pandas .diff 按行的多索引值

Question

In [66]: t1
Out[69]: 
job_date  branch_id
2018-05   1            0.618980
          2            0.600590
          3            0.603486
          4            0.043931
          5            0.588168
          6            0.381518
          7            0.357035
2018-06   1            0.690575
          2            0.700900
          3            0.571556
          4            0.351935
          5            0.626428
          6            0.461813
          7            0.329663
Name: utilization, dtype: float64

In [86]: t1.index
Out[86]: 
MultiIndex(levels=[[2018-05, 2018-06], [1, 2, 3, 4, 5, 6, 7]],
           labels=[[0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1], [0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6]],
           names=['job_date', 'branch_id'])

如何通过索引值对行进行差异化？

所以(2018-05, 1) 和 (2018-06, 1)之间的差异 应该是0.690575-0.618980=0.071595

如果我做 t1.diff()，我会得到逐行比较，这不是我想要的

In [87]: t1.diff()
Out[87]: 
job_date  branch_id
2018-05   1                 NaN
          2           -0.018390
          3            0.002895
          4           -0.559554
          5            0.544237
          6           -0.206651
          7           -0.024483
2018-06   1            0.333540
          2            0.010325
          3           -0.129345
          4           -0.219621
          5            0.274494
          6           -0.164615
          7           -0.132150

现在我正在做这个

In [49]: t1.unstack(level=0)['utilization'].diff(axis=1)
Out[49]: 
job_date   2018-05   2018-06
branch_id                   
1              NaN  0.071595
2              NaN  0.100310
3              NaN -0.031930
4              NaN  0.308003
5              NaN  0.038260
6              NaN  0.080295
7              NaN -0.027372

有没有不拆的方法？？

Answer 1

一种可能的解决方案是移动MultiIndex一个月并减去，如果每个之间的差异Period相同，则它有效 - 这里是一个月：

a = df.index.get_level_values(0).to_period('M')
b = df.index.get_level_values(1)
mux1 = pd.MultiIndex.from_arrays([a,b], names=df.index.names)
mux2 = pd.MultiIndex.from_arrays([a + 1, b], names=df.index.names)

df = df.set_index(mux1)
df1 = df.set_index(mux2)

df['utilization'] = df.sub(df1)
print (df)
                    utilization
job_date branch_id             
2018-05  1                  NaN
         2                  NaN
         3                  NaN
         4                  NaN
         5                  NaN
         6                  NaN
         7                  NaN
2018-06  1             0.071595
         2             0.100310
         3            -0.031930
         4             0.308004
         5             0.038260
         6             0.080295
         7            -0.027372