如何冻结 Python pandas 数据帧中 NaN 之间序列中的第一个数字
How to freeze first numbers in sequences between NaNs in Python pandas dataframe
有没有一种 Pythonic 的方法,在时间序列数据框中,按列向下选择序列中的第一个数字,然后将其向前推直到下一个 NaN,然后取下一个非 NaN 数字并将那个向下推直到下一个 NaN,依此类推(保留索引和 NaN)。
例如,我想转换这个数据帧:
DF = pd.DataFrame(data={'A':[np.nan,1,3,5,7,np.nan,2,4,6,np.nan], 'B':[8,6,4,np.nan,np.nan,9,7,3,np.nan,3], 'C':[np.nan,np.nan,4,2,6,np.nan,1,5,2,8]})
A B C
0 NaN 8.0 NaN
1 1.0 6.0 NaN
2 3.0 4.0 4.0
3 5.0 NaN 2.0
4 7.0 NaN 6.0
5 NaN 9.0 NaN
6 2.0 7.0 1.0
7 4.0 3.0 5.0
8 6.0 NaN 2.0
9 NaN 3.0 8.0
到这个数据框:
Result = pd.DataFrame(data={'A':[np.nan,1,1,1,1,np.nan,2,2,2,np.nan], 'B':[8,8,8,np.nan,np.nan,9,9,9,np.nan,3], 'C':[np.nan,np.nan,4,4,4,np.nan,1,1,1,1]})
A B C
0 NaN 8.0 NaN
1 1.0 8.0 NaN
2 1.0 8.0 4.0
3 1.0 NaN 4.0
4 1.0 NaN 4.0
5 NaN 9.0 NaN
6 2.0 9.0 1.0
7 2.0 9.0 1.0
8 2.0 NaN 1.0
9 NaN 3.0 1.0
我知道我可以使用循环来遍历列来执行此操作,但希望能提供一些帮助,帮助您了解如何在非常大的数据帧上以更高效的 Pythonic 方式执行此操作。谢谢。
IIUC:
# where DF is not NaN
mask = DF.notna()
Result = (DF.shift(-1) # fill the original NaN's with their next value
.mask(mask) # replace all the original non-NaN with NaN
.ffill() # forward fill
.fillna(DF.iloc[0]) # starting of the the columns with a non-NaN
.where(mask) # replace the original NaN's back
)
输出:
A B C
0 NaN 8.0 NaN
1 1.0 8.0 NaN
2 1.0 8.0 4.0
3 1.0 NaN 4.0
4 1.0 NaN 4.0
5 NaN 9.0 NaN
6 2.0 9.0 1.0
7 2.0 9.0 1.0
8 2.0 NaN 1.0
9 NaN 3.0 1.0
有没有一种 Pythonic 的方法,在时间序列数据框中,按列向下选择序列中的第一个数字,然后将其向前推直到下一个 NaN,然后取下一个非 NaN 数字并将那个向下推直到下一个 NaN,依此类推(保留索引和 NaN)。
例如,我想转换这个数据帧:
DF = pd.DataFrame(data={'A':[np.nan,1,3,5,7,np.nan,2,4,6,np.nan], 'B':[8,6,4,np.nan,np.nan,9,7,3,np.nan,3], 'C':[np.nan,np.nan,4,2,6,np.nan,1,5,2,8]})
A B C
0 NaN 8.0 NaN
1 1.0 6.0 NaN
2 3.0 4.0 4.0
3 5.0 NaN 2.0
4 7.0 NaN 6.0
5 NaN 9.0 NaN
6 2.0 7.0 1.0
7 4.0 3.0 5.0
8 6.0 NaN 2.0
9 NaN 3.0 8.0
到这个数据框:
Result = pd.DataFrame(data={'A':[np.nan,1,1,1,1,np.nan,2,2,2,np.nan], 'B':[8,8,8,np.nan,np.nan,9,9,9,np.nan,3], 'C':[np.nan,np.nan,4,4,4,np.nan,1,1,1,1]})
A B C
0 NaN 8.0 NaN
1 1.0 8.0 NaN
2 1.0 8.0 4.0
3 1.0 NaN 4.0
4 1.0 NaN 4.0
5 NaN 9.0 NaN
6 2.0 9.0 1.0
7 2.0 9.0 1.0
8 2.0 NaN 1.0
9 NaN 3.0 1.0
我知道我可以使用循环来遍历列来执行此操作,但希望能提供一些帮助,帮助您了解如何在非常大的数据帧上以更高效的 Pythonic 方式执行此操作。谢谢。
IIUC:
# where DF is not NaN
mask = DF.notna()
Result = (DF.shift(-1) # fill the original NaN's with their next value
.mask(mask) # replace all the original non-NaN with NaN
.ffill() # forward fill
.fillna(DF.iloc[0]) # starting of the the columns with a non-NaN
.where(mask) # replace the original NaN's back
)
输出:
A B C
0 NaN 8.0 NaN
1 1.0 8.0 NaN
2 1.0 8.0 4.0
3 1.0 NaN 4.0
4 1.0 NaN 4.0
5 NaN 9.0 NaN
6 2.0 9.0 1.0
7 2.0 9.0 1.0
8 2.0 NaN 1.0
9 NaN 3.0 1.0