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python - Pandas：在一系列可用值之前/之后估算给定数量的缺失值

问题描述

假设我有一个时间序列，其中我通常有某个连续年份的可用数据，但在该跨度之前和之后缺少值，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

有没有办法根据可用值中看到​​的趋势来填充（给定数量的）缺失值？

假设我想在每个方向上最多填充 2 个值，结果必须如下所示：

   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：有没有办法确保仅在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（或更一般地说，我只想在每个方向上填充最多 2 个值）条款，仅当 n + m 可用时才填写 n) ?

标签： pythonpandasimputation

解决方案

感谢@olv1do 向我展示了interpolate()可以满足我的需求。

使用 interpolate 和.first_valid_indexand.last_valid_index允许实现所需的行为：

#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

对于问题中的示例：

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

如果可用的值少于 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，spline如果值不像我的示例中那样完全线性，该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

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