np.nan, np.isnan, None, pd.isnull, pd.isna finishing and summary

np.nan

np.nan
not a member	nan的含义是“not a number“,它是对array中缺失的数值的占位符
type 为 float	As the missing value,np.nan的type是float
可参与计算	np.nan的type是float,所以np.nan可参与计算,And the dominance of,Is the calculated values arenp.nan
不等于自身	如果用np.nan==np.nan,会发现返回的结果为False,如果用np.nan in array,也不能发现array中是否包含np.nan
np.isnan判断是否存在np.nan	np.nanBecause there is no equal to the concept of,So to judge whether there isnp.nan,只可以用np.isnan来判断
np.isnanCan be judged only numeric	如果用np.isnan对其余type的array判断,会报错,np.isnanCan be judged only Numbersarray

忽略np.nan的numpy计算
nansum()
nanmax()
nanmin()
nanargmax()
nanargmin()

None

None
NoneType	None的type为NoneType,是python内置的
不可参与计算	None不可参与计算,否则报错
None is None 或者 None==None	Both the return value isTrue

pd.isnull 和 pd.isna

pd.isnull pd.isna
pd.isnull 和 pd.isna是同一个东西	用命令pd.isnull==pd.isna,Will find that the return value asTrue
None和np.nan在pandasIs the default	None和np.nan都会被pd.isnull或者pd.isna判断为True
可以判断ListOr a single value	pd.isnull和pd.isnaThe judgment of the object can be a value or alist
pandasThe default values are not involved in calculation	当使用df.min()When computing,Will find that the default values are not involved in calculation

代码示例

• 只可以用np.isnan判断array中是否包含np.nan

a = np.array([1,2,3,4,np.nan])
a == np.nan
>>> array([False, False, False, False, False])
np.nan in a
>>> False
np.isnan(a)
>>> array([False, False, False, False,  True])

• np.nan可参与计算,np.nan占主导地位

a = np.array([1,2,3,4,np.nan])
a.max()
>>> np.nan
a.min()
>>> np.nan
b = np.array([1,np.nan,3,4,5])
a+b
>>> array([ 2., nan,  6.,  8., nan])

• np.isnanOnly can the numericarray判断,当arrayFor the rest of the element in the string is,array会将np.nan转换成’nan’
  
• 在pandas中,np.nan和NoneIs treated as the default

df = pd.DataFrame({
    'col1':['a','b','c','d'], 'col2':[1,2,3,4]})
df.loc[1,'col1'] = np.nan
df.loc[2,'col1'] = None
df
>>> col1	col2
0	a	1
1	NaN	2
2	None	3
3	d	4

pd.isna(df)
>>>	col1	col2
0	False	False
1	True	False
2	True	False
3	False	False

• pandasThe default values are not involved in calculation

df = pd.DataFrame(np.arange(8).reshape(2,4))
df.iloc[0,1] = None
df.iloc[0,2] = np.nan
df
>>> 0	1	2	3
0	0	NaN	NaN	3
1	4	5.0	6.0	7

df.min(axis=1)
>>> 0    0.0
1    4.0
dtype: float64