Pandas 根据 DataFrame 中的另一列填充 NA 的增量值

Question

I have a dataframe with sessions for each user. One of the column is sessions till now. Some of these sessions have null values. I believe I could use fillna and transform methods to appropriately fill the dataframe.

import pandas as pd

df = pd.DataFrame({'user': [A, A, A, A, A, B, B, B, B, C, C, C, C, C],  'sessions': [28, NaN, NaN, NaN , 32, NaN, NaN,NaN,12, NaN,15, NaN, 17,NaN]})

Expected Output DataFrame:

df_out = pd.DataFrame({'user': [A, A, A, A, A, B, B, B, B, C, C, C, C, C],  'sessions': [28, 29, 30, 31 , 32, 9, 10, 11,12, 14,15,16,17,18]})

Tried Code:

df['sessions'] = df['sessions'].fillna(df.groupby('user')['sessions'].transform('mean'))

this works if I were to fill mean and this is as far as I could think. Please suggest a few approaches.

PS - The starting value of the session is not 1. I am doing it from a snapshot at some point of time. I do not have data going back till session number 1 for every user.

Answer 1

Assuming there is no mismatch between the not NaN values, you could do the following:

def fun(x):
    _, diff = (~x.reset_index().isna()).idxmax()  # find the absolute position of the first non NaN

    start = x[(~x.isna()).idxmax()] - diff  # find the start value

    result = pd.RangeIndex(start, start + len(x))  # generate range based on first value and length of group

    return pd.Series(data=result.values, index=x.index)  # return series


df['count'] = df.groupby('user').sessions.apply(fun)

print(df)

Output

   user  sessions  count
0     A      28.0     28
1     A       NaN     29
2     A       NaN     30
3     A       NaN     31
4     A      32.0     32
5     B       NaN      9
6     B       NaN     10
7     B       NaN     11
8     B      12.0     12
9     C       NaN     14
10    C      15.0     15
11    C       NaN     16
12    C      17.0     17
13    C       NaN     18

The first line of the function fun , is equivalent to:

diff = (~x.reset_index().isna()).idxmax()[1]

Basically find the index position in a normalized (starting from 0) index.

Answer 2

Use cumsum with fillna(1) for each group:

df.groupby('user',sort=False)['sessions'].apply(lambda x: x.fillna(1).cumsum()).reset_index()

Answer 3

You may re-construct sessions by using groupby cumcount and first

s = df.groupby('user').sessions.cumcount()
s1 = (df.sessions - s).groupby(df.user).transform('first')

df['sessions'] = s1 + s

In [867]: df
Out[867]:
   user  sessions
0     A      28.0
1     A      29.0
2     A      30.0
3     A      31.0
4     A      32.0
5     B       9.0
6     B      10.0
7     B      11.0
8     B      12.0
9     C      14.0
10    C      15.0
11    C      16.0
12    C      17.0
13    C      18.0