Conditionally offseting values by group with Pandas

user_9209546 0
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I am looking for a more efficient and maintainable way to offset values conditionally by group. Easiest to show an example.

Value is always non-negative for Offset == False and always negative for Offset == True. What I'm looking to do is "collapse" positive Values (flooring at 0) against negative ones by Label.

Note Label + Offset combined are always unique. Since Offset is Boolean, you can only have a maximum of 2 rows per Label.

Example 1

df = pd.DataFrame({'Label': ['L1', 'L2', 'L3', 'L3'],
                   'Offset': [False, False, False, True],
                   'Value': [100, 100, 50, -100]})

# input
#   Label Offset  Value
# 0    L1  False    100
# 1    L2  False    100
# 2    L3  False     50
# 3    L3   True   -100

Desired output:

  Label Offset  Value
0    L1  False    100
1    L2  False    100
2    L3  False      0
3    L3   True    -50

Example 2

df = pd.DataFrame({'Label': ['L1', 'L2', 'L3', 'L3'],
                   'Offset': [False, False, False, True],
                   'Value': [100, 100, 100, -50]})

# input
#   Label Offset  Value
# 0    L1  False    100
# 1    L2  False    100
# 2    L3  False    100
# 3    L3   True    -50

Desired output:

  Label Offset  Value
0    L1  False    100
1    L2  False    100
2    L3  False     50
3    L3   True      0

Current inefficient solution

My current solution is a manual loop which is slow and difficult to maintain:

for label in df['Label'].unique():
    mask = df['Label'] == label
    if len(df.loc[mask]) == 2:
        val_false = df.loc[~df['Offset'] & mask, 'Value'].iloc[0]
        val_true = df.loc[df['Offset'] & mask, 'Value'].iloc[0]
        if val_false > abs(val_true):
            df.loc[~df['Offset'] & mask, 'Value'] += val_true
            df.loc[df['Offset'] & mask, 'Value'] = 0
        else:
            df.loc[~df['Offset'] & mask, 'Value'] = 0
            df.loc[df['Offset'] & mask, 'Value'] += val_false

I'm looking for a vectorised, or at least partially vectorised, solution to improve performance and get rid of this mess.

pandas-groupby pandas dataframe python performance
3 Answers

Maybe:

label_sums = df.Value.groupby(df.Label).transform(sum)
df["new_sum"] = label_sums.where(np.sign(label_sums) == np.sign(df.Value), 0)

which gives me

In [42]: df
Out[42]: 
  Label  Offset  Value  new_sum
0    L1   False    100      100
1    L2   False    100      100
2    L3   False     50        0
3    L3    True   -100      -50
4    L4   False    100      100
5    L5   False    100      100
6    L6   False    100       50
7    L6    True    -50        0
user_4873390

This is the best I've got: create an auxiliary column to find where to display the aggregate and then set the other members of the group to 0

df['aux'] = abs(df['Value'])
idx = abs(df.groupby(['Label'])['aux'].transform(max)) == abs(df['aux'])
df['aux2'] = False
df.loc[idx,'aux2'] = True
df  = df.join(df.groupby('Label').Value.sum(), on='Label', rsuffix = 'jpp')
df.loc[df['aux2']==False, 'Valuejpp'] = 0
df = df.drop(['aux', 'aux2','Value'], axis = 1)

Result

      Label  Offset  Valuejpp
0    L1   False       100
1    L2   False       100
2    L3   False         0
3    L3    True       -50
user_97541690

Data From DSM

df1=df.copy()

df.loc[df.Offset,'Value']=df.Value.abs()

s1=(df.groupby('Label').Value.diff().lt(0)).groupby(df['Label']).transform('any')
s2=df.groupby('Label').Value.transform('count')

s3=df1.groupby('Label').Value.transform('sum')
np.where(s2<=1,df1.Value,np.where(s1,s3*(-df1.Offset),s3*df1.Offset))

Out[757]: array([100, 100,   0, -50, 100, 100,  50,   0], dtype=int64)
user_79645270
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