首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >Pandas高级教程之:处理缺失数据

Pandas高级教程之:处理缺失数据

原创
作者头像
程序那些事
修改2021-06-24 10:19:41
8500
修改2021-06-24 10:19:41
举报
文章被收录于专栏:程序那些事程序那些事

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

整数类型的缺失值

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

Datetimes 类型的缺失值

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

None 和 np.nan 的转换

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

缺失值的计算

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

使用fillna填充NaN数据

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')

使用dropna删除包含NA的数据

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)

插值interpolation

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

使用replace替换值

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

本文已收录于 http://www.flydean.com/07-python-pandas-missingdata/ 最通俗的解读,最深刻的干货,最简洁的教程,众多你不知道的小技巧等你来发现! 欢迎关注我的公众号:「程序那些事」,懂技术,更懂你!

原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。

如有侵权,请联系 cloudcommunity@tencent.com 删除。

原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。

如有侵权,请联系 cloudcommunity@tencent.com 删除。

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
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
相关产品与服务
数据万象
数据万象(Cloud Infinite,CI)是依托腾讯云对象存储的数据处理平台,涵盖图片处理、内容审核、媒体处理、AI 识别、文档预览等功能,为客户提供一站式的专业数据处理解决方案,满足您多种业务场景的需求。
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档