Pandas one liner to filter rows by nunique count on a specific column

Question

In pandas, I regularly use the following to filter a dataframe by number of occurrences

df = df.groupby('A').filter(lambda x: len(x) >= THRESHOLD)

Assume df has another column 'B' and I want to filter the dataframe this time by the count of unique values on that column, I would expect something like

df = df.groupby('A').filter(lambda x: len(np.unique(x['B'])) >= THRESHOLD2)

But that doesn't seem to work, what would be the right approach?

Answer 1

It should working nice with nunique:

df = pd.DataFrame({'B':list('abccee'),
                   'E':[5,3,6,9,2,4],
                   'A':list('aabbcc')})

print (df)
   A  B  E
0  a  a  5
1  a  b  3
2  b  c  6
3  b  c  9
4  c  e  2
5  c  e  4

THRESHOLD2 = 2
df1 = df.groupby('A').filter(lambda x: x['B'].nunique() >= THRESHOLD2)
print (df1)
   A  B  E
0  a  a  5
1  a  b  3

But if need faster solution use transform and filter by boolean indexing:

df2 = df[df.groupby('A')['B'].transform('nunique') >= THRESHOLD2]
print (df2)
   A  B  E
0  a  a  5
1  a  b  3

Timings:

np.random.seed(123)
N = 1000000
L = list('abcde') 
df = pd.DataFrame({'B': np.random.choice(L, N, p=(0.75,0.0001,0.0005,0.0005,0.2489)),
                   'A':np.random.randint(10000,size=N)})
df = df.sort_values(['A','B']).reset_index(drop=True)
print (df)

THRESHOLD2 = 3

In [403]: %timeit df.groupby('A').filter(lambda x: x['B'].nunique() >= THRESHOLD2)
1 loop, best of 3: 3.05 s per loop

In [404]: %timeit df[df.groupby('A')['B'].transform('nunique')>= THRESHOLD2]
1 loop, best of 3: 558 ms per loop

Caveat

The results do not address performance given the number of groups, which will affect timings a lot for some of these solutions.