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

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Question :

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

I have just noticed this:

df[df.condition1 & df.condition2]
df[(df.condition1) & (df.condition2)]

Why does the output of these two lines differ?


I cannot share the exact data but I am gonna try to provide as much detail as I can:

df[df.col1 == False & df.col2.isnull()] # returns 33 rows and the rule `df.col2.isnull()` is not in effect
df[(df.col1 == False) & (df.col2.isnull())] # returns 29 rows and both conditions are applied correctly 

Thanks to @jezrael and @ayhan, here is what happened, and let me use the example provided by @jezael:

df = pd.DataFrame({'col1':[True, False, False, False],
                   'col2':[4, np.nan, np.nan, 1]})

print (df)
    col1  col2
0   True   4.0
1  False   NaN
2  False   NaN
3  False   1.0

If we take a look at row 3:

    col1  col2
3  False   1.0

and the way I wrote the condition:

df.col1 == False & df.col2.isnull() # is equivalent to False == False & False

Because the & sign has higher priority than ==, without brackets False == False & False is equivalent of:

False == (False & False)
print(False == (False & False)) # prints True

With brackets:

print((False == False) & False) # prints False

I think it is a bit easier to illustrate this problem with numbers:

print(5 == 5 & 1) # prints False, because 5 & 1 returns 1 and 5==1 returns False
print(5 == (5 & 1)) # prints False, same reason as above
print((5 == 5) & 1) # prints 1, because 5 == 5 returns True, and True & 1 returns 1

So lessons learned: always add brackets!!!

I wish I can split the answer points to both @jezrael and @ayhan 🙁

Asked By: Cheng

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Answer #1:

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
Answered By: Cheng

Answer #2:

You are right, it is difference and I think there is problem with priority of operators – check docs:

df = pd.DataFrame({'col1':[True, False, False, False],
                   'col2':[4, np.nan, np.nan, 1]})

print (df)
    col1  col2
0   True   4.0
1  False   NaN
2  False   NaN
3  False   1.0

# operator & precedence
print (df[df.col1 == False & df.col2.isnull()])
    col1  col2
1  False   NaN
2  False   NaN
3  False   1.0

# operator == precedence bacause in brackets
print (df[(df.col1 == False) & (df.col2.isnull())])
    col1  col2
1  False   NaN
2  False   NaN

It seems I found it in docs – 6.16. Operator precedence where see & have higher priority as ==:

Operator                                Description

lambda                                  Lambda expression
ifelse                               Conditional expression
or                                      Boolean OR
and                                     Boolean AND
not x                                   Boolean NOT
in, not in, is, is not,                 Comparisons, including membership tests    
<, <=, >, >=, !=, ==                    and identity tests
|                                       Bitwise OR
^                                       Bitwise XOR
&                                       Bitwise AND

(expressions...), [expressions...],     Binding or tuple display, list display,       
{key: value...}, {expressions...}       dictionary display, set display
Answered By: ayhan

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