Filter out nan rows in a specific column

Use dropna:

df = df.dropna(subset=['Col2'])
print (df)
  Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

Another solution - boolean indexing with notnull:

df = df[df['Col2'].notnull()]
print (df)
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

What is same as:

df = df[~df['Col2'].isnull()]
print (df)
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

you can use DataFrame.dropna() method:

In [202]: df.dropna(subset=['Col2'])
Out[202]:
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

or (in this case) less idiomatic Series.notnull():

In [204]: df.loc[df.Col2.notnull()]
Out[204]:
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

or using DataFrame.query() method:

In [205]: df.query("Col2 == Col2")
Out[205]:
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

numexpr solution:

In [241]: import numexpr as ne

In [242]: col = df.Col2

In [243]: df[ne.evaluate("col == col")]
Out[243]:
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

Using numpy's isnan to mask and construct a new dataframe

m = ~np.isnan(df.Col2.values)
pd.DataFrame(df.values[m], df.index[m], df.columns)

   Col1  Col2  Col3
1   2.0   5.0   4.0
2   3.0   3.0   NaN

---

Timing
Bigger Data

np.random.seed([3,1415])
df = pd.DataFrame(np.random.choice([np.nan, 1], size=(10000, 10))).add_prefix('Col')

%%timeit
m = ~np.isnan(df.Col2.values)
pd.DataFrame(df.values[m], df.index[m], df.columns)
1000 loops, best of 3: 326 µs per loop

%timeit df.query("Col2 == Col2")
1000 loops, best of 3: 1.48 ms per loop

%timeit df.loc[df.Col2.notnull()]
1000 loops, best of 3: 417 µs per loop

%timeit df[~df['Col2'].isnull()]
1000 loops, best of 3: 385 µs per loop

%timeit df.dropna(subset=['Col2'])
1000 loops, best of 3: 913 µs per loop