pandas logical and operator with and without brackets produces different results [duplicate]

There is no difference between df[condition1 & condition2] and df[(condition1) & (condition2)]. The difference arises when you write an expression and the operator & takes precedence:

df = pd.DataFrame(np.random.randint(0, 10, size=(5, 3)), columns=list('abc'))    
df
Out: 
   a  b  c
0  5  0  3
1  3  7  9
2  3  5  2
3  4  7  6
4  8  8  1

condition1 = df['a'] > 3
condition2 = df['b'] < 5

df[condition1 & condition2]
Out: 
   a  b  c
0  5  0  3

df[(condition1) & (condition2)]
Out: 
   a  b  c
0  5  0  3

However, if you type it like this you’ll see an error:

df[df['a'] > 3 & df['b'] < 5]
Traceback (most recent call last):

  File "<ipython-input-7-9d4fd21246ca>", line 1, in <module>
    df[df['a'] > 3 & df['b'] < 5]

  File "/home/ayhan/anaconda3/lib/python3.5/site-packages/pandas/core/generic.py", line 892, in __nonzero__
    .format(self.__class__.__name__))

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

This is because 3 & df['b'] is evaluated first (this corresponds to False & df.col2.isnull() in your example). So you need to group the conditions in parentheses:

df[(df['a'] > 3) & (df['b'] < 5)]
Out[8]: 
   a  b  c
0  5  0  3