Get and Modify column in groups on rows that meet a condition - pandas

Question

I have this DataFrame:
df = pd.DataFrame({'day': [1, 1, 1, 2, 2, 2, 3, 3, 3], 'hour': [10, 10, 10, 11, 11, 11, 12, 12, 12], 'sales': [0, 40, 30, 10, 80, 70, 0, 0, 20]})
day hour sales
0 1 10 0
1 1 10 40
2 1 10 30
3 2 11 10
4 2 11 80
5 2 11 70
6 3 12 0
7 3 12 0
8 3 12 20
And I would like to filter to get the first entry of each day that has volume greater than 0. And as an additional thing I would like to change the 'sales' column for these to 9.
So to get something like this:
day hour sales
0 1 10 0
1 1 9 40
2 1 10 30
3 2 9 10
4 2 11 80
5 2 11 70
6 3 12 0
7 3 12 0
8 3 9 20
I only came up with this iterative solution. But is there a solution, how I can apply it in a more functional way?
# Group by day:
groups = df.groupby(by=['day'])
# Get all indices of first non-zero sales entry per day:
indices = []
for name, group in groups:
group = group[group['sales'] > 0]
indices.append(group.index.to_list()[0])
# Change their values:
df.iloc[indices, df.columns.get_loc('hour')] = 9

Answer 1

You can create a group of df['day'] after checking of sales is greater than 0 , then get idxmax and filter out groups which doesnot have any value greater than 0 by using any , then assign with loc[]
g = df['sales'].gt(0).groupby(df['day'])
idx = g.idxmax()
df.loc[idx[g.any()],'hour']=9
print(df)
day hour sales
0 1 10 0
1 1 9 40
2 1 10 30
3 2 9 10
4 2 11 80
5 2 11 70
6 3 12 0
7 3 12 0
8 3 9 20

Answer 2

Create a mask m that groups by day as well as rows where the sales are not 0.
Then, use this mask as well as df['sales'] > 0 to change those specific rows to 9 with np.where()
df = pd.DataFrame({'day': [1, 1, 1, 2, 2, 2, 3, 3, 3],
'hour': [10, 10, 10, 11, 11, 11, 12, 12, 12],
'sales': [0, 40, 30, 10, 80, 70, 0, 0, 20]})
m = df.groupby(['day', df['sales'].ne(0)])['sales'].transform('first')
df['hour'] = np.where((df['sales'] == m) & (df['sales'] > 0), 9, df['hour'])
df
Out[37]:
day hour sales
0 1 10 0
1 1 9 40
2 1 10 30
3 2 9 10
4 2 11 80
5 2 11 70
6 3 12 0
7 3 12 0
8 3 9 20