Python 熊猫：在一系列可用值之前/之后输入给定数量的缺失值

假设我有一个时间序列，我通常有连续几年的数据，但在这段时间前后缺少值，如下所示：

df = pd.DataFrame({'year': ["2000","2001","2002", "2003","2004", "2005","2006", "2007"], 'cakes eaten': [np.nan, np.nan, np.nan, 3, 4, 5, np.nan, np.nan]})
print(df)

   year  cakes eaten
0  2000          NaN
1  2001          NaN
2  2002          NaN
3  2003          3.0
4  2004          4.0
5  2005          5.0
6  2006          NaN
7  2007          NaN

   year  cakes eaten
0  2000          NaN
1  2001          1.0
2  2002          2.0
3  2003          3.0
4  2004          4.0
5  2005          5.0
6  2006          6.0
7  2007          7.0

是否有办法根据可用值中的趋势来填充（给定数量的）缺失值

假设我想在每个方向上填充最多2个值，结果必须如下所示：

df = pd.DataFrame({'year': ["2000","2001","2002", "2003","2004", "2005","2006", "2007"], 'cakes eaten': [np.nan, np.nan, np.nan, 3, 4, 5, np.nan, np.nan]})
print(df)

   year  cakes eaten
0  2000          NaN
1  2001          NaN
2  2002          NaN
3  2003          3.0
4  2004          4.0
5  2005          5.0
6  2006          NaN
7  2007          NaN

   year  cakes eaten
0  2000          NaN
1  2001          1.0
2  2002          2.0
3  2003          3.0
4  2004          4.0
5  2005          5.0
6  2006          6.0
7  2007          7.0

---

另外：是否有办法确保只有当有足够的可用值时才执行此插补，例如，如果至少有3个可用值，我只想在每个方向上最多填充2个值（或者更一般地说，如果n+m可用，则仅填充n）？

我会使用上面提到的插值（）。您可以使用各种方法来产生不同的结果。我使用

krogh

方法得到了一条线性趋势线<在两个方向填充趋势时，需要代码>限制\方向='both'：

test_dict  = {'col': [np.nan, np.nan,np.nan, np.nan, np.nan, 4, 5, 6 ,np.nan]}
df = pd.DataFrame(test_dict)
df['trend'] = df['col'].interpolate(method='krogh', limit_direction='both')

    col trend
0   NaN -1.0
1   NaN 0.0
2   NaN 1.0
3   NaN 2.0
4   NaN 3.0
5   4.0 4.0
6   5.0 5.0
7   6.0 6.0
8   NaN 7.0

---

完成后，您可以删除0以下不需要的

趋势值。
感谢@olv1do向我展示了我想要的功能

使用interpolate和
。first\u valid\u index
和
。last\u valid\u index
可实现所需的行为：

#impute n values in both directions if at least m values are available
def interpolate(data, n, m):
  first_valid = data['cakes eaten'].first_valid_index()
  last_valid = data['cakes eaten'].last_valid_index()

  if(abs(first_valid - last_valid) + 1 >= m):
    data['imputed'] = data['cakes eaten'].interpolate(method='spline',order = 1, limit_direction='both', limit = n)
  return data

关于问题中的示例：

df = pd.DataFrame({'year': ["2000","2001","2002", "2003","2004", "2005","2006", "2007"], 'cakes eaten': [np.nan, np.nan, np.nan, 3, 4, 5, np.nan, np.nan]})
interpolate(df, 2,3)

year    cakes eaten     imputed
0   2000    NaN     NaN
1   2001    NaN     1.0
2   2002    NaN     2.0
3   2003    3.0     3.0
4   2004    4.0     4.0
5   2005    5.0     5.0
6   2006    NaN     6.0
7   2007    NaN     7.0

如果可用值少于m，则不执行任何操作：

df = pd.DataFrame({'year': ["2000","2001","2002", "2003","2004", "2005","2006", "2007"], 'cakes eaten': [np.nan, np.nan, np.nan, 3, 4,  np.nan, np.nan, np.nan]})
interpolate(df, 2,3)

    year    cakes eaten
0   2000    NaN
1   2001    NaN
2   2002    NaN
3   2003    3.0
4   2004    4.0
5   2005    NaN
6   2006    NaN
7   2007    NaN

此外，如果值不像我的示例中那样完全线性，则
样条线
方法也可以很好地工作：

df = pd.DataFrame({'year': ["2000","2001","2002", "2003","2004", "2005","2006", "2007"], 'cakes eaten': [np.nan, np.nan, 1, 4, 2,  3, np.nan, np.nan]})
interpolate(df, 1,4)

    year    cakes eaten     imputed
0   2000    NaN     NaN
1   2001    NaN     1.381040
2   2002    1.0     1.000000
3   2003    4.0     4.000000
4   2004    2.0     2.000000
5   2005    3.0     3.000000
6   2006    NaN     3.433167
7   2007    NaN     NaN

您将如何识别可用值中的趋势？感谢您将我重新指向插值函数，它似乎确实能够实现我想要的功能。对于我上面发布的示例，Krogh工作得非常好，但是如果趋势不是完全线性的，则会产生一些非常奇怪的值。但是，我发现使用
order=2
的
spline
方法效果更好