Python——np.nan, None的判断和比较

Posted on

Python值的判断与比较： np.nan, None

None

1
2
3
4
5
6
7
8
type(None)
# Output: <type 'NoneType'>

None is None
# Output: True

None == None
# Output: True

np.nan

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
type(np.nan)
# Output: <type 'float'>

np.nan == np.nan
# Output: False 

np.nan is np.nan
# Output: True 

np.nan != np.nan
# Output: True 

np.nan > np.nan
# Output: False

np.nan < np.nan
# Output: False

np.nan 与 None

1
2
3
4
5
6
7
8
None == np.nan
# Output: False 

None != np.nan
# Output: True

None is np.nan
# Output: False

与数字的比较

1
2
3
4
5
6
7
8
np.nan > 10
# Output: False 

np.nan < 10
# Output: False 

np.nan == 10
# Output: False

总结

  • 【Python2和Python3表现相同】
    np.nan 只有在np.nan != np.nan或者np.nan is np.nan时为True， 其他情况下和数字比较(包括和自身)都为False
  • 【Python2和Python3表现相异】
    Python3与Python2在直接使用np.nan时表现正常,但是当涉及到DataFrame的NaN时表现不同

特殊情况

DataFrame中的NaN与数字比较时会出现有时候为True有时候为False的情况

  • 这种情况出现在Python3中，当NaN 与数字比较时
    • 此时对于列属性类型为数值型，那么返回False
    • 否则返回True
  • Python2中NaN和数字的就是np.nan和数字比较的结果，都为False

Python2与Python3比较

  • 代码示例：
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
# Python3：
import pandas as pd
df = pd.DataFrame([['a',2,3], ['a',3,4], ['a',8,9]], index=['a', 'b', 'c'])
df = df.reindex(['a', 'b', 'c', 'd', 'e', 'f'])
print(df)
# Output: 
	     0    1    2
	a    a  2.0  3.0
	b    a  3.0  4.0
	c    a  8.0  9.0
	d  NaN  NaN  NaN
	e  NaN  NaN  NaN
	f  NaN  NaN  NaN

print(df > 5)
# Output: 
	      0      1      2
	a  True  False  False
	b  True  False  False
	c  True   True   True
	d  True  False  False
	e  True  False  False
	f  True  False  False

print(df.values)
# Output:
	[['a' 2.0 3.0]
	 ['a' 3.0 4.0]
	 ['a' 8.0 9.0]
	 [nan nan nan]
	 [nan nan nan]
	 [nan nan nan]]
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
# Python2：
import pandas as pd
df = pd.DataFrame([['a',2,3], ['a',3,4], ['a',8,9]], index=['a', 'b', 'c'])
df = df.reindex(['a', 'b', 'c', 'd', 'e', 'f'])
print(df)
# Output: 
	     0    1    2
	a    a  2.0  3.0
	b    a  3.0  4.0
	c    a  8.0  9.0
	d  NaN  NaN  NaN
	e  NaN  NaN  NaN
	f  NaN  NaN  NaN

print(df > 5)
# Output: 
	       0      1      2
	a   True  False  False
	b   True  False  False
	c   True   True   True
	d  False  False  False
	e  False  False  False
	f  False  False  False

print(df.values)
# Output: 
	[['a' 2.0 3.0]
	 ['a' 3.0 4.0]
	 ['a' 8.0 9.0]
	 [nan nan nan]
	 [nan nan nan]
	 [nan nan nan]]