如何用数据帧的 NaN 删除第一行和最后一行,并将剩余的 NaN 替换为上下值的平均值?

How to delete the first and last rows with NaN of a dataframe and replace the remaining NaN with the average of the values below and above?

让我们以这个数据帧为例:

df = pd.DataFrame(dict(Col1=[np.nan,1,1,2,3,8,7], Col2=[1,1,np.nan,np.nan,3,np.nan,4], Col3=[1,1,np.nan,5,1,1,np.nan]))

   Col1  Col2  Col3
0   NaN   1.0   1.0
1   1.0   1.0   1.0
2   1.0   NaN   NaN
3   2.0   NaN   5.0
4   3.0   3.0   1.0
5   8.0   NaN   1.0
6   7.0   4.0   NaN

我想先删除第一行和最后一行,直到第一行和最后一行不再有 NaN。

中间预期输出:

   Col1  Col2  Col3
1   1.0   1.0   1.0
2   1.0   NaN   NaN
3   2.0   NaN   5.0
4   3.0   3.0   1.0

然后,我想将剩余的 NaN 替换为最近的不是 NaN 的值,以及上面的那个。

最终预期输出:

   Col1  Col2  Col3
0   1.0   1.0   1.0
1   1.0   2.0   3.0
2   2.0   2.0   5.0
3   3.0   3.0   1.0

我知道我可以通过

在我的数据框中找到 NaN 的位置
df.isna()

但我无法解决我的问题。请问我该怎么办?

我的做法:

# identify the rows with some NaN
s = df.notnull().all(1)

# remove those with NaN at beginning and at the end:
new_df = df.loc[s.idxmax():s[::-1].idxmax()]

# average:
new_df = (new_df.ffill()+ new_df.bfill())/2

输出:

   Col1  Col2  Col3
1   1.0   1.0   1.0
2   1.0   2.0   3.0
3   2.0   2.0   5.0
4   3.0   3.0   1.0

另一种选择是使用 DataFrame.interpolateround:

nans = df.notna().all(axis=1).cumsum().drop_duplicates()
low, high = nans.idxmin(), nans.idxmax()

df.loc[low+1: high].interpolate().round()

   Col1  Col2  Col3
1   1.0   1.0   1.0
2   1.0   2.0   3.0
3   2.0   2.0   5.0
4   3.0   3.0   1.0