CODEWITHSUNDEEP
pythonpandas
weirdpotatoes
weirdpotatoes

Reputation: 83

Deleting rows which sum to zero in 1 column but are otherwise duplicates in pandas

I have a pandas dataframe of the following structure:

df = pd.DataFrame({'ID':['A001', 'A001', 'A001', 'A002', 'A002', 'A003', 'A003', 'A004', 'A004', 'A004', 'A005', 'A005'],
                   'Val1':[2, 2, 2, 5, 6, 8, 8, 3, 3, 3, 7, 7],
                   'Val2':[100, -100, 50, -40, 40, 60, -50, 10, -10, 10, 15, 15]})

    ID    Val1  Val2
 0  A001     2   100
 1  A001     2  -100
 2  A001     2    50
 3  A002     5   -40
 4  A002     6    40
 5  A003     8    60
 6  A003     8   -50
 7  A004     3    10
 8  A004     3   -10
 9  A004     3    10
10  A005     7    15
11  A005     7    15

I want to remove duplicate rows where ID and Val1 are duplicates, and where Val2 sums to zero across two rows. The positive/negative Val2 rows may not be consecutive either, even under a groupby

In the above sample data, rows 0 and 1, as well as 7, 8, 9 fulfill these criteria. I'd want to remove [0, 1], and either [7, 8] or [8, 9].

Another constraint here is that there could be entirely duplicate rows ([10, 11]). In this case, I want to keep both rows.

The desired output is thus:

    ID    Val1  Val2
 2  A001     2    50
 3  A002     5   -40
 4  A002     6    40
 5  A003     8    60
 6  A003     8   -50
 9  A004     3    10
10  A005     7    15
11  A005     7    15

Short of iterating over each row and looking for other rows which fit the criteria, I'm out of ideas for a more "pythonic" way to do this. Any help is much appreciated.

Upvotes: 8

Views: 1022

Answers (5)

ResidentSleeper
ResidentSleeper

Reputation: 2495

Use groupby and cumsum to find which index of Val2 sums to zero

s = df.groupby(['ID', 'Val1']).Val2.cumsum() == 0

n = np.where(s==1)[0]

to_remove = np.concatenate((n, (n-1))) 

new_df = df[~df.index.isin(to_remove)]

new_df 

      ID  Val1  Val2
2   A001     2    50
3   A002     5   -40
4   A002     6    40
5   A003     8    60
6   A003     8   -50
9   A004     3    10
10  A005     7    15
11  A005     7    15

Upvotes: 1

sammywemmy
sammywemmy

Reputation: 28709

I put some comments in the code, so hopefully, my line of thought should be clear :

cond = df.assign(temp=df.Val2.abs())
# a way to get the same values (differentiated by their sign)
# to follow each other
cond = cond.sort_values(["ID", "Val1", "temp"])

# cumsum should yield a zero for numbers that are different
# only by their sign
cond["check"] = cond.groupby(["ID", "temp"]).Val2.cumsum()
cond["check"] = np.where(cond.check != 0, np.nan, cond.check)

# the backward fill here allows us to assign an identifier
# to the two values that summed to zero
cond["check"] = cond["check"].bfill(limit=1)

# this is where we implement your other condition
# essentially, it looks for rows that are duplicates
# and rows that any two rows sum to zero
cond.loc[
    ~(cond.duplicated(["ID", "Val1"], keep=False) & (cond.check == 0)),
    ["ID", "Val1", "Val2"],
]



     ID Val1    Val2
2   A001    2   50
3   A002    5   -40
4   A002    6   40
6   A003    8   -50
5   A003    8   60
9   A004    3   10

Upvotes: 2

Serge de Gosson de Varennes
Serge de Gosson de Varennes

Reputation: 11512

I believe there might be a less "brute force" method than this one, but it has the merit of being transparent.

import pandas as pd

df = pd.DataFrame({'ID':['A001', 'A001', 'A001', 'A002', 'A002', 'A003', 'A003', 'A004', 'A004', 'A004'],
                   'Val1':[2, 2, 2, 5, 6, 8, 8, 3, 3, 3],
                   'Val2':[100, -100, 50, 40, 45, 60, -50, 10, -10, 10]})
df['Val3'] = df['Val2'].abs()
df2        = df.drop_duplicates()
df2        = df.drop(['Val2'], axis = 1)
df3        = df2.drop_duplicates()
result     = pd.merge(df3, df, left_index=True, right_index=True, how='inner')
results    = result.drop(['ID_x', 'Val1_x', 'Val3_x', 'Val3_y'],axis = 1)

Upvotes: 0

ky_aaaa
ky_aaaa

Reputation: 320

Use drop_duplicates method to remove duplicates.

Here is the sample of code:

>>> df = pd.DataFrame({'ID':['A001', 'A001', 'A001', 'A002', 'A002', 'A003', 'A003', 'A004', 'A004', 'A004'],
...                    'Val1':[2, 2, 2, 5, 6, 8, 8, 3, 3, 3],
...                    'Val2':[100, -100, 50, 40, 45, 60, -50, 10, -10, 10]})
>>> 
>>> df.drop_duplicates(subset="Val2", keep= "last", inplace = True)
>>> df
     ID  Val1  Val2
0  A001     2   100
1  A001     2  -100
2  A001     2    50
3  A002     5    40
4  A002     6    45
5  A003     8    60
6  A003     8   -50
8  A004     3   -10
9  A004     3    10
>>> df.drop_duplicates(subset="Val1", keep= "last", inplace = True)
>>> df
     ID  Val1  Val2
2  A001     2    50
3  A002     5    40
4  A002     6    45
6  A003     8   -50
9  A004     3    10
>>> ~

If you can explain more clearly what does it mean: and where Val2 sums to zero across two rows. ?

This can help you further to get to the full solution.

Upvotes: 0

tgrandje
tgrandje

Reputation: 2534

What about :

temp = df.groupby('ID')[['Val2']].rolling(2).sum()
ix = temp[temp.Val2==0].index
ar = np.array([x[1] for x in ix.values])
ix2 = ar.tolist() + (ar-1).tolist()
df.drop(ix2, inplace=True)
df.drop_duplicates(['ID', 'Val1'], keep='first', inplace=True)

But this answers refers to your "textual" answer : rows 8 & 9 'Val2' are actually summing to zero (which is not what you the "desired output" you posted)...

Upvotes: 0

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