Pandas: Selecting rows based on value counts of a particular column

Question

Whats the simplest way of selecting all rows from a panda dataframe, who\'s sym occurs exactly twice in the entire table? For example, in the table below, I would like to se

Answer 1

You can use map, which should be faster than using groupby and transform:

df[df['sym'].map(df['sym'].value_counts()) == 2]

e.g.

%%timeit
df[df['sym'].map(df['sym'].value_counts()) == 2]
Out[1]:
1.83 ms ± 23.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%%timeit
df[df.groupby("sym")["sym"].transform('size') == 2]
Out[2]:
2.08 ms ± 41.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 2

I think you can use groupby by column sym and filter values with length == 2:

print df.groupby("sym").filter(lambda x: len(x) == 2)
      price sym
1  0.400157   b
2  0.978738   b
7 -0.151357   e
8 -0.103219   e

Second solution use isin with boolean indexing:

s = df.sym.value_counts()

print s[s == 2].index
Index([u'e', u'b'], dtype='object')

print df[df.sym.isin(s[s == 2].index)]
      price sym
1  0.400157   b
2  0.978738   b
7 -0.151357   e
8 -0.103219   e

And fastest solution with transform and boolean indexing:

print (df[df.groupby("sym")["sym"].transform('size') == 2])
    price sym
1 -1.2940   b
2  1.8423   b
7  0.6280   e
8  0.5361   e