文章目录
- 简介
- NaN的例子
- 整数类型的缺失值
- Datetimes 类型的缺失值
- None 和 np.nan 的转换
- 缺失值的计算
- 使用fillna填充NaN数据
- 使用dropna删除包含NA的数据
- 插值interpolation
- 使用replace替换值
简介
在数据处理中，Pandas会将无法解析的数据或者缺失的数据使用NaN来表示。虽然所有的数据都有了相应的表示，但是NaN很明显是无法进行数学运算的。
本文将会讲解Pandas对于NaN数据的处理方法。
NaN的例子上面讲到了缺失的数据会被表现为NaN，我们来看一个具体的例子：
我们先来构建一个DF：
In [1]: df = pd.DataFrame(np.random.randn(5, 3), index=['a', 'c', 'e', 'f', 'h'],
...: columns=['one', 'two', 'three'])
...:
In [2]: df['four'] = 'bar'
In [3]: df['five'] = df['one'] > 0
In [4]: df
Out[4]:
one two three four five
a 0.469112 -0.282863 -1.509059 bar True
c -1.135632 1.212112 -0.173215 bar False
e 0.119209 -1.044236 -0.861849 bar True
f -2.104569 -0.494929 1.071804 bar False
h 0.721555 -0.706771 -1.039575 bar True
上面DF只有acefh这几个index，我们重新index一下数据：
In [5]: df2 = df.reindex(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])
In [6]: df2
Out[6]:
one two three four five
a 0.469112 -0.282863 -1.509059 bar True
b NaN NaN NaN NaN NaN
c -1.135632 1.212112 -0.173215 bar False
d NaN NaN NaN NaN NaN
e 0.119209 -1.044236 -0.861849 bar True
f -2.104569 -0.494929 1.071804 bar False
g NaN NaN NaN NaN NaN
h 0.721555 -0.706771 -1.039575 bar True
数据缺失，就会产生很多NaN。
为了检测是否NaN，可以使用isna()或者notna() 方法。
In [7]: df2['one']
Out[7]:
a 0.469112
b NaN
c -1.135632
d NaN
e 0.119209
f -2.104569
g NaN
h 0.721555
Name: one, dtype: float64
In [8]: pd.isna(df2['one'])
Out[8]:
a False
b True
c False
d True
e False
f False
g True
h False
Name: one, dtype: bool
In [9]: df2['four'].notna()
Out[9]:
a True
b False
c True
d False
e True
f True
g False
h True
Name: four, dtype: bool
注意在Python中None是相等的：
In [11]: None == None # noqa: E711
Out[11]: True
但是np.nan是不等的：
In [12]: np.nan == np.nan
Out[12]: False
NaN默认是float类型的，如果是整数类型，我们可以强制进行转换：
In [14]: pd.Series([1, 2, np.nan, 4], dtype=pd.Int64Dtype())
Out[14]:
0 1
1 2
2 <NA>
3 4
dtype: Int64
时间类型的缺失值使用NaT来表示：
In [15]: df2 = df.copy()
In [16]: df2['timestamp'] = pd.Timestamp('20120101')
In [17]: df2
Out[17]:
one two three four five timestamp
a 0.469112 -0.282863 -1.509059 bar True 2012-01-01
c -1.135632 1.212112 -0.173215 bar False 2012-01-01
e 0.119209 -1.044236 -0.861849 bar True 2012-01-01
f -2.104569 -0.494929 1.071804 bar False 2012-01-01
h 0.721555 -0.706771 -1.039575 bar True 2012-01-01
In [18]: df2.loc[['a', 'c', 'h'], ['one', 'timestamp']] = np.nan
In [19]: df2
Out[19]:
one two three four five timestamp
a NaN -0.282863 -1.509059 bar True NaT
c NaN 1.212112 -0.173215 bar False NaT
e 0.119209 -1.044236 -0.861849 bar True 2012-01-01
f -2.104569 -0.494929 1.071804 bar False 2012-01-01
h NaN -0.706771 -1.039575 bar True NaT
In [20]: df2.dtypes.value_counts()
Out[20]:
float64 3
datetime64[ns] 1
bool 1
object 1
dtype: int64
对于数字类型的，如果赋值为None，那么会转换为相应的NaN类型：
In [21]: s = pd.Series([1, 2, 3])
In [22]: s.loc[0] = None
In [23]: s
Out[23]:
0 NaN
1 2.0
2 3.0
dtype: float64
如果是对象类型，使用None赋值，会保持原样：
In [24]: s = pd.Series(["a", "b", "c"])
In [25]: s.loc[0] = None
In [26]: s.loc[1] = np.nan
In [27]: s
Out[27]:
0 None
1 NaN
2 c
dtype: object
缺失值的数学计算还是缺失值：
In [28]: a
Out[28]:
one two
a NaN -0.282863
c NaN 1.212112
e 0.119209 -1.044236
f -2.104569 -0.494929
h -2.104569 -0.706771
In [29]: b
Out[29]:
one two three
a NaN -0.282863 -1.509059
c NaN 1.212112 -0.173215
e 0.119209 -1.044236 -0.861849
f -2.104569 -0.494929 1.071804
h NaN -0.706771 -1.039575
In [30]: a + b
Out[30]:
one three two
a NaN NaN -0.565727
c NaN NaN 2.424224
e 0.238417 NaN -2.088472
f -4.209138 NaN -0.989859
h NaN NaN -1.413542
但是在统计中会将NaN当成0来对待。
In [31]: df
Out[31]:
one two three
a NaN -0.282863 -1.509059
c NaN 1.212112 -0.173215
e 0.119209 -1.044236 -0.861849
