Groupby Rows and Sum
I have the following dataframe:
print(inventory_df)
dt_op Prod_1 Prod_2 ... Prod_n
10/09/18 0 8 0
10/09/18 5 0 2
11/09/18 4 0 0
11/09/18 0 10 0
...
And I would like to get:
print(final_df)
dt_op Prod_1 Prod_2 ... Prod_n
10/09/18 5 8 2
11/09/18 4 10 0
...
I tried with:
final_df = inventory_df.drop_duplicates(subset=None, keep='first', inplace=False)
But it does not produce the desired output. How can I create final_df?
python pandas
asked Nov 20 '18 at 10:39
Alessandro Ceccarelli
247211
add a comment |
I have the following dataframe:
print(inventory_df)
dt_op Prod_1 Prod_2 ... Prod_n
10/09/18 0 8 0
10/09/18 5 0 2
11/09/18 4 0 0
11/09/18 0 10 0
...
And I would like to get:
print(final_df)
dt_op Prod_1 Prod_2 ... Prod_n
10/09/18 5 8 2
11/09/18 4 10 0
...
I tried with:
final_df = inventory_df.drop_duplicates(subset=None, keep='first', inplace=False)
But it does not produce the desired output. How can I create final_df?
python pandas
asked Nov 20 '18 at 10:39
Alessandro Ceccarelli
247211
do you want to drop all zeros from your dataframe even that they are assigned to different entries?
– CIsForCookies
Nov 20 '18 at 10:43
Possible duplicate of Pandas group-by and sum
– jpp
Nov 20 '18 at 15:28
add a comment |
I have the following dataframe:
print(inventory_df)
dt_op Prod_1 Prod_2 ... Prod_n
10/09/18 0 8 0
10/09/18 5 0 2
11/09/18 4 0 0
11/09/18 0 10 0
...
And I would like to get:
print(final_df)
dt_op Prod_1 Prod_2 ... Prod_n
10/09/18 5 8 2
11/09/18 4 10 0
...
I tried with:
final_df = inventory_df.drop_duplicates(subset=None, keep='first', inplace=False)
But it does not produce the desired output. How can I create final_df?
python pandas
asked Nov 20 '18 at 10:39
Alessandro Ceccarelli
247211
I have the following dataframe:
print(inventory_df)
dt_op Prod_1 Prod_2 ... Prod_n
10/09/18 0 8 0
10/09/18 5 0 2
11/09/18 4 0 0
11/09/18 0 10 0
...
And I would like to get:
print(final_df)
dt_op Prod_1 Prod_2 ... Prod_n
10/09/18 5 8 2
11/09/18 4 10 0
...
I tried with:
final_df = inventory_df.drop_duplicates(subset=None, keep='first', inplace=False)
But it does not produce the desired output. How can I create final_df?
python pandas
python pandas
asked Nov 20 '18 at 10:39
Alessandro Ceccarelli
247211
asked Nov 20 '18 at 10:39
Alessandro Ceccarelli
247211
asked Nov 20 '18 at 10:39
Alessandro Ceccarelli
247211
asked Nov 20 '18 at 10:39
Alessandro Ceccarelli
247211
asked Nov 20 '18 at 10:39
Alessandro Ceccarelli
247211
247211
do you want to drop all zeros from your dataframe even that they are assigned to different entries?
– CIsForCookies
Nov 20 '18 at 10:43
Possible duplicate of Pandas group-by and sum
– jpp
Nov 20 '18 at 15:28
add a comment |
do you want to drop all zeros from your dataframe even that they are assigned to different entries?
– CIsForCookies
Nov 20 '18 at 10:43
Possible duplicate of Pandas group-by and sum
– jpp
Nov 20 '18 at 15:28
do you want to drop all zeros from your dataframe even that they are assigned to different entries?
– CIsForCookies
Nov 20 '18 at 10:43
do you want to drop all zeros from your dataframe even that they are assigned to different entries?
– CIsForCookies
Nov 20 '18 at 10:43
Possible duplicate of Pandas group-by and sum
– jpp
Nov 20 '18 at 15:28
Possible duplicate of Pandas group-by and sum
– jpp
Nov 20 '18 at 15:28
add a comment |
2 Answers
2
active
oldest
votes
Just simulated the Stated DataFrame, you asked about the groupby + sum() across the rows.
