Question

我有如下数据：

{'grp': {0: 828893, 1: 828893, 2: 828893, 3: 828893, 4: 828893, 5: 828893, 6: 828893, 7: 828893, 8: 828893, 9: 828893, 10: 828893, 11: 828893, 12: 828893, 13: 828893, 14: 828893, 15: 828893, 16: 828893, 17: 828893, 18: 828893, 19: 828893, 20: 828893, 21: 828893, 22: 828893, 23: 828893, 24: 828893}, 'grp2': {0: nan, 1: nan, 2: nan, 3: nan, 4: '1', 5: '1', 6: '1', 7: '1', 8: '1', 9: '1', 10: nan, 11: nan, 12: '2', 13: '2', 14: '2', 15: '2', 16: nan, 17: nan, 18: nan, 19: '3', 20: nan, 21: '4', 22: '4', 23: '4', 24: '4'}, 'val1': {0: -50.0, 1: -50.0, 2: -50.0, 3: -50.0, 4: 7.600000000000001, 5: 54.599999999999994, 6: 38.599999999999994, 7: 50.599999999999994, 8: 91.0, 9: 100.80000000000001, 10: 19.200000000000003, 11: -50.0, 12: -50.0, 13: 69.6, 14: 42.0, 15: 90.19999999999999, 16: -50.0, 17: -50.0, 18: 47.599999999999994, 19: 98.80000000000001, 20: 27.599999999999994, 21: 11.799999999999997, 22: nan, 23: 13.0, 24: 0.0}, 'val2': {0: 0.0, 1: 0.0, 2: 0.0, 3: 0.0, 4: 30.1, 5: 21.5, 6: 20.7, 7: 4.2, 8: 5.0, 9: 21.6, 10: 85.1, 11: 0.0, 12: 0.0, 13: 36.4, 14: 56.6, 15: 51.2, 16: 0.0, 17: 0.0, 18: 58.5, 19: 42.2, 20: 76.1, 21: 68.7, 22: nan, 23: 90.3, 24: 95.3}}

我想先按列grp和grp2对其进行分组，然后创建一个新列val1_b和val2_b定义为最后一次观察与第一次观察之间的差异（在一组）分别来自val1和val2。 R中的代码类似于：

ex %>% 
  group_by(grp, grp2) %>% 
  mutate(val1_b = last(val1) - first(val1),
         val2_b = last(val2) - first(val2)) %>%
  ungroup()

但是我需要在Python中完成。我能得到的最接近的是：

pd.DataFrame(ex).groupby(['grp', 'grp2'])['val1'].apply(lambda x: x.iat[-1] - x.iat[0])

但是这仅适用于一列，结果是汇总的，而不是添加到现有的数据框中。因此，如何计算一组中最后几观察与第一观察之间的差异，并将其作为新列添加到现有数据框中？

Answer 1

将GroupBy.transform与GroupBy.first一起使用， GroupBy.last，一种可能的解决方案，其中有DataFrame.add_prefix和DataFrame.join用于新列：

df = pd.DataFrame(ex)
#columns for processing defined after groupby
g = df.groupby(['grp', 'grp2'])['val1', 'val2']
out = df.join((g.transform('last') - g.transform('first')).add_prefix('new_'))

可能没有评论join，就像评论中提到的@ Wen-Ben一样（谢谢）：

df[['new_val1',  'new_val2']] = g.transform('last') - g.transform('first')

print (out)
       grp grp2   val1  val2  new_val1  new_val2
0   828893  NaN  -50.0   0.0       NaN       NaN
1   828893  NaN  -50.0   0.0       NaN       NaN
2   828893  NaN  -50.0   0.0       NaN       NaN
3   828893  NaN  -50.0   0.0       NaN       NaN
4   828893    1    7.6  30.1      93.2      -8.5
5   828893    1   54.6  21.5      93.2      -8.5
6   828893    1   38.6  20.7      93.2      -8.5
7   828893    1   50.6   4.2      93.2      -8.5
8   828893    1   91.0   5.0      93.2      -8.5
9   828893    1  100.8  21.6      93.2      -8.5
10  828893  NaN   19.2  85.1       NaN       NaN
11  828893  NaN  -50.0   0.0       NaN       NaN
12  828893    2  -50.0   0.0     140.2      51.2
13  828893    2   69.6  36.4     140.2      51.2
14  828893    2   42.0  56.6     140.2      51.2
15  828893    2   90.2  51.2     140.2      51.2
16  828893  NaN  -50.0   0.0       NaN       NaN
17  828893  NaN  -50.0   0.0       NaN       NaN
18  828893  NaN   47.6  58.5       NaN       NaN
19  828893    3   98.8  42.2       0.0       0.0
20  828893  NaN   27.6  76.1       NaN       NaN
21  828893    4   11.8  68.7     -11.8      26.6
22  828893    4    NaN   NaN     -11.8      26.6
23  828893    4   13.0  90.3     -11.8      26.6
24  828893    4    0.0  95.3     -11.8      26.6

Answer 2

您的意思是R中的mutate，这里pandas中的是transform

df=pd.DataFrame(ex)
g=df.groupby(['grp', 'grp2'])
df['val1_b']=g['val1'].transform('first')-g['val1'].transform('last')
df['val2_b']=g['val2'].transform('first')-g['val1'].transform('last')

如何在大熊猫中的组中创建最后值和第一个值之间的差异的列

2 个答案: