基于现有列在 pandas 数据框中创建新列

Question

我有一个 Python 字典如下：

ref_dict = {
"Company1" :["C1_Dev1","C1_Dev2","C1_Dev3","C1_Dev4","C1_Dev5",],
"Company2" :["C2_Dev1","C2_Dev2","C2_Dev3","C2_Dev4","C2_Dev5",],
"Company3" :["C3_Dev1","C3_Dev2","C3_Dev3","C3_Dev4","C3_Dev5",],
 }

我有一个名为df的 Pandas 数据框，其中一列如下所示：

    DESC_DETAIL
0   Probably task Company2 C2_Dev5
1   File system C3_Dev1
2   Weather subcutaneous Company2
3   Company1 Travesty C1_Dev3
4   Does not match anything 
...........

我的目标是在此数据框中添加两个额外的列，并将这些列命名为COMPANY和DEVICE 。 COMPANY列的每一行中的值将是字典中的公司键，如果它存在于DESC_DETAIL列中，或者相应的设备存在于DESC_DETAIL列中。 DEVICE列中的值将只是DESC_DETAIL列中的设备字符串。 如果未找到匹配项，则对应的行为空。 因此最终的 output 将如下所示：

     DESC_DETAIL                        COMPANY         DEVICE
 0   Probably task Company2 C2_Dev5     Company2        C2_Dev5
 1   File system C3_Dev1                Company3        C3_Dev1
 2   Weather subcutaneous Company2      Company2        NaN
 3   Company1 Travesty C1_Dev3          Company1        C1_Dev3
 4   Does not match anything            NaN             NaN

我的尝试：

for key, value in ref_dict.items():
    df['COMPANY'] = df.apply(lambda row: key if row['DESC_DETAIL'].isin(key) else Nan, axis=1)

这显然是错误的并且不起作用。 我如何使它工作？

Answer 1

您可以使用正则表达式模式通过str.extract提取值：

import re

s = pd.Series(ref_dict).explode()

# extract company
df['COMPANY'] = df['DESC_DETAIL'].str.extract(
    f"({'|'.join(s.index.unique())})", flags=re.IGNORECASE)

# extract device
df['DEVICE'] = df['DESC_DETAIL'].str.extract(
    f"({'|'.join(s)})", flags=re.IGNORECASE)

# fill missing company values based on device
df['COMPANY'] = df['COMPANY'].fillna(
    df['DEVICE'].str.lower().map(dict(zip(s.str.lower(), s.index))))

df

Output：

                      DESC_DETAIL   COMPANY   DEVICE
0  Probably task Company2 C2_Dev5  Company2  C2_Dev5
1             File system C3_Dev1  Company3  C3_Dev1
2   Weather subcutaneous Company2  Company2      NaN
3       Company1 Travesty C1_Dev3  Company1  C1_Dev3
4         Does not match anything       NaN      NaN

Answer 2

您还需要一个设备到公司字典，您可以从ref_dict轻松构建它，如下所示：

dev_to_company_dict = {v:l[0] for l in zip(ref_dict.keys(), ref_dict.values()) for v in l[1]}

然后很容易做到这一点：

df['COMPANY'] = df['DESC_DETAIL'].apply(lambda det : ''.join(set(re.split("\\s+", det)).intersection(ref_dict.keys())))
df['COMPANY'].replace('', np.nan, inplace=True)
df['DEVICE'] = df['DESC_DETAIL'].apply(lambda det : ''.join(set(re.split("\\s+", det)).intersection(dev_to_company_dict.keys())))
df['DEVICE'].replace('', np.nan, inplace=True)
df['COMPANY'] = df['COMPANY'].fillna(df['DEVICE'].map(dev_to_company_dict))

Output：

                       DESC_DETAIL   COMPANY     DEVICE
0   Probably task Company2 C2_Dev5  Company2    C2_Dev5
1   File system C3_Dev1             Company3    C3_Dev1
2   Weather subcutaneous Company2   Company2        NaN
3   Company1 Travesty C1_Dev3       Company1    C1_Dev3
4   Does not match anything              NaN        NaN