在以下数据帧中，我想将系列"resource"正确地拆分为不同的组件，这些组件以字符"_"分隔：
| 资源 |
|---|
| MTUG1_ABO_DPP_1 |
| MTUG1_ABO_DPP_2 |
| MTUG1_ABO_DPP_3 |
| MTUG1_ABO_DPP_4 |
| MTUG1_ABO_DPP_5 |
| MTUG1_ABO_DPU_1 |
| MTUG1_ABO_DPU_2 |
| MTUG1_ABO_DPU_3 |
| MTUG1_ABO_UUB_VDU1_1 |
| MTUG1_ABO_UUB_VDU1_2 |
| MTUG1_ABO_UUB_VDU1_3 |
| MTUG1_ABO_UUB_VDU2_1 |
| MTUG1_ABO_UUB_VDU2_2 |
您可以使用布尔索引分别应用这两种情况，在布尔索引中检查vdu_num是否为NaN
df.loc[~vdu_num.isna(), 'vm']=df['Unit'].str.cat(unit_num, sep="_").str.cat(vdu_num, sep="_")
df.loc[vdu_num.isna(), 'vm']=df['Unit'].str.cat(unit_num, sep="_")
产生
resource Node Unit unit_num vdu_num vm
-- -------------------- ------ ------ ---------- --------- ----------
0 MTUG1_ABO_DPP_1 MTUG1 DPP 1 nan DPP_1
1 MTUG1_ABO_DPP_2 MTUG1 DPP 2 nan DPP_2
2 MTUG1_ABO_DPP_3 MTUG1 DPP 3 nan DPP_3
3 MTUG1_ABO_DPP_4 MTUG1 DPP 4 nan DPP_4
4 MTUG1_ABO_DPP_5 MTUG1 DPP 5 nan DPP_5
5 MTUG1_ABO_DPU_1 MTUG1 DPU 1 nan DPU_1
6 MTUG1_ABO_DPU_2 MTUG1 DPU 2 nan DPU_2
7 MTUG1_ABO_DPU_3 MTUG1 DPU 3 nan DPU_3
8 MTUG1_ABO_UUB_VDU1_1 MTUG1 UUB VDU1 1 UUB_VDU1_1
9 MTUG1_ABO_UUB_VDU1_2 MTUG1 UUB VDU1 2 UUB_VDU1_2
10 MTUG1_ABO_UUB_VDU1_3 MTUG1 UUB VDU1 3 UUB_VDU1_3
11 MTUG1_ABO_UUB_VDU2_1 MTUG1 UUB VDU2 1 UUB_VDU2_1
12 MTUG1_ABO_UUB_VDU2_2 MTUG1 UUB VDU2 2 UUB_VDU2_2
您可以简化拆分(只做一次(，然后也可以使用where取unit_num或vdu_num，具体取决于非null：
df2 = (
df['resource']
.str.split('_', expand=True)[[0, 2, 3, 4]]
.set_axis('Node Unit unit_num vdu_num'.split(), axis=1)
)
df2['vm'] = df2['Unit'].str.cat(
df2['unit_num'].where(df2['vdu_num'].isnull(), df2['vdu_num']), sep='_')
或者，如果您喜欢覆盖原始df:
df = pd.concat([
df['resource'],
df['resource']
.str.split('_', expand=True)[[0, 2, 3, 4]]
.set_axis('Node Unit unit_num vdu_num'.split(), axis=1)
], axis=1)
df['vm'] = df['Unit'].str.cat(
df['unit_num'].where(df['vdu_num'].isnull(), df['vdu_num']), sep='_')