CODEWITHSUNDEEP
pythonpandasdataframe
Ankita Patnaik
Ankita Patnaik

Reputation: 271

Pandas: remove group from the data when a value in the group contains similar value

df = pd.DataFrame({"name":["A", "A", "B" ,"B", "C", "C"],
                   "nickname":["X","Y","X","Z","Y", "Y"]})

How can I group df by "name" and drop those groups that contains only 'Y'? In my case 'C' should be dropped.

I am using below code but it is not working:

df_new = df.groupby('name').filter(lambda x: all(x['nickname'] != 'Y'))

In case, Y occurs in any other 'name' with some other nickname then that name should be retained. Kindly help.

Upvotes: 1

Views: 140

Answers (4)

jezrael
jezrael

Reputation: 862521

Here groupby is not necessary. You can use boolean indexing:

df = df[df['name'].isin(df.loc[df['nickname'].ne('Y'), 'name'].unique())]
print (df)
  name nickname
0    A        X
1    A        Y
2    B        X
3    B        Z

Explanation:

First compare by ne for not equal values:

print (df['nickname'].ne('Y'))
0     True
1    False
2     True
3     True
4    False
5    False
Name: nickname, dtype: bool

Then select column name bu boolean mask:

print (df.loc[df['nickname'].ne('Y'), 'name'])
0    A
2    B
3    B
Name: name, dtype: object

For better performance get unique values:

print(df.loc[df['nickname'].ne('Y'), 'name'].unique())
['A' 'B']

And filter by isin for final mask:

print (df['name'].isin(df.loc[df['nickname'].ne('Y'), 'name'].unique()))
0     True
1     True
2     True
3     True
4    False
5    False
Name: name, dtype: bool

Performance:

Depends of number of rows, number of unique groups and number of matched values - best test in your real data:

np.random.seed(123)
N = 100000

df = pd.DataFrame({'name': np.random.randint(1000,size=N).astype(str),
                   'nickname':np.random.randint(200,size=N).astype(str)})
#print (df)

In [152]: %timeit df[df.nickname.ne('Y').groupby(df.name).transform('sum').astype(bool)]
27.6 ms ± 292 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [153]: %timeit df[~df.nickname.eq('Y').groupby(df.name).transform('all')]
27.3 ms ± 162 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [154]: %timeit df[df['name'].isin(df.loc[df['nickname'].ne('Y'), 'name'].unique())]
28.9 ms ± 189 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [155]: %timeit df[~df.assign(mask=df.nickname.eq('Y')).groupby('name').mask.transform('all')]
30.3 ms ± 469 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [156]: %timeit df[df.groupby('name')['nickname'].transform('unique').astype(str) !="['Y']"]
15.6 s ± 233 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [157]: %timeit df.groupby('name').filter(lambda x: any(x['nickname'] != 'Y'))
408 ms ± 29.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Upvotes: 2

cs95
cs95

Reputation: 402353

You probably want groupby and transform, or some derivative of the same thing.

df[~df.nickname.eq('Y').groupby(df.name).transform('all')]
# Or,
# df[~df.assign(mask=df.nickname.eq('Y')).groupby('name').mask.transform('all')]

  name nickname
0    A        X
1    A        Y
2    B        X
3    B        Z

An even faster groupby-related solution involves... counting!

df[df.nickname.ne('Y').groupby(df.name).transform('sum').astype(bool)]

  name nickname
0    A        X
1    A        Y
2    B        X
3    B        Z

Upvotes: 1

Joe
Joe

Reputation: 12417

I think that maybe you needed in your solution any instead of all:

df_new = df.groupby('name').filter(lambda x: any(x['nickname'] != 'Y'))

Output:

  name nickname
0    A        X
1    A        Y
2    B        X
3    B        Z

Upvotes: 0

Mohamed Thasin ah
Mohamed Thasin ah

Reputation: 11192

simply use this,

temp= df.groupby('name')['nickname'].transform('unique').astype(str)
df=df[temp!="['Y']"]
print df

O/P

  name nickname
0    A        X
1    A        Y
2    B        X
3    B        Z

Upvotes: 0

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