使用和不使用括号的pandas逻辑和运算符会产生不同的结果

Question

我刚刚注意到了这一点:

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

为什么这两行的输出不同？

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我无法分享确切的数据,但我会尝试尽可能多地提供详细信息:

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 

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感谢@jezrael和@ayhan,这里发生了什么,让我使用@jezael提供的示例:

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

如果我们看看第3行:

    col1  col2
3  False   1.0

以及我写条件的方式:

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

因为&标志的优先级高于==,没有括号False == False & False相当于:

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

带括号:

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

我认为用数字来说明这个问题要容易一些:

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

所以经验教训:总是添加括号!

我希望我能将答案分为@jezrael和@ayhan :(

Answer 1

df[condition1 & condition2] 和之间没有区别df[(condition1) & (condition2)].当您编写表达式并且运算符&优先时会出现差异:

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

但是,如果你这样输入,你会看到一个错误:

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().

这是因为3 & df['b']首先评估(这False & df.col2.isnull()在您的示例中对应).所以你需要在括号中对条件进行分组:

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