Question :
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 🙁
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
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
if – else 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