Reproduced DataFrame:
>>> df
dt_op Prod_1 Prod_2 Prod_n
0 10/09/18 0 8 0
1 10/09/18 5 0 2
2 11/09/18 4 0 0
Using groupby around the columns axis=1(of dimension 1, which is what used to be columns) or simply df.groupby('dt_op').sum :
>>> df.groupby('dt_op').sum(axis=1)
Prod_1 Prod_2 Prod_n
dt_op
10/09/18 5 8 2
11/09/18 4 0 0
However, you are looking for the literal sum() of rows across the columns:
>>> df['new_sum'] = df.sum(axis=1)
>>> df
dt_op Prod_1 Prod_2 Prod_n new_sum
0 10/09/18 0 8 0 8
1 10/09/18 5 0 2 7
2 11/09/18 4 0 0 4
answered Nov 20 '18 at 15:21
pygo
2,1511617
add a comment |
You can use pandas groupby function with sum():
In [412]: inventory_df
Out[412]:
dt_op Prod_1 Prod_2
0 10/09/18 0 8
1 10/09/18 5 0
2 11/09/18 4 0
3 11/09/18 0 10
In [413]: inventory_df.groupby('dt_op').sum()
Out[413]:
Prod_1 Prod_2
dt_op
10/09/18 5 8
11/09/18 4 10
answered Nov 20 '18 at 10:44
Mayank Porwal
4,4991624
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Just simulated the Stated DataFrame, you asked about the groupby + sum() across the rows.
Reproduced DataFrame:
>>> df
dt_op Prod_1 Prod_2 Prod_n
0 10/09/18 0 8 0
1 10/09/18 5 0 2
2 11/09/18 4 0 0
Using groupby around the columns axis=1(of dimension 1, which is what used to be columns) or simply df.groupby('dt_op').sum :
>>> df.groupby('dt_op').sum(axis=1)
Prod_1 Prod_2 Prod_n
dt_op
10/09/18 5 8 2
11/09/18 4 0 0
However, you are looking for the literal sum() of rows across the columns:
>>> df['new_sum'] = df.sum(axis=1)
>>> df
dt_op Prod_1 Prod_2 Prod_n new_sum
0 10/09/18 0 8 0 8
1 10/09/18 5 0 2 7
2 11/09/18 4 0 0 4
answered Nov 20 '18 at 15:21
pygo
2,1511617
add a comment |
Just simulated the Stated DataFrame, you asked about the groupby + sum() across the rows.
Reproduced DataFrame:
>>> df
dt_op Prod_1 Prod_2 Prod_n
0 10/09/18 0 8 0
1 10/09/18 5 0 2
2 11/09/18 4 0 0
Using groupby around the columns axis=1(of dimension 1, which is what used to be columns) or simply df.groupby('dt_op').sum :
>>> df.groupby('dt_op').sum(axis=1)
Prod_1 Prod_2 Prod_n
dt_op
10/09/18 5 8 2
11/09/18 4 0 0
However, you are looking for the literal sum() of rows across the columns:
>>> df['new_sum'] = df.sum(axis=1)
>>> df
dt_op Prod_1 Prod_2 Prod_n new_sum
0 10/09/18 0 8 0 8
1 10/09/18 5 0 2 7
2 11/09/18 4 0 0 4
answered Nov 20 '18 at 15:21
pygo
2,1511617
add a comment |
Just simulated the Stated DataFrame, you asked about the groupby + sum() across the rows.
Reproduced DataFrame:
>>> df
dt_op Prod_1 Prod_2 Prod_n
0 10/09/18 0 8 0
1 10/09/18 5 0 2
2 11/09/18 4 0 0
Using groupby around the columns axis=1(of dimension 1, which is what used to be columns) or simply df.groupby('dt_op').sum :
>>> df.groupby('dt_op').sum(axis=1)
Prod_1 Prod_2 Prod_n
dt_op
10/09/18 5 8 2
11/09/18 4 0 0
However, you are looking for the literal sum() of rows across the columns:
>>> df['new_sum'] = df.sum(axis=1)
>>> df
dt_op Prod_1 Prod_2 Prod_n new_sum
0 10/09/18 0 8 0 8
1 10/09/18 5 0 2 7
2 11/09/18 4 0 0 4
answered Nov 20 '18 at 15:21
pygo
2,1511617
Just simulated the Stated DataFrame, you asked about the groupby + sum() across the rows.
