Machine_number Machine_Running_Hours
0 1.0 424.0
1 2.0 458.0
2 3.0 465.0
3 4.0 446.0
4 5.0 466.0
5 6.0 466.0
6 7.0 445.0
7 8.0 466.0
8 9.0 447.0
9 10.0 469.0
10 11.0 467.0
11 12.0 449.0
12 13.0 436.0
13 14.0 465.0
14 15.0 463.0
15 16.0 372.0
16 17.0 460.0
17 18.0 450.0
18 19.0 467.0
19 20.0 463.0
20 21.0 205.0
我正在尝试根据机器编号进行分类。与Machine_number一样，1到5将是一组。然后在一组中为6到10，依此类推。
答案 0 :(得分：2)
df['g'] = df.Machine_number.sub(1).floordiv(5)
#same as //
#df['g'] = df.Machine_number.sub(1) // 5
print (df)
Machine_number Machine_Running_Hours g
0 1.0 424.0 -0.0
1 2.0 458.0 0.0
2 3.0 465.0 0.0
3 4.0 446.0 0.0
4 5.0 466.0 0.0
5 6.0 466.0 1.0
6 7.0 445.0 1.0
7 8.0 466.0 1.0
8 9.0 447.0 1.0
9 10.0 469.0 1.0
10 11.0 467.0 2.0
11 12.0 449.0 2.0
12 13.0 436.0 2.0
13 14.0 465.0 2.0
14 15.0 463.0 2.0
15 16.0 372.0 3.0
16 17.0 460.0 3.0
17 18.0 450.0 3.0
18 19.0 467.0 3.0
19 20.0 463.0 3.0
20 21.0 205.0 4.0
如果需要存储在字典中，请groupby使用dict comprehension：
dfs = {i:g for i, g in df.groupby(df.Machine_number.astype(int).sub(1).floordiv(5))}
print (dfs)
{0: Machine_number Machine_Running_Hours
0 1.0 424.0
1 2.0 458.0
2 3.0 465.0
3 4.0 446.0
4 5.0 466.0, 1: Machine_number Machine_Running_Hours
5 6.0 466.0
6 7.0 445.0
7 8.0 466.0
8 9.0 447.0
9 10.0 469.0, 2: Machine_number Machine_Running_Hours
10 11.0 467.0
11 12.0 449.0
12 13.0 436.0
13 14.0 465.0
14 15.0 463.0, 3: Machine_number Machine_Running_Hours
15 16.0 372.0
16 17.0 460.0
17 18.0 450.0
18 19.0 467.0
19 20.0 463.0, 4: Machine_number Machine_Running_Hours
20 21.0 205.0}
print (dfs[0])
Machine_number Machine_Running_Hours
0 1.0 424.0
1 2.0 458.0
2 3.0 465.0
3 4.0 446.0
4 5.0 466.0
print (dfs[1])
Machine_number Machine_Running_Hours
5 6.0 466.0
6 7.0 445.0
7 8.0 466.0
8 9.0 447.0
9 10.0 469.0