Working with missing data

In this section, we will discuss missing (also referred to as NA) values in pandas.

See the cookbook for some advanced strategies

Missing data basics

When / why does data become missing?

Some might quibble over our usage of missing. By ?missing? we simply mean null or ?not present for whatever reason?. Many data sets simply arrive with missing data, either because it exists and was not collected or it never existed. For example, in a collection of financial time series, some of the time series might start on different dates. Thus, values prior to the start date would generally be marked as missing.

In pandas, one of the most common ways that missing data is introduced into a data set is by reindexing. For example

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

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

Values considered ?missing?

As data comes in many shapes and forms, pandas aims to be flexible with regard to handling missing data. While NaN is the default missing value marker for reasons of computational speed and convenience, we need to be able to easily detect this value with data of different types: floating point, integer, boolean, and general object. In many cases, however, the Python None will arise and we wish to also consider that ?missing? or ?null?.

To make detecting missing values easier (and across different array dtypes), pandas provides the isnull() and notnull() functions, which are also methods on Series and DataFrame objects:

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.isnull(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'].notnull()
Out[9]: 
a     True
b    False
c     True
d    False
e     True
f     True
g    False
h     True
Name: four, dtype: bool

In [10]: df2.isnull()
Out[10]: 
     one    two  three   four   five
a  False  False  False  False  False
b   True   True   True   True   True
c  False  False  False  False  False
d   True   True   True   True   True
e  False  False  False  False  False
f  False  False  False  False  False
g   True   True   True   True   True
h  False  False  False  False  False

In [11]: None == None
Out[11]: True

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

In [13]: df2['one'] == np.nan
Out[13]: 
a    False
b    False
c    False
d    False
e    False
f    False
g    False
h    False
Name: one, dtype: bool

Datetimes

For datetime64[ns] types, NaT represents missing values. This is a pseudo-native sentinel value that can be represented by numpy in a singular dtype (datetime64[ns]). pandas objects provide intercompatibility between NaT and NaN.

In [14]: df2 = df.copy()

In [15]: df2['timestamp'] = pd.Timestamp('20120101')

In [16]: df2
Out[16]: 
        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 [17]: df2.ix[['a','c','h'],['one','timestamp']] = np.nan

In [18]: df2
Out[18]: 
        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 [19]: df2.get_dtype_counts()
Out[19]: 
bool              1
datetime64[ns]    1
float64           3
object            1
dtype: int64

Inserting missing data

You can insert missing values by simply assigning to containers. The actual missing value used will be chosen based on the dtype.

For example, numeric containers will always use NaN regardless of the missing value type chosen:

In [20]: s = pd.Series([1, 2, 3])

In [21]: s.loc[0] = None

In [22]: s
Out[22]: 
0    NaN
1    2.0
2    3.0
dtype: float64

Likewise, datetime containers will always use NaT.

For object containers, pandas will use the value given:

In [23]: s = pd.Series(["a", "b", "c"])

In [24]: s.loc[0] = None

In [25]: s.loc[1] = np.nan

In [26]: s
Out[26]: 
0    None
1     NaN
2       c
dtype: object

Calculations with missing data

Missing values propagate naturally through arithmetic operations between pandas objects.

In [27]: a
Out[27]: 
        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 [28]: b
Out[28]: 
        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 [29]: a + b
Out[29]: 
        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

The descriptive statistics and computational methods discussed in the data structure overview (and listed here and here) are all written to account for missing data. For example:

• When summing data, NA (missing) values will be treated as zero
• If the data are all NA, the result will be NA
• Methods like cumsum and cumprod ignore NA values, but preserve them in the resulting arrays

In [30]: df
Out[30]: 
        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 [31]: df['one'].sum()
Out[31]: -1.9853605075978744

In [32]: df.mean(1)
Out[32]: 
a   -0.895961
c    0.519449
e   -0.595625
f   -0.509232
h   -0.873173
dtype: float64

In [33]: df.cumsum()
Out[33]: 
        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

NA values in GroupBy

NA groups in GroupBy are automatically excluded. This behavior is consistent with R, for example:

In [34]: df
Out[34]: 
        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 [35]: df.groupby('one').mean()
Out[35]: 
                two     three
one                          
-2.104569 -0.494929  1.071804
 0.119209 -1.044236 -0.861849

See the groupby section here for more information.

