问题
Say I have a DataFrame like
A B
0 0.1880 0.345
1 0.2510 0.585
2 NaN NaN
3 NaN NaN
4 NaN 1.150
5 0.2300 1.210
6 0.1670 1.290
7 0.0835 1.400
8 0.0418 NaN
9 0.0209 NaN
10 NaN NaN
11 NaN NaN
12 NaN NaN
I want a new DataFrame of the same shape where each entry represents the number of NaNs counted up to its position started from the last valid value as follows
A B
0 0 0
1 0 0
2 1 1
3 2 2
4 3 0
5 0 0
6 0 0
7 0 0
8 0 1
9 0 2
10 1 3
11 2 4
12 3 5
I wonder if this can be done efficiently by utilizing some of the Pandas/Numpy functions?
回答1:
You can use:
a = df.isnull()
b = a.cumsum()
df1 = b.sub(b.mask(a).ffill().fillna(0).astype(int))
print (df1)
A B
0 0 0
1 0 0
2 1 1
3 2 2
4 3 0
5 0 0
6 0 0
7 0 0
8 0 1
9 0 2
10 1 3
11 2 4
12 3 5
For better understanding:
#add NaN where True in a
a2 = b.mask(a)
#forward filling NaN
a3 = b.mask(a).ffill()
#replace NaN to 0, cast to int
a4 = b.mask(a).ffill().fillna(0).astype(int)
#substract b to a4
a5 = b.sub(b.mask(a).ffill().fillna(0).astype(int))
df1 = pd.concat([a,b,a2, a3, a4, a5], axis=1,
keys=['a','b','where','ffill nan','substract','output'])
print (df1)
a b where ffill nan substract output
A B A B A B A B A B A B
0 False False 0 0 0.0 0.0 0.0 0.0 0 0 0 0
1 False False 0 0 0.0 0.0 0.0 0.0 0 0 0 0
2 True True 1 1 NaN NaN 0.0 0.0 0 0 1 1
3 True True 2 2 NaN NaN 0.0 0.0 0 0 2 2
4 True False 3 2 NaN 2.0 0.0 2.0 0 2 3 0
5 False False 3 2 3.0 2.0 3.0 2.0 3 2 0 0
6 False False 3 2 3.0 2.0 3.0 2.0 3 2 0 0
7 False False 3 2 3.0 2.0 3.0 2.0 3 2 0 0
8 False True 3 3 3.0 NaN 3.0 2.0 3 2 0 1
9 False True 3 4 3.0 NaN 3.0 2.0 3 2 0 2
10 True True 4 5 NaN NaN 3.0 2.0 3 2 1 3
11 True True 5 6 NaN NaN 3.0 2.0 3 2 2 4
12 True True 6 7 NaN NaN 3.0 2.0 3 2 3 5
来源:https://stackoverflow.com/questions/43517953/fast-way-to-get-the-number-of-nans-in-a-column-counted-from-the-last-valid-value