I am quite new to pandas, but I use python at a good level.

I have a pandas dataframe which is organized as follows

idrun    idbasin    time    q
-192540      1        0     0
-192540      1        1     0.5
...
-192540      2        0     0
-192540      2        1     1
...
-192540      3        0     0
-192540      3        1     1
...
-192541      1        0     0
-192541      1        1     0.5
...
-192541      2        0     0
-192541      2        1     1
...
-192541      3        0     0
-192541      3        1     1
...

It is a fairly large dataframe (7 columns and ~600k rows).

What I would like to do is: given a tuple containing values referring to the idbasin column (e.g. (1,2)), if the idrun value is the same

1. sum the q column of the referred idbasin values, i.e. for the example it would be (1,2)
2. remove the rows corresponding to that idrun value and the tuple-specified idbasin values
3. insert the summed values with idbasin equal to the first number of the tuple.

Referring to my example df, the results would be

idrun    idbasin    time    q
-192540      1        0     0
-192540      1        1     1.5
...
-192540      3        0     0
-192540      3        1     1
...
-192541      1        0     0
-192541      1        1     1.5
...
-192541      3        0     0
-192541      3        1     1
...

My solution would to use groupby to turn the df to a dict and then do the operation with one or two for loops, but I understand that iterating in pandas is not the optimal solution, so I believe there could be a "pandas" solution using the df.

CodePudding user response：

You can replace values of tuple by first value of tuple in Series.mask and then aggregate sum:

tup = (1, 2)

df['idbasin'] = df['idbasin'].mask(df['idbasin'].isin(tup), tup[0])
#alternative
#df['idbasin'] = np.where(df['idbasin'].isin(tup), tup[0], df['idbasin'])
df = df.groupby(['idrun', 'idbasin','time'], as_index=False)['q'].sum()
print (df)
    idrun  idbasin  time    q
0 -192541        1     0  0.0
1 -192541        1     1  1.5
2 -192541        3     0  0.0
3 -192541        3     1  1.0
4 -192540        1     0  0.0
5 -192540        1     1  1.5
6 -192540        3     0  0.0
7 -192540        3     1  1.0