计算大熊猫每组的数值差异

Question

我的Dataframe有以下结构：

patient_id  |  timestamp  |  measurement
A           |  2014-10-10 |  5.7
A           |  2014-10-11 |  6.3
B           |  2014-10-11 |  6.1
B           |  2014-10-10 |  4.1

我想计算一个delta (差)之间的每一个测量的每一个病人。

结果应该是：

patient_id  |  timestamp  |  measurement  |    delta
A           |  2014-10-10 |  5.7          |     NaN
A           |  2014-10-11 |  6.3          |     0.6
B           |  2014-10-11 |  6.1          |     2.0
B           |  2014-10-10 |  4.1          |     NaN

怎么才能在熊猫身上做到最优雅呢？

Answer 1

在“度量值”列上调用transform并传递方法diff，transform返回一系列与原始df对齐的索引：

In [4]:

df['delta'] = df.groupby('patient_id')['measurement'].transform(pd.Series.diff)
df
Out[4]:
  patient_id   timestamp  measurement  delta
0          A  2014-10-10          5.7    NaN
1          A  2014-10-11          6.3    0.6
2          B  2014-10-10          4.1    NaN
3          B  2014-10-11          6.1    2.0

编辑

如果您打算对transform的结果应用一些排序，那么首先对df进行排序：

In [10]:

df['delta'] = df.sort(columns=['patient_id', 'timestamp']).groupby('patient_id')['measurement'].transform(pd.Series.diff)
df
Out[10]:
  patient_id   timestamp  measurement  delta
0          A  2014-10-10          5.7    NaN
1          A  2014-10-11          6.3    0.6
2          B  2014-10-11          6.1    2.0
3          B  2014-10-10          4.1    NaN