我有数据框df
import pandas as pd
b=np.array([0,1,2,2,0,1,2,2,3,4,4,4,5,6,0,1,0,0]).reshape(-1,1)
c=np.array(['a','a','a','a','b','b','b','b','b','b','b','b','b','b','c','c','d','e']).reshape(-1,1)
df = pd.DataFrame(np.hstack([b,c]),columns=['Start','File'])
df
Out[22]:
Start File
0 0 a
1 1 a
2 2 a
3 2 a
4 0 b
5 1 b
6 2 b
7 2 b
8 3 b
9 4 b
10 4 b
11 4 b
12 5 b
13 6 b
14 0 c
15 1 c
16 0 d
17 0 e
我想使用index_File重命名索引 为了使0_a,1_a,... 17_e为indeces
答案 0 :(得分:4)
您可以在set_index
inplace=True
df.set_index(df.File.radd(df.index.astype(str) + '_'))
Start File
File
0_a 0 a
1_a 1 a
2_a 2 a
3_a 2 a
4_b 0 b
5_b 1 b
6_b 2 b
7_b 2 b
8_b 3 b
9_b 4 b
10_b 4 b
11_b 4 b
12_b 5 b
13_b 6 b
14_c 0 c
15_c 1 c
16_d 0 d
17_e 0 e
以更多代码字符为代价,我们可以加快这一点,并处理不必要的索引名称
df.set_index(df.File.values.__radd__(df.index.astype(str) + '_'))
Start File
0_a 0 a
1_a 1 a
2_a 2 a
3_a 2 a
4_b 0 b
5_b 1 b
6_b 2 b
7_b 2 b
8_b 3 b
9_b 4 b
10_b 4 b
11_b 4 b
12_b 5 b
13_b 6 b
14_c 0 c
15_c 1 c
16_d 0 d
17_e 0 e
答案 1 :(得分:3)
您可以直接分配到索引,首先使用astype
将默认索引转换为str
,然后照常连接str:
In[41]:
df.index = df.index.astype(str) + '_' + df['File']
df
Out[41]:
Start File
File
0_a 0 a
1_a 1 a
2_a 2 a
3_a 2 a
4_b 0 b
5_b 1 b
6_b 2 b
7_b 2 b
8_b 3 b
9_b 4 b
10_b 4 b
11_b 4 b
12_b 5 b
13_b 6 b
14_c 0 c
15_c 1 c
16_d 0 d
17_e 0 e