Cleaning / filling missing data

pandas objects are equipped with various data manipulation methods for dealing with missing data.

Filling missing values: fillna

The fillna function can ?fill in? NA values with non-null data in a couple of ways, which we illustrate:

Replace NA with a scalar value

In [36]: df2
Out[36]: 
        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 [37]: df2.fillna(0)
Out[37]: 
        one       two     three four   five  timestamp
a  0.000000 -0.282863 -1.509059  bar   True 1970-01-01
c  0.000000  1.212112 -0.173215  bar  False 1970-01-01
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  0.000000 -0.706771 -1.039575  bar   True 1970-01-01

In [38]: df2['four'].fillna('missing')
Out[38]: 
a    bar
c    bar
e    bar
f    bar
h    bar
Name: four, dtype: object

Fill gaps forward or backward

Using the same filling arguments as reindexing, we can propagate non-null values forward or backward:

In [39]: df
Out[39]: 
        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 [40]: df.fillna(method='pad')
Out[40]: 
        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

Limit the amount of filling

If we only want consecutive gaps filled up to a certain number of data points, we can use the limit keyword:

In [41]: df
Out[41]: 
   one       two     three
a  NaN -0.282863 -1.509059
c  NaN  1.212112 -0.173215
e  NaN       NaN       NaN
f  NaN       NaN       NaN
h  NaN -0.706771 -1.039575

In [42]: df.fillna(method='pad', limit=1)
Out[42]: 
   one       two     three
a  NaN -0.282863 -1.509059
c  NaN  1.212112 -0.173215
e  NaN  1.212112 -0.173215
f  NaN       NaN       NaN
h  NaN -0.706771 -1.039575

To remind you, these are the available filling methods:

With time series data, using pad/ffill is extremely common so that the ?last known value? is available at every time point.

The ffill() function is equivalent to fillna(method='ffill') and bfill() is equivalent to fillna(method='bfill')

Filling with a PandasObject

You can also fillna using a dict or Series that is alignable. The labels of the dict or index of the Series must match the columns of the frame you wish to fill. The use case of this is to fill a DataFrame with the mean of that column.

In [43]: dff = pd.DataFrame(np.random.randn(10,3), columns=list('ABC'))

In [44]: dff.iloc[3:5,0] = np.nan

In [45]: dff.iloc[4:6,1] = np.nan

In [46]: dff.iloc[5:8,2] = np.nan

In [47]: dff
Out[47]: 
          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 [48]: dff.fillna(dff.mean())
Out[48]: 
          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 [49]: dff.fillna(dff.mean()['B':'C'])
Out[49]: 
          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

Same result as above, but is aligning the ?fill? value which is a Series in this case.

In [50]: dff.where(pd.notnull(dff), dff.mean(), axis='columns')
Out[50]: 
          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

Dropping axis labels with missing data: dropna

You may wish to simply exclude labels from a data set which refer to missing data. To do this, use the dropna method:

In [51]: df
Out[51]: 
   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 [52]: df.dropna(axis=0)
Out[52]: 
Empty DataFrame
Columns: [one, two, three]
Index: []

In [53]: df.dropna(axis=1)
Out[53]: 
        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 [54]: df['one'].dropna()
Out[54]: Series([], Name: one, dtype: float64)

Series.dropna is a simpler method as it only has one axis to consider. DataFrame.dropna has considerably more options than Series.dropna, which can be examined in the API.

Interpolation

Both Series and Dataframe objects have an interpolate method that, by default, performs linear interpolation at missing datapoints.