f -2.104569 -0.494929 1.071804
h NaN -0.706771 -1.039575
In [32]: df['one'].sum()
Out[32]: -1.9853605075978744
In [33]: df.mean(1)
Out[33]:
a -0.895961
c 0.519449
e -0.595625
f -0.509232
h -0.873173
dtype: float64
如果是在cumsum或者cumprod中，默认是会跳过NaN，如果不想统计NaN，可以加上参数skipna=False
In [34]: df.cumsum()
Out[34]:
one two three
a NaN -0.282863 -1.509059
c NaN 0.929249 -1.682273
e 0.119209 -0.114987 -2.544122
f -1.985361 -0.609917 -1.472318
h NaN -1.316688 -2.511893
In [35]: df.cumsum(skipna=False)
Out[35]:
one two three
a NaN -0.282863 -1.509059
c NaN 0.929249 -1.682273
e NaN -0.114987 -2.544122
f NaN -0.609917 -1.472318
h NaN -1.316688 -2.511893
数据分析中，如果有NaN数据，那么需要对其进行处理，一种处理方法就是使用fillna来进行填充。
下面填充常量：
In [42]: df2
Out[42]:
one two three four five timestamp
a NaN -0.282863 -1.509059 bar True NaT
c NaN 1.212112 -0.173215 bar False NaT
e 0.119209 -1.044236 -0.861849 bar True 2012-01-01
f -2.104569 -0.494929 1.071804 bar False 2012-01-01
h NaN -0.706771 -1.039575 bar True NaT
In [43]: df2.fillna(0)
Out[43]:
one two three four five timestamp
a 0.000000 -0.282863 -1.509059 bar True 0
c 0.000000 1.212112 -0.173215 bar False 0
e 0.119209 -1.044236 -0.861849 bar True 2012-01-01 00:00:00
f -2.104569 -0.494929 1.071804 bar False 2012-01-01 00:00:00
h 0.000000 -0.706771 -1.039575 bar True 0
还可以指定填充方法，比如pad：
In [45]: df
Out[45]:
one two three
a NaN -0.282863 -1.509059
c NaN 1.212112 -0.173215
e 0.119209 -1.044236 -0.861849
f -2.104569 -0.494929 1.071804
h NaN -0.706771 -1.039575
In [46]: df.fillna(method='pad')
Out[46]:
one two three
a NaN -0.282863 -1.509059
c NaN 1.212112 -0.173215
e 0.119209 -1.044236 -0.861849
f -2.104569 -0.494929 1.071804
h -2.104569 -0.706771 -1.039575
可以指定填充的行数：
In [48]: df.fillna(method='pad', limit=1)
fill方法统计：
| 方法名 | 描述 |
|---|---|
| pad / ffill | 向前填充 |
| bfill / backfill | 向后填充 |
可以使用PandasObject来填充：
In [53]: dff
Out[53]:
A B C
0 0.271860 -0.424972 0.567020
1 0.276232 -1.087401 -0.673690
2 0.113648 -1.478427 0.524988
3 NaN 0.577046 -1.715002
4 NaN NaN -1.157892
5 -1.344312 NaN NaN
6 -0.109050 1.643563 NaN
7 0.357021 -0.674600 NaN
8 -0.968914 -1.294524 0.413738
9 0.276662 -0.472035 -0.013960
In [54]: dff.fillna(dff.mean())
Out[54]:
A B C
0 0.271860 -0.424972 0.567020
1 0.276232 -1.087401 -0.673690
2 0.113648 -1.478427 0.524988
3 -0.140857 0.577046 -1.715002
4 -0.140857 -0.401419 -1.157892
5 -1.344312 -0.401419 -0.293543
6 -0.109050 1.643563 -0.293543
7 0.357021 -0.674600 -0.293543
8 -0.968914 -1.294524 0.413738
9 0.276662 -0.472035 -0.013960
In [55]: dff.fillna(dff.mean()['B':'C'])
Out[55]:
A B C
0 0.271860 -0.424972 0.567020
1 0.276232 -1.087401 -0.673690
2 0.113648 -1.478427 0.524988
3 NaN 0.577046 -1.715002
4 NaN -0.401419 -1.157892
5 -1.344312 -0.401419 -0.293543
6 -0.109050 1.643563 -0.293543
7 0.357021 -0.674600 -0.293543
8 -0.968914 -1.294524 0.413738
9 0.276662 -0.472035 -0.013960
上面操作等同于：
In [56]: dff.where(pd.notna(dff), dff.mean(), axis='columns')
除了fillna来填充数据之外，还可以使用dropna删除包含na的数据。
In [57]: df
Out[57]:
one two three
a NaN -0.282863 -1.509059
c NaN 1.212112 -0.173215
e NaN 0.000000 0.000000
f NaN 0.000000 0.000000
h NaN -0.706771 -1.039575
In [58]: df.dropna(axis=0)
Out[58]:
Empty DataFrame
Columns: [one, two, three]
Index: []
In [59]: df.dropna(axis=1)
Out[59]:
two three
a -0.282863 -1.509059
c 1.212112 -0.173215
e 0.000000 0.000000
f 0.000000 0.000000
h -0.706771 -1.039575
In [60]: df['one'].dropna()
Out[60]: Series([], Name: one, dtype: float64)
数据分析时候，为了数据的平稳，我们需要一些插值运算interpolate() ，使用起来很简单：
In [61]: ts
Out[61]:
2000-01-31 0.469112
2000-02-29 NaN
2000-03-31 NaN
2000-04-28 NaN
2000-05-31 NaN
...
