Reputation: 3355
I have a dataframe like this, I want to calculate and add a new column which follows the formula: Value = A(where Time=1) + A(where Time=3)
, I don't want to use A (where Time=5).
Type subType Time A Value
X a 1 3 =3+9=12
X a 3 9
X a 5 9
X b 1 4 =4+5=9
X b 3 5
X b 5 0
Y a 1 1 =1+2=3
Y a 3 2
Y a 5 3
Y b 1 4 =4+5=9
Y b 3 5
Y b 5 2
I know how to do by selecting the cell needed for the formula, but is there any other better ways to perform the calculation? I suspect I need to add a condition but not sure how, any suggestion?
Upvotes: 0
Views: 120
Reputation: 30920
Use Series.eq
with DataFrame.groupby
and Series.cumsum
to create groups and add.
c1 = df.Time.eq(1)
c3 = df.Time.eq(3)
df['Value'] = (df.loc[c1|c3]
.groupby(c1.cumsum())
.A
.transform('sum')
.loc[c1])
print(df)
or if you want to identify it based on the non-equivalence with 5:
c = df['Time'].eq(5)
df['value'] = (df['A'].mask(c)
.groupby(c.cumsum())
.transform('sum')
.where(c.shift(fill_value = True))
)
#Another option is map
c = df['Time'].eq(5)
c_cumsum = c.cumsum()
df['value'] = (c_cumsum.map(df['A'].mask(c)
.groupby(c_cumsum)
.sum())
.where(c.shift(fill_value = True)))
Output
Type subType Time A Value
0 X a 1 3 12.0
1 X a 3 9 NaN
2 X a 5 9 NaN
3 X b 1 4 9.0
4 X b 3 5 NaN
5 X b 5 0 NaN
6 Y a 1 1 3.0
7 Y a 3 2 NaN
8 Y a 5 3 NaN
9 Y b 1 4 9.0
10 Y b 3 5 NaN
11 Y b 5 2 NaN
MISSING VALUES
c = df['Time'].eq(5)
df['value'] = (df['A'].mask(c)
.groupby(c.cumsum())
.transform('sum')
)
#or method 1
#c1 = df.Time.eq(1)
#c3 = df.Time.eq(3)
#df['Value'] = (df.loc[c1|c3]
# .groupby(c1.cumsum())
# .A
# .transform('sum')
# )
print(df)
Output
Type subType Time A value
0 X a 1 3 12.0
1 X a 3 9 12.0
2 X a 5 9 9.0
3 X b 1 4 9.0
4 X b 3 5 9.0
5 X b 5 0 3.0
6 Y a 1 1 3.0
7 Y a 3 2 3.0
8 Y a 5 3 9.0
9 Y b 1 4 9.0
10 Y b 3 5 9.0
11 Y b 5 2 0.0
or filling all except where Time is 5
c = df['Time'].eq(5)
df['value'] = (df['A'].mask(c)
.groupby(c.cumsum())
.transform('sum').mask(c))
#c1 = df.Time.eq(1)
#c3 = df.Time.eq(3)
#or method 1
#df['Value'] = (df.loc[c1|c3]
# .groupby(c1.cumsum())
# .A
# .transform('sum')
# .loc[c1|c3])
print(df)
Type subType Time A value
0 X a 1 3 12.0
1 X a 3 9 12.0
2 X a 5 9 NaN
3 X b 1 4 9.0
4 X b 3 5 9.0
5 X b 5 0 NaN
6 Y a 1 1 3.0
7 Y a 3 2 3.0
8 Y a 5 3 NaN
9 Y b 1 4 9.0
10 Y b 3 5 9.0
11 Y b 5 2 NaN
Why not use apply here?
Even in a small data frame it is already slower
%%timeit
(
df.groupby(by=['Type','subType'])
.apply(lambda x: x.loc[x.Time!=5].A.sum()) # sum time each group exclu
.to_frame('Value').reset_index()
.pipe(lambda x: pd.merge(df, x, on=['Type', 'subType'], how='left'))
)
13.6 ms ± 2.67 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
c = df['Time'].eq(5)
df['value'] = (df['A'].mask(c)
.groupby(c.cumsum())
.transform('sum')
.where(c.shift(fill_value = True))
)
3.67 ms ± 118 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Upvotes: 3
Reputation: 19947
You can use groupby to sum A when Time is not 5. Then merge it back with the original df.
(
df.groupby(by=['Type','subType'])
.apply(lambda x: x.loc[x.Time!=5].A.sum()) # sum time each group exclu
.to_frame('Value').reset_index()
.pipe(lambda x: pd.merge(df, x, on=['Type', 'subType'], how='left'))
)
Type subType Time A Value
0 X a 1 3 12.0
1 X a 3 9 12.0
2 X a 5 9 12.0
3 X b 1 4 9.0
4 X b 3 5 9.0
5 X b 5 0 9.0
6 Y a 1 1 3.0
7 Y a 3 2 3.0
8 Y a 5 3 3.0
9 Y b 1 4 9.0
10 Y b 3 5 9.0
11 Y b 5 2 9.0
Upvotes: 3
Reputation: 4215
Answer using only indexing and conditions:
df.loc[df['Time'] == 1,'Value'] = (df[df['Time'] == 1].reset_index()+df[df['Time'] == 3].reset_index())['A'].values
df
Type subType Time A Value
0 X a 1 3 12.0
1 X a 3 9 NaN
2 X a 5 9 NaN
3 X b 1 4 9.0
4 X b 3 5 NaN
5 X b 5 0 NaN
6 Y a 1 1 3.0
7 Y a 3 2 NaN
8 Y a 5 3 NaN
9 Y b 1 4 9.0
10 Y b 3 5 NaN
11 Y b 5 2 NaN
Upvotes: 0