pandas：根据其他列将多行中一个单元格的值替换为一个特定行

Question

my aim:

     uniqueIdentity    beginTime    progrNumber
0   2018-02-07-6253554  17:40:29    1
1   2018-02-07-6253554  17:40:29    2
2   2018-02-07-6253554  17:40:29    3
3   2018-02-07-6253554  17:40:29    4
4   2018-02-07-6253554  17:40:29    5
5   2018-02-07-5555333  17:48:29    2
6   2018-02-07-5555333  17:48:29    3
7   2018-02-07-5555333  17:48:29    4
8   2018-02-07-2345622  18:40:29    1
9   2018-02-07-2345622  18:40:29    2
10  2018-02-07-2345622  18:40:29    3
11  2018-02-07-2345622  18:40:29    4

my dataset now:

     uniqueIdentity    beginTime    progrNumber
0   2018-02-07-6253554  17:40:29    1
1   2018-02-07-6253554  17:41:15    2
2   2018-02-07-6253554  17:41:55    3
3   2018-02-07-6253554  17:42:54    4
4   2018-02-07-6253554  17:43:29    5
5   2018-02-07-5555333  17:49:15    2
6   2018-02-07-5555333  17:49:55    3
7   2018-02-07-5555333  17:50:54    4
8   2018-02-07-2345622  18:40:29    1
9   2018-02-07-2345622  18:41:15    2
10  2018-02-07-2345622  18:41:55    3
11  2018-02-07-2345622  18:42:54    4

That means: for rows having same 'uniqueIdentity', the 'beginTime' should be replaced by the value of cell which having the same'uniqueIdentity' and 'progrNumber' is the min 'progrNumber'.

Answer 1

As you mention in the comments, the lowest progrNumber will also be the lowest beginTime . This means you can just take the lowest beginTime per uniqueIdentity using groupby and transform .

Note if beginTime is of type string, this will only work if it has consistent formatting. (eg '09:40:20' instead of '9:40:20')

df['beginTime'] = df.groupby('uniqueIdentity').beginTime.transform('min')

        uniqueIdentity beginTime progrNumber
0   2018-02-07-6253554  17:40:29           1
1   2018-02-07-6253554  17:40:29           2
2   2018-02-07-5555333  17:48:29           3
3   2018-02-07-5555333  17:48:29           4
4   2018-02-07-6253554  17:40:29           3
5   2018-02-07-6253554  17:40:29           4
6   2018-02-07-5555333  17:48:29           1
7   2018-02-07-5555333  17:48:29           2
8   2018-02-07-2345622  18:40:29           1
9   2018-02-07-2345622  18:40:29           3
10  2018-02-07-2345622  18:40:29           4

Answer 2

Here's another option using a left join and some renaming

    # find rows where progrNumber is 1 
    df_prog1=df[df.progrNumber==1]
    # do a left join on the original 
    df=df.merge(df_prog1,on='uniqueIdentity',how='left',suffixes=('','_y'))
    # keep only the beginTime from the right frame 
    df=df[['uniqueIdentity','beginTime_y','progrNumber']]
    # rename columns
    df=df.rename(columns={'beginTime_y':'beginTime'})
    print(df)

Results in:

        uniqueIdentity beginTime  progrNumber
0   2018-02-07-6253554  17:40:29            1
1   2018-02-07-6253554  17:40:29            2
2   2018-02-07-6253554  17:40:29            3
3   2018-02-07-6253554  17:40:29            4
4   2018-02-07-5555333  17:48:29            1
5   2018-02-07-5555333  17:48:29            2
6   2018-02-07-5555333  17:48:29            3
7   2018-02-07-5555333  17:48:29            4
8   2018-02-07-2345622  18:40:29            1
9   2018-02-07-2345622  18:40:29            2
10  2018-02-07-2345622  18:40:29            3
11  2018-02-07-2345622  18:40:29            4

if you're not sure which record within a uniqueIdentity will have the minimum time, you can use a groupby instead of selecting where progrNumber==1 :

    df_prog1=df.groupby('uniqueIdentity')['beginTime'].min().reset_index()

And do the left join as above.

Answer 3

If the first beginTime for each user will always correspond to the minimum program number for each user, you can do:

d = df.groupby('uniqueIdentity')['beginTime'].first().to_dict()
df['beginTime'] = df['uniqueIdentity'].map(d)

To be more explicit about getting the time where the program number is minimum (regardless of its position), you replace d in the above with:

d = df.groupby('uniqueIdentity').apply(lambda x: x['beginTime'][x['progrNumber'].idxmin()]).to_dict()

These two yield the same result for your example data, but they will differ if there are users where the first beginTime (or minimum beginTime per Hugolmn) does not correspond to the minimum progrNumber for the user

Answer 4

Using groupby and map

The hypothesis is that beginTime will always be minimal for a minimal progrNumber . This condition is true based on the question's comments.

In this answer, I collect the minimum beginTime of each uniqueIdentity and then map it to the original DataFrame based on uniqueIdentity .

times = df.groupby('uniqueIdentity').beginTime.min()
df['beginTime'] = df.uniqueIdentity.map(times)

Answer 5

If we cannot assume that the min progrNumber is also the min beginTime , a more sophisiticated approach is required:

df['beginTime'] = (
     df.groupby('uniqueIdentity', as_index=False, group_keys=False)
       .apply(lambda s: pd.Series(s[s.progrNumber==s.progrNumber.min()]
              .beginTime.item(), index=s.index)
       )
)

df
#    uniqueIdentity beginTime   progrNumber
# 0  2018-02-07-6253554 17:40:29    1
# 1  2018-02-07-6253554 17:40:29    2
# 2  2018-02-07-6253554 17:40:29    3
# 3  2018-02-07-6253554 17:40:29    4
# 4  2018-02-07-6253554 17:40:29    5
# 5  2018-02-07-5555333 17:49:15    2
# 6  2018-02-07-5555333 17:49:15    3
# 7  2018-02-07-5555333 17:49:15    4
# 8  2018-02-07-2345622 18:40:29    1
# 9  2018-02-07-2345622 18:40:29    2
# 10 2018-02-07-2345622 18:40:29    3
# 11 2018-02-07-2345622 18:40:29    4

If you don't want a oneliner, an approach with map would be ideal

mapping  = (
     df.groupby('uniqueIdentity')
       .apply(lambda s: s[s.progrNumber==s.progrNumber.min()].beginTime.iloc[0])
 )

 df['beingTime'] = df.uniqueIdentity.map(mapping)

note: You can replace the iloc[0] by item() if you guarantee that only one value has the min progrNumber