In [55]: ts
Out[55]: 
2000-01-31    0.469112
2000-02-29         NaN
2000-03-31         NaN
2000-04-28         NaN
2000-05-31         NaN
2000-06-30         NaN
2000-07-31         NaN
                ...   
2007-10-31   -3.305259
2007-11-30   -5.485119
2007-12-31   -6.854968
2008-01-31   -7.809176
2008-02-29   -6.346480
2008-03-31   -8.089641
2008-04-30   -8.916232
Freq: BM, dtype: float64

In [56]: ts.count()
Out[56]: 61

In [57]: ts.interpolate().count()
Out[57]: 100

In [58]: ts.interpolate().plot()
Out[58]: <matplotlib.axes._subplots.AxesSubplot at 0x7ff2667af150>

Index aware interpolation is available via the method keyword:

In [59]: ts2
Out[59]: 
2000-01-31    0.469112
2000-02-29         NaN
2002-07-31   -5.689738
2005-01-31         NaN
2008-04-30   -8.916232
dtype: float64

In [60]: ts2.interpolate()
Out[60]: 
2000-01-31    0.469112
2000-02-29   -2.610313
2002-07-31   -5.689738
2005-01-31   -7.302985
2008-04-30   -8.916232
dtype: float64

In [61]: ts2.interpolate(method='time')
Out[61]: 
2000-01-31    0.469112
2000-02-29    0.273272
2002-07-31   -5.689738
2005-01-31   -7.095568
2008-04-30   -8.916232
dtype: float64

For a floating-point index, use method='values':

In [62]: ser
Out[62]: 
0.0      0.0
1.0      NaN
10.0    10.0
dtype: float64

In [63]: ser.interpolate()
Out[63]: 
0.0      0.0
1.0      5.0
10.0    10.0
dtype: float64

In [64]: ser.interpolate(method='values')
Out[64]: 
0.0      0.0
1.0      1.0
10.0    10.0
dtype: float64

You can also interpolate with a DataFrame:

In [65]: 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 [66]: df
Out[66]: 
     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 [67]: df.interpolate()
Out[67]: 
     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

The method argument gives access to fancier interpolation methods. If you have scipy installed, you can set pass the name of a 1-d interpolation routine to method. You?ll want to consult the full scipy interpolation documentation and reference guide for details. The appropriate interpolation method will depend on the type of data you are working with.

• If you are dealing with a time series that is growing at an increasing rate, method='quadratic' may be appropriate.
• If you have values approximating a cumulative distribution function, then method='pchip' should work well.
• To fill missing values with goal of smooth plotting, use method='akima'.

In [68]: df.interpolate(method='barycentric')
Out[68]: 
      A       B
0  1.00   0.250
1  2.10  -7.660
2  3.53  -4.515
3  4.70   4.000
4  5.60  12.200
5  6.80  14.400

In [69]: df.interpolate(method='pchip')
Out[69]: 
         A          B
0  1.00000   0.250000
1  2.10000   0.672808
2  3.43454   1.928950
3  4.70000   4.000000
4  5.60000  12.200000
5  6.80000  14.400000

In [70]: df.interpolate(method='akima')
Out[70]: 
          A          B
0  1.000000   0.250000
1  2.100000  -0.873316
2  3.406667   0.320034
3  4.700000   4.000000
4  5.600000  12.200000
5  6.800000  14.400000

When interpolating via a polynomial or spline approximation, you must also specify the degree or order of the approximation:

In [71]: df.interpolate(method='spline', order=2)
Out[71]: 
          A          B
0  1.000000   0.250000
1  2.100000  -0.428598
2  3.404545   1.206900
3  4.700000   4.000000
4  5.600000  12.200000
5  6.800000  14.400000

In [72]: df.interpolate(method='polynomial', order=2)
Out[72]: 
          A          B
0  1.000000   0.250000
1  2.100000  -4.161538
2  3.547059  -2.911538
3  4.700000   4.000000
4  5.600000  12.200000
5  6.800000  14.400000

Compare several methods:

In [73]: np.random.seed(2)

In [74]: ser = pd.Series(np.arange(1, 10.1, .25)**2 + np.random.randn(37))

In [75]: bad = np.array([4, 13, 14, 15, 16, 17, 18, 20, 29])

In [76]: ser[bad] = np.nan

In [77]: methods = ['linear', 'quadratic', 'cubic']

In [78]: df = pd.DataFrame({m: ser.interpolate(method=m) for m in methods})

In [79]: df.plot()
Out[79]: <matplotlib.axes._subplots.AxesSubplot at 0x7ff2666771d0>

Another use case is interpolation at new values. Suppose you have 100 observations from some distribution. And let?s suppose that you?re particularly interested in what?s happening around the middle. You can mix pandas? reindex and interpolate methods to interpolate at the new values.

In [80]: ser = pd.Series(np.sort(np.random.uniform(size=100)))

# interpolate at new_index
In [81]: new_index = ser.index | pd.Index([49.25, 49.5, 49.75, 50.25, 50.5, 50.75])

In [82]: interp_s = ser.reindex(new_index).interpolate(method='pchip')

In [83]: interp_s[49:51]
Out[83]: 
49.00    0.471410
49.25    0.476841
49.50    0.481780
49.75    0.485998
50.00    0.489266
50.25    0.491814
50.50    0.493995
50.75    0.495763
51.00    0.497074
dtype: float64

Interpolation Limits

Like other pandas fill methods, interpolate accepts a limit keyword argument. Use this argument to limit the number of consecutive interpolations, keeping NaN values for interpolations that are too far from the last valid observation:

In [84]: ser = pd.Series([np.nan, np.nan, 5, np.nan, np.nan, np.nan, 13])

In [85]: ser.interpolate(limit=2)
Out[85]: 
0     NaN
1     NaN
2     5.0
3     7.0
4     9.0
5     NaN
6    13.0
dtype: float64

By default, limit applies in a forward direction, so that only NaN values after a non-NaN value can be filled. If you provide 'backward' or 'both' for the limit_direction keyword argument, you can fill NaN values before non-NaN values, or both before and after non-NaN values, respectively:

In [86]: ser.interpolate(limit=1)  # limit_direction == 'forward'
Out[86]: 
0     NaN
1     NaN
2     5.0
3     7.0
4     NaN
5     NaN
6    13.0
dtype: float64

In [87]: ser.interpolate(limit=1, limit_direction='backward')
Out[87]: 
0     NaN
1     5.0
2     5.0
3     NaN
4     NaN
5    11.0
6    13.0
dtype: float64

In [88]: ser.interpolate(limit=1, limit_direction='both')
Out[88]: 
0     NaN
1     5.0
2     5.0
3     7.0
4     NaN
5    11.0
6    13.0
dtype: float64

Replacing Generic Values

Often times we want to replace arbitrary values with other values. New in v0.8 is the replace method in Series/DataFrame that provides an efficient yet flexible way to perform such replacements.

For a Series, you can replace a single value or a list of values by another value:

In [89]: ser = pd.Series([0., 1., 2., 3., 4.])

In [90]: ser.replace(0, 5)
Out[90]: 
0    5.0
1    1.0
2    2.0
3    3.0
4    4.0
dtype: float64

You can replace a list of values by a list of other values:

In [91]: ser.replace([0, 1, 2, 3, 4], [4, 3, 2, 1, 0])
Out[91]: 
0    4.0
1    3.0
2    2.0
3    1.0
4    0.0
dtype: float64

You can also specify a mapping dict:

In [92]: ser.replace({0: 10, 1: 100})
Out[92]: 
0     10.0
1    100.0
2      2.0
3      3.0
4      4.0
dtype: float64

For a DataFrame, you can specify individual values by column:

In [93]: df = pd.DataFrame({'a': [0, 1, 2, 3, 4], 'b': [5, 6, 7, 8, 9]})

