null

有人可以帮我吗？

Answer 1

您可以将numpy.select与~一起使用，以反转boolena蒙版：

message1 = 'name column cannot be null'
message4 = 'Failed due to wrong plan_year'


df['Failed message'] = np.select([~m1, ~m4], [message1, message4], default='OK')
print (df)
   plan_year                          name metal_level_name  \
0      20118  Gold Heritage Plus 1500 - 02             Gold   
1       2018                           NaN         Platinum   
2       2018  Gold Heritage Plus 2000 - 01             Gold   

                  Failed message  
0  Failed due to wrong plan_year  
1     name column cannot be null  
2                             OK

df1 = df[df['Failed message'] != 'OK']
print (df1)
   plan_year                          name metal_level_name  \
0      20118  Gold Heritage Plus 1500 - 02             Gold   
1       2018                           NaN         Platinum   

                  Failed message  
0  Failed due to wrong plan_year  
1     name column cannot be null

编辑：对于多个错误消息，请使用concat创建新的DataFrame，然后使用按列名使用dot的列名对它进行矩阵化，最后使用{{3} }：

print (df)
   plan_year                          name metal_level_name
0      20118  Gold Heritage Plus 1500 - 02             Gold
1       2018                           NaN         Platinum
2       2018  Gold Heritage Plus 2000 - 01             Gold
1      20148                           NaN         Platinum

message1 = 'name column cannot be null'
message4 = 'Failed due to wrong plan_year'

df1 = pd.concat([~m1, ~m4], axis=1, keys=[message1, message4])
print (df1)
   name column cannot be null  Failed due to wrong plan_year
0                       False                           True
1                        True                          False
2                       False                          False
1                        True                           True


df['Failed message'] = df1.dot(df1.columns + ', ').str.rstrip(', ')
print (df)

   plan_year                          name metal_level_name  \
0      20118  Gold Heritage Plus 1500 - 02             Gold   
1       2018                           NaN         Platinum   
2       2018  Gold Heritage Plus 2000 - 01             Gold   
1      20148                           NaN         Platinum   

                                      Failed message  
0                      Failed due to wrong plan_year  
1                         name column cannot be null  
2                                                     
1  name column cannot be null, Failed due to wron...

df1 = df[df['Failed message'] != '']
print (df1)
   plan_year                          name metal_level_name  \
0      20118  Gold Heritage Plus 1500 - 02             Gold   
1       2018                           NaN         Platinum   
1      20148                           NaN         Platinum   

                                      Failed message  
0                      Failed due to wrong plan_year  
1                         name column cannot be null  
1  name column cannot be null, Failed due to wron...

在熊猫中找出并记录失败的验证条件

1 个答案: