Pandas advanced tutorial: dealing with missing data

Introduction

In data processing, Pandas will use NaN to represent unparseable data or missing data. Although all data have corresponding representations, NaN obviously cannot perform mathematical operations.

This article will explain how Pandas handles NaN data.

NaN example

As mentioned above, missing data will be represented as NaN. Let’s look at a specific example:

Let's first build a 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

The above DF only has a few indexes of acefh, let's reindex the data:

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

If the data is missing, a lot of NaN will be generated.

In order to detect whether it is NaN, you can use the isna() or notna() method.

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

Note that None is equivalent in Python:

In [11]: None == None                                                 # noqa: E711
Out[11]: True

But np.nan is not equal:

In [12]: np.nan == np.nan
Out[12]: False

Missing value of integer type

NaN is a float type by default. If it is an integer type, we can force a conversion:

In [14]: pd.Series([1, 2, np.nan, 4], dtype=pd.Int64Dtype())
Out[14]: 
0       1
1       2
2    <NA>
3       4
dtype: Int64

Missing values of type Datetimes

The missing value of the time type is represented by NaT:

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

Conversion of None and np.nan

For numeric types, if the assignment is None, it will be converted to the corresponding NaN type:

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

If it is an object type, use None to assign a value, it will remain as it is:

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

Calculation of missing values

The mathematical calculation of missing values is still missing values:

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

However, NaN is treated as 0 in statistics.

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

If it is in cumsum or cumprod, NaN will be skipped by default. If you don’t want to count NaN, you can add the parameter 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

Use fillna to fill NaN data

In data analysis, if there is NaN data, then it needs to be processed. One processing method is to use fillna for filling.

Fill in the constant below:

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

You can also specify the padding method, such as 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

You can specify the number of rows to fill:

In [48]: df.fillna(method='pad', limit=1)

Fill method statistics:

Method name	description
pad / ffill	Forward padding
bfill / backfill	Fill back

You can use PandasObject to fill:

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

The above operation is equivalent to:

In [56]: dff.where(pd.notna(dff), dff.mean(), axis='columns')

Use dropna to delete data containing NA

In addition to fillna to fill data, you can also use dropna to delete data containing na.

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)

Interpolation

During data analysis, in order to stabilize the data, we need some interpolation operations interpolate(), which is very simple to use:

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

The interpolation function can also add parameters to specify the interpolation method, such as interpolation by time:

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

Interpolate according to the float value of index:

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

In addition to interpolating Series, you can also interpolate 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 also receives the limit parameter, which can specify the number of interpolations.

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

Use replace to replace values

Replace can replace constants or 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

You can replace specific values in DF:

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

You can use interpolation substitution:

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

This article has been included in http://www.flydean.com/07-python-pandas-missingdata/

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