In [94]: df.replace({'a': 0, 'b': 5}, 100)
Out[94]: 
     a    b
0  100  100
1    1    6
2    2    7
3    3    8
4    4    9

Instead of replacing with specified values, you can treat all given values as missing and interpolate over them:

In [95]: ser.replace([1, 2, 3], method='pad')
Out[95]: 
0    0.0
1    0.0
2    0.0
3    0.0
4    4.0
dtype: float64

String/Regular Expression Replacement

Replace the ?.? with nan (str -> str)

In [96]: d = {'a': list(range(4)), 'b': list('ab..'), 'c': ['a', 'b', np.nan, 'd']}

In [97]: df = pd.DataFrame(d)

In [98]: df.replace('.', np.nan)
Out[98]: 
   a    b    c
0  0    a    a
1  1    b    b
2  2  NaN  NaN
3  3  NaN    d

Now do it with a regular expression that removes surrounding whitespace (regex -> regex)

In [99]: df.replace(r'\s*\.\s*', np.nan, regex=True)
Out[99]: 
   a    b    c
0  0    a    a
1  1    b    b
2  2  NaN  NaN
3  3  NaN    d

Replace a few different values (list -> list)

In [100]: df.replace(['a', '.'], ['b', np.nan])
Out[100]: 
   a    b    c
0  0    b    b
1  1    b    b
2  2  NaN  NaN
3  3  NaN    d

list of regex -> list of regex

In [101]: df.replace([r'\.', r'(a)'], ['dot', '\1stuff'], regex=True)
Out[101]: 
   a       b       c
0  0  stuff  stuff
1  1       b       b
2  2     dot     NaN
3  3     dot       d

Only search in column 'b' (dict -> dict)

In [102]: df.replace({'b': '.'}, {'b': np.nan})
Out[102]: 
   a    b    c
0  0    a    a
1  1    b    b
2  2  NaN  NaN
3  3  NaN    d

Same as the previous example, but use a regular expression for searching instead (dict of regex -> dict)

In [103]: df.replace({'b': r'\s*\.\s*'}, {'b': np.nan}, regex=True)
Out[103]: 
   a    b    c
0  0    a    a
1  1    b    b
2  2  NaN  NaN
3  3  NaN    d

You can pass nested dictionaries of regular expressions that use regex=True

In [104]: df.replace({'b': {'b': r''}}, regex=True)
Out[104]: 
   a  b    c
0  0  a    a
1  1       b
2  2  .  NaN
3  3  .    d

or you can pass the nested dictionary like so

In [105]: df.replace(regex={'b': {r'\s*\.\s*': np.nan}})
Out[105]: 
   a    b    c
0  0    a    a
1  1    b    b
2  2  NaN  NaN
3  3  NaN    d

You can also use the group of a regular expression match when replacing (dict of regex -> dict of regex), this works for lists as well

In [106]: df.replace({'b': r'\s*(\.)\s*'}, {'b': r'\1ty'}, regex=True)
Out[106]: 
   a    b    c
0  0    a    a
1  1    b    b
2  2  .ty  NaN
3  3  .ty    d

You can pass a list of regular expressions, of which those that match will be replaced with a scalar (list of regex -> regex)

In [107]: df.replace([r'\s*\.\s*', r'a|b'], np.nan, regex=True)
Out[107]: 
   a   b    c
0  0 NaN  NaN
1  1 NaN  NaN
2  2 NaN  NaN
3  3 NaN    d

All of the regular expression examples can also be passed with the to_replace argument as the regex argument. In this case the value argument must be passed explicitly by name or regex must be a nested dictionary. The previous example, in this case, would then be

In [108]: df.replace(regex=[r'\s*\.\s*', r'a|b'], value=np.nan)
Out[108]: 
   a   b    c
0  0 NaN  NaN
1  1 NaN  NaN
2  2 NaN  NaN
3  3 NaN    d

This can be convenient if you do not want to pass regex=True every time you want to use a regular expression.

