How to replace NaNs by preceding values in pandas

问题:

Suppose I have a DataFrame with some NaNs:

>>> import pandas as pd
>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df
    0   1   2
0   1   2   3
1   4 NaN NaN
2 NaN NaN   9

What I need to do is replace every NaN with the first non-NaN value in the same column above it. It is assumed that the first row will never contain a NaN. So for the previous example the result would be

   0  1  2
0  1  2  3
1  4  2  3
2  4  2  9

I can just loop through the whole DataFrame column-by-column, element-by-element and set the values directly, but is there an easy (optimally a loop-free) way of achieving this?

回答1:

You could use the fillna method on the DataFrame and specify the method as ffill (forward fill):

>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df.fillna(method=\'ffill\')
   0  1  2
0  1  2  3
1  4  2  3
2  4  2  9

This method...

propagate[s] last valid observation forward to next valid

To go the opposite way, there\'s also a bfill method.

This method doesn\'t modify the DataFrame inplace - you\'ll need to rebind the returned DataFrame to a variable or else specify inplace=True:

df.fillna(method=\'ffill\', inplace=True)

回答2:

You can use pandas.DataFrame.fillna with the method=\'ffill\' option. \'ffill\' stands for \'forward fill\' and will propagate last valid observation forward. The alternative is \'bfill\' which works the same way, but backwards.

import pandas as pd

df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
df = df.fillna(method=\'ffill\')

print(df)
#   0  1  2
#0  1  2  3
#1  4  2  3
#2  4  2  9

There is also a direct synonym function for this, pandas.DataFrame.ffill, to make things simpler.

回答3:

The accepted answer is perfect. I had a related but slightly different situation where I had to fill in forward but only within groups. In case someone has the same need, know that fillna works on a DataFrameGroupBy object.

>>> example = pd.DataFrame({\'number\':[0,1,2,nan,4,nan,6,7,8,9],\'name\':list(\'aaabbbcccc\')})
>>> example
  name  number
0    a     0.0
1    a     1.0
2    a     2.0
3    b     NaN
4    b     4.0
5    b     NaN
6    c     6.0
7    c     7.0
8    c     8.0
9    c     9.0
>>> example.groupby(\'name\')[\'number\'].fillna(method=\'ffill\') # fill in row 5 but not row 3
0    0.0
1    1.0
2    2.0
3    NaN
4    4.0
5    4.0
6    6.0
7    7.0
8    8.0
9    9.0
Name: number, dtype: float64

回答4:

One thing that I noticed when trying this solution is that if you have N/A at the start or the end of the array, ffill and bfill don\'t quite work. You need both.

In [224]: df = pd.DataFrame([None, 1, 2, 3, None, 4, 5, 6, None])

In [225]: df.ffill()
Out[225]:
     0
0  NaN
1  1.0
...
7  6.0
8  6.0

In [226]: df.bfill()
Out[226]:
     0
0  1.0
1  1.0
...
7  6.0
8  NaN

In [227]: df.bfill().ffill()
Out[227]:
     0
0  1.0
1  1.0
...
7  6.0
8  6.0

回答5:

ffill now has it\'s own method pd.DataFrame.ffill

df.ffill()

     0    1    2
0  1.0  2.0  3.0
1  4.0  2.0  3.0
2  4.0  2.0  9.0

回答6:

In my case, we have time series from different devices but some devices could not send any value during some period. So we should create NA values for every device and time period and after that do fillna.

df = pd.DataFrame([[\"device1\", 1, \'first val of device1\'], [\"device2\", 2, \'first val of device2\'], [\"device3\", 3, \'first val of device3\']])
df.pivot(index=1, columns=0, values=2).fillna(method=\'ffill\').unstack().reset_index(name=\'value\')

Result:

        0   1   value
0   device1     1   first val of device1
1   device1     2   first val of device1
2   device1     3   first val of device1
3   device2     1   None
4   device2     2   first val of device2
5   device2     3   first val of device2
6   device3     1   None
7   device3     2   None
8   device3     3   first val of device3

回答7:

Only one column version

• Fill NAN with last valid value

df[column_name].fillna(method=\'ffill\', inplace=True)

• Fill NAN with next valid value

df[column_name].fillna(method=\'backfill\', inplace=True)