Reproduced DataFrame:
>>> df
dt_op Prod_1 Prod_2 Prod_n
0 10/09/18 0 8 0
1 10/09/18 5 0 2
2 11/09/18 4 0 0
Using groupby around the columns axis=1(of dimension 1, which is what used to be columns) or simply df.groupby('dt_op').sum :
>>> df.groupby('dt_op').sum(axis=1)
Prod_1 Prod_2 Prod_n
dt_op
10/09/18 5 8 2
11/09/18 4 0 0
However, you are looking for the literal sum() of rows across the columns:
>>> df['new_sum'] = df.sum(axis=1)
>>> df
dt_op Prod_1 Prod_2 Prod_n new_sum
0 10/09/18 0 8 0 8
1 10/09/18 5 0 2 7
2 11/09/18 4 0 0 4
answered Nov 20 '18 at 15:21
pygo
2,1511617
edited Nov 20 '18 at 15:36
answered Nov 20 '18 at 15:21
pygo
2,1511617
answered Nov 20 '18 at 15:21
pygo
2,1511617
answered Nov 20 '18 at 15:21
pygo
2,1511617
2,1511617
add a comment |
add a comment |
You can use pandas groupby function with sum():
In [412]: inventory_df
Out[412]:
dt_op Prod_1 Prod_2
0 10/09/18 0 8
1 10/09/18 5 0
2 11/09/18 4 0
3 11/09/18 0 10
In [413]: inventory_df.groupby('dt_op').sum()
Out[413]:
Prod_1 Prod_2
dt_op
10/09/18 5 8
11/09/18 4 10
answered Nov 20 '18 at 10:44
Mayank Porwal
4,4991624
add a comment |
You can use pandas groupby function with sum():
In [412]: inventory_df
Out[412]:
dt_op Prod_1 Prod_2
0 10/09/18 0 8
1 10/09/18 5 0
2 11/09/18 4 0
3 11/09/18 0 10
In [413]: inventory_df.groupby('dt_op').sum()
Out[413]:
Prod_1 Prod_2
dt_op
10/09/18 5 8
11/09/18 4 10
answered Nov 20 '18 at 10:44
Mayank Porwal
4,4991624
add a comment |
You can use pandas groupby function with sum():
In [412]: inventory_df
Out[412]:
dt_op Prod_1 Prod_2
0 10/09/18 0 8
1 10/09/18 5 0
2 11/09/18 4 0
3 11/09/18 0 10
In [413]: inventory_df.groupby('dt_op').sum()
Out[413]:
Prod_1 Prod_2
dt_op
10/09/18 5 8
11/09/18 4 10
answered Nov 20 '18 at 10:44
Mayank Porwal
4,4991624
You can use pandas groupby function with sum():
In [412]: inventory_df
Out[412]:
dt_op Prod_1 Prod_2
0 10/09/18 0 8
1 10/09/18 5 0
2 11/09/18 4 0
3 11/09/18 0 10
In [413]: inventory_df.groupby('dt_op').sum()
Out[413]:
Prod_1 Prod_2
dt_op
10/09/18 5 8
11/09/18 4 10
answered Nov 20 '18 at 10:44
Mayank Porwal
4,4991624
edited Nov 20 '18 at 12:23
answered Nov 20 '18 at 10:44
Mayank Porwal
4,4991624
answered Nov 20 '18 at 10:44
Mayank Porwal
4,4991624
answered Nov 20 '18 at 10:44
Mayank Porwal
4,4991624
4,4991624
add a comment |
add a comment |
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do you want to drop all zeros from your dataframe even that they are assigned to different entries?
– CIsForCookies
Nov 20 '18 at 10:43
Possible duplicate of Pandas group-by and sum
– jpp
Nov 20 '18 at 15:28