Numeric Replacement

Similar to DataFrame.fillna

In [109]: df = pd.DataFrame(np.random.randn(10, 2))

In [110]: df[np.random.rand(df.shape[0]) > 0.5] = 1.5

In [111]: df.replace(1.5, np.nan)
Out[111]: 
          0         1
0 -0.844214 -1.021415
1  0.432396 -0.323580
2  0.423825  0.799180
3  1.262614  0.751965
4       NaN       NaN
5       NaN       NaN
6 -0.498174 -1.060799
7  0.591667 -0.183257
8  1.019855 -1.482465
9       NaN       NaN

Replacing more than one value via lists works as well

In [112]: df00 = df.values[0, 0]

In [113]: df.replace([1.5, df00], [np.nan, 'a'])
Out[113]: 
          0         1
0         a -1.021415
1  0.432396 -0.323580
2  0.423825  0.799180
3   1.26261  0.751965
4       NaN       NaN
5       NaN       NaN
6 -0.498174 -1.060799
7  0.591667 -0.183257
8   1.01985 -1.482465
9       NaN       NaN

In [114]: df[1].dtype
Out[114]: dtype('float64')

You can also operate on the DataFrame in place

In [115]: df.replace(1.5, np.nan, inplace=True)

s = pd.Series([True, False, True])
s.replace({'a string': 'new value', True: False})  # raises

TypeError: Cannot compare types 'ndarray(dtype=bool)' and 'str'

In [116]: s = pd.Series([True, False, True])

In [117]: s.replace('a string', 'another string')
Out[117]: 
0     True
1    False
2     True
dtype: bool

Missing data casting rules and indexing

While pandas supports storing arrays of integer and boolean type, these types are not capable of storing missing data. Until we can switch to using a native NA type in NumPy, we?ve established some ?casting rules? when reindexing will cause missing data to be introduced into, say, a Series or DataFrame. Here they are:

For example:

In [118]: s = pd.Series(np.random.randn(5), index=[0, 2, 4, 6, 7])

In [119]: s > 0
Out[119]: 
0    True
2    True
4    True
6    True
7    True
dtype: bool

In [120]: (s > 0).dtype
Out[120]: dtype('bool')

In [121]: crit = (s > 0).reindex(list(range(8)))

In [122]: crit
Out[122]: 
0    True
1     NaN
2    True
3     NaN
4    True
5     NaN
6    True
7    True
dtype: object

In [123]: crit.dtype
Out[123]: dtype('O')

Ordinarily NumPy will complain if you try to use an object array (even if it contains boolean values) instead of a boolean array to get or set values from an ndarray (e.g. selecting values based on some criteria). If a boolean vector contains NAs, an exception will be generated:

In [124]: reindexed = s.reindex(list(range(8))).fillna(0)

In [125]: reindexed[crit]
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-125-2da204ed1ac7> in <module>()
----> 1 reindexed[crit]

/home/joris/scipy/pandas/pandas/core/series.pyc in __getitem__(self, key)
    639             key = list(key)
    640 
--> 641         if com.is_bool_indexer(key):
    642             key = check_bool_indexer(self.index, key)
    643 

/home/joris/scipy/pandas/pandas/core/common.pyc in is_bool_indexer(key)
    199             if not lib.is_bool_array(key):
    200                 if isnull(key).any():
--> 201                     raise ValueError('cannot index with vector containing '
    202                                      'NA / NaN values')
    203                 return False

ValueError: cannot index with vector containing NA / NaN values

However, these can be filled in using fillna and it will work fine:

In [126]: reindexed[crit.fillna(False)]
Out[126]: 
0    0.126504
2    0.696198
4    0.697416
6    0.601516
7    0.003659
dtype: float64

In [127]: reindexed[crit.fillna(True)]
Out[127]: 
0    0.126504
1    0.000000
2    0.696198
3    0.000000
4    0.697416
5    0.000000
6    0.601516
7    0.003659
dtype: float64