How to mask values in column based on a condition per group

Question

I have pandas DataFrame like this:

data = {'ID_1':['A', 'A','A', 'B', 'B', 'B'],
        'ID_2':[1, 2, 2, 1, 1, 2],
        'DATE':['2021-11-21', '2021-12-19', '2021-09-05', '2021-11-07', '2021-12-05','2021-12-26'],
        'VALUE': [0.5, 0.5, 0.5, 0.6, 0.6, 0.6]}
 
df = pd.DataFrame(data)

And I would like to leave value in column 'VALUE' only for lowest date from column 'DATE' for subset of 'ID_1' and 'ID_2'
Desired output look like this:

data = {'ID_1':['A', 'A','A', 'B', 'B', 'B'],
        'ID_2':[1, 2, 2, 1, 1, 2],
        'DATE':['2021-11-21', '2021-12-19', '2021-09-05', '2021-11-07', '2021-12-05','2021-12-26'],
        'VALUE': [0.5, np.NaN, 0.5, 0.6, np.NaN, 0.6]}
 
df = pd.DataFrame(data)

What I tried is to create function which twice grouping this dataframe but I'm ending with ValueError Length of values (2) does not match length of index (1)

My function:

def foo(val):
    
    def add_mask(val):
        val.reset_index(inplace=True)
        min_date = val['DATE'].min()
        mask = val.DATE == min_date
        return val[mask]
    
    return val.groupby('ID_1').apply(add_mask)

test = df.groupby('ID_2').apply(foo)

Answer 1

You can groupby "ID_1" and "ID_2" and transform the min of "DATE" for each group for the DataFrame. Then use eq to identify the rows where the group mins exist. Finally, use where to assign NaN values to "VALUE"s that are not min:

df['VALUE'] = df['VALUE'].where(df.groupby(['ID_1','ID_2'])['DATE'].transform('min').eq(df['DATE']))

Output:

  ID_1  ID_2        DATE  VALUE
0    A     1  2021-11-21    0.5
1    A     2  2021-12-19    NaN
2    A     2  2021-09-05    0.5
3    B     1  2021-11-07    0.6
4    B     1  2021-12-05    NaN
5    B     2  2021-12-26    0.6

Function foo doesn't work because you never use mask you create in it to modify "VALUE" in each group. If you replace

return val[mask]

with

val['VALUE'] = val['VALUE'].where(mask)
return val

it will produce the expected outcome (you'll need to fix the index but the general structure will be what you expect).

Answer 2

Many elegant answers, but here is how I will go about it;

grp = df.groupby(["ID_1", "ID_2"])
grp

def change(df):
    df.loc[df.DATE != df.DATE.min(), 'VALUE'] = np.nan
    return df

grp.apply(change)

Results in:

    ID_1    ID_2    DATE    VALUE
0   A   1   2021-11-21  0.5
1   A   2   2021-12-19  NaN
2   A   2   2021-09-05  0.5
3   B   1   2021-11-07  0.6
4   B   1   2021-12-05  NaN
5   B   2   2021-12-26  0.6

Answer 3

Just another way to do it:

df['DATE'] = df.groupby(['ID_1','ID_2']).DATE.transform(lambda x: (x==min(x))*x).replace('', np.NaN)

Using the boolean (x==min(x)):

df['is_min'] = df.groupby(['ID_1','ID_2']).DATE.transform(lambda x: x==min(x))
#
#   ID_1  ID_2        DATE  VALUE  is_min
# 0    A     1  2021-11-21    0.5    True
# 1    A     2  2021-12-19    0.5   False
# 2    A     2  2021-09-05    0.5    True
# 3    B     1  2021-11-07    0.6    True
# 4    B     1  2021-12-05    0.6   False
# 5    B     2  2021-12-26    0.6    True