IIUC，您需要反转diff和fillna：

df['desired'] = df['col'].diff(-1).fillna(df['col'])

输出：

idx  col  desired
0    0   47     14.0
1    1   33     10.0
2    2   23    -10.0
3    3   33      1.0
4    4   32      1.0
5    5   31      9.0
6    6   22     17.0
7    7    5      5.0

a = {'idx': range(8),
     'col': [47,33,23,33,32,31,22,5],
     }

df = pd.DataFrame(a)

df['col'] - df['col'].shift(-1, fill_value=0)

0    14
1    10
2   -10
3     1
4     1
5     9
6    17
7     5
Name: col, dtype: int64

df['desired'] = df['col'] - df['col'].shift(-1, fill_value=0)

   idx  col  desired
0    0   47       14
1    1   33       10
2    2   23      -10
3    3   33        1
4    4   32        1
5    5   31        9
6    6   22       17
7    7    5        5

与使用numpy的@mozway相同的解决方案（更快）：

import numpy as np

df['desired'] = -np.diff(df['col'], append=0)

输出：

>>> df
   idx  col  desired
0    0   47       14
1    1   33       10
2    2   23      -10
3    3   33        1
4    4   32        1
5    5   31        9
6    6   22       17
7    7    5        5

对于10 k条记录：

# @mozway
>>> %timeit df['col'].diff(-1).fillna(df['col'])
281 µs ± 14 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

# @GodIsOne
>>> %timeit df['col'] - df['col'].shift(-1, fill_value=0)
144 µs ± 4.18 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

# @Corralien
>>> %timeit (-np.diff(df['col'], append=0))
32.7 µs ± 951 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)