简介
在数据处理中，Pandas 会将无奈解析的数据或者缺失的数据应用 NaN 来示意。尽管所有的数据都有了相应的示意，然而 NaN 很显著是无奈进行数学运算的。
本文将会解说 Pandas 对于 NaN 数据的解决办法。
NaN 的例子
下面讲到了缺失的数据会被体现为 NaN，咱们来看一个具体的例子：
咱们先来构建一个 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下面 DF 只有 acefh 这几个 index，咱们从新 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，能够应用 isna() 或者 notna() 办法。
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留神在 Python 中 None 是相等的：
In [11]: None == None # noqa: E711
Out[11]: True然而 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，咱们能够强制进行转换：
In [14]: pd.Series([1, 2, np.nan, 4], dtype=pd.Int64Dtype())
Out[14]:
0 1
1 2
2 <NA>
3 4
dtype: Int64Datetimes 类型的缺失值
工夫类型的缺失值应用 NaT 来示意：
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: int64None 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，应用 None 赋值，会放弃原样：
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然而在统计中会将 NaN 当成 0 来看待。
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，能够加上参数 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应用 fillna 填充 NaN 数据
数据分析中，如果有 NaN 数据，那么须要对其进行解决，一种解决办法就是应用 fillna 来进行填充。
上面填充常量：
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还能够指定填充办法，比方 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能够指定填充的行数：
In [48]: df.fillna(method='pad', limit=1)fill 办法统计：
| 办法名 | 形容 |
|---|---|
| pad / ffill | 向前填充 |
| bfill / backfill | 向后填充 |
能够应用 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下面操作等同于：
In [56]: dff.where(pd.notna(dff), dff.mean(), axis='columns')应用 dropna 删除蕴含 NA 的数据
除了 fillna 来填充数据之外，还能够应用 dropna 删除蕴含 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，为了数据的安稳，咱们须要一些插值运算 interpolate()，应用起来很简略：
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: float64In [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插值函数还能够增加参数，指定插值的办法，比方按工夫插值：
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，还能够插值 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 14.40interpolate 还接管 limit 参数，能够指定插值的个数。
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应用 replace 替换值
replace 能够替换常量，也能够替换 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: float64In [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能够替换 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能够应用插值替换：
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/
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