2007-12-31 -6.950267
2008-01-31 -7.904475
2008-02-29 -6.441779
2008-03-31 -8.184940
2008-04-30 -9.011531
Freq: BM, Length: 100, dtype: float64
In [64]: ts.interpolate()
Out[64]:
2000-01-31 0.469112
2000-02-29 0.434469
2000-03-31 0.399826
2000-04-28 0.365184
2000-05-31 0.330541
...
2007-12-31 -6.950267
2008-01-31 -7.904475
2008-02-29 -6.441779
2008-03-31 -8.184940
2008-04-30 -9.011531
Freq: BM, Length: 100, dtype: float64
插值函数还可以添加参数，指定插值的方法，比如按时间插值：
In [67]: ts2
Out[67]:
2000-01-31 0.469112
2000-02-29 NaN
2002-07-31 -5.785037
2005-01-31 NaN
2008-04-30 -9.011531
dtype: float64
In [68]: ts2.interpolate()
Out[68]:
2000-01-31 0.469112
2000-02-29 -2.657962
2002-07-31 -5.785037
2005-01-31 -7.398284
2008-04-30 -9.011531
dtype: float64
In [69]: ts2.interpolate(method='time')
Out[69]:
2000-01-31 0.469112
2000-02-29 0.270241
2002-07-31 -5.785037
2005-01-31 -7.190866
2008-04-30 -9.011531
dtype: float64
按index的float value进行插值：
In [70]: ser
Out[70]:
0.0 0.0
1.0 NaN
10.0 10.0
dtype: float64
In [71]: ser.interpolate()
Out[71]:
0.0 0.0
1.0 5.0
10.0 10.0
dtype: float64
In [72]: ser.interpolate(method='values')
Out[72]:
0.0 0.0
1.0 1.0
10.0 10.0
dtype: float64
除了插值Series，还可以插值DF：
In [73]: df = pd.DataFrame({'A': [1, 2.1, np.nan, 4.7, 5.6, 6.8],
....: 'B': [.25, np.nan, np.nan, 4, 12.2, 14.4]})
....:
In [74]: df
Out[74]:
A B
0 1.0 0.25
1 2.1 NaN
2 NaN NaN
3 4.7 4.00
4 5.6 12.20
5 6.8 14.40
In [75]: df.interpolate()
Out[75]:
A B
0 1.0 0.25
1 2.1 1.50
2 3.4 2.75
3 4.7 4.00
4 5.6 12.20
5 6.8 14.40
interpolate还接收limit参数，可以指定插值的个数。
In [95]: ser.interpolate(limit=1)
Out[95]:
0 NaN
1 NaN
2 5.0
3 7.0
4 NaN
5 NaN
6 13.0
7 13.0
8 NaN
dtype: float64
replace可以替换常量，也可以替换list：
In [102]: ser = pd.Series([0., 1., 2., 3., 4.])
In [103]: ser.replace(0, 5)
Out[103]:
0 5.0
1 1.0
2 2.0
3 3.0
4 4.0
dtype: float64
In [104]: ser.replace([0, 1, 2, 3, 4], [4, 3, 2, 1, 0])
Out[104]:
0 4.0
1 3.0
2 2.0
3 1.0
4 0.0
dtype: float64
可以替换DF中特定的数值：
In [106]: df = pd.DataFrame({'a': [0, 1, 2, 3, 4], 'b': [5, 6, 7, 8, 9]})
In [107]: df.replace({'a': 0, 'b': 5}, 100)
Out[107]:
a b
0 100 100
1 1 6
2 2 7
3 3 8
4 4 9
可以使用插值替换：
In [108]: ser.replace([1, 2, 3], method='pad')
Out[108]:
0 0.0
1 0.0
2 0.0
3 0.0
4 4.0
dtype: float64