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# Working with missing data

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

Note

The choice of using `NaN` internally to denote missing data was largely for simplicity and performance reasons. It differs from the MaskedArray approach of, for example, `scikits.timeseries`. We are hopeful that NumPy will soon be able to provide a native NA type solution (similar to R) performant enough to be used 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 = DataFrame(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  0.059117  1.138469 -2.400634  bar   True
    c -0.280853  0.025653 -1.386071  bar  False
    e  0.863937  0.252462  1.500571  bar   True
    f  1.053202 -2.338595 -0.374279  bar   True
    h -2.359958 -1.157886 -0.551865  bar  False

    [5 rows x 5 columns]

    In [5]: df2 = df.reindex(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])

    In [6]: df2
    Out[6]:
            one       two     three four   five
    a  0.059117  1.138469 -2.400634  bar   True
    b       NaN       NaN       NaN  NaN    NaN
    c -0.280853  0.025653 -1.386071  bar  False
    d       NaN       NaN       NaN  NaN    NaN
    e  0.863937  0.252462  1.500571  bar   True
    f  1.053202 -2.338595 -0.374279  bar   True
    g       NaN       NaN       NaN  NaN    NaN
    h -2.359958 -1.157886 -0.551865  bar  False

    [8 rows x 5 columns]

### 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”.

Until recently, for legacy reasons `inf` and `-inf` were also considered to be “null” in computations. This is no longer the case by default; use the `mode.use_inf_as_null` option to recover it.

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

    In [7]: df2['one']
    Out[7]:
    a    0.059117
    b         NaN
    c   -0.280853
    d         NaN
    e    0.863937
    f    1.053202
    g         NaN
    h   -2.359958
    Name: one, dtype: float64

    In [8]: 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

Summary: `NaN` and `None` (in object arrays) are considered missing by the `isnull` and `notnull` functions. `inf` and `-inf` are no longer considered missing by default.

## 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 [10]: df2 = df.copy()

    In [11]: df2['timestamp'] = Timestamp('20120101')

    In [12]: df2
    Out[12]:
            one       two     three four   five  timestamp
    a  0.059117  1.138469 -2.400634  bar   True 2012-01-01
    c -0.280853  0.025653 -1.386071  bar  False 2012-01-01
    e  0.863937  0.252462  1.500571  bar   True 2012-01-01
    f  1.053202 -2.338595 -0.374279  bar   True 2012-01-01
    h -2.359958 -1.157886 -0.551865  bar  False 2012-01-01

    [5 rows x 6 columns]

    In [13]: df2.ix[['a','c','h'],['one','timestamp']] = np.nan

    In [14]: df2
    Out[14]:
            one       two     three four   five  timestamp
    a       NaN  1.138469 -2.400634  bar   True        NaT
    c       NaN  0.025653 -1.386071  bar  False        NaT
    e  0.863937  0.252462  1.500571  bar   True 2012-01-01
    f  1.053202 -2.338595 -0.374279  bar   True 2012-01-01
    h       NaN -1.157886 -0.551865  bar  False        NaT

    [5 rows x 6 columns]

    In [15]: df2.get_dtype_counts()
    Out[15]:
    bool              1
    datetime64[ns]    1
    float64           3
    object            1
    dtype: int64

## Calculations with missing data

Missing values propagate naturally through arithmetic operations between pandas objects.

    In [16]: a
    Out[16]:
            one       two
    a       NaN  1.138469
    c       NaN  0.025653
    e  0.863937  0.252462
    f  1.053202 -2.338595
    h  1.053202 -1.157886

    [5 rows x 2 columns]

    In [17]: b
    Out[17]:
            one       two     three
    a       NaN  1.138469 -2.400634
    c       NaN  0.025653 -1.386071
    e  0.863937  0.252462  1.500571
    f  1.053202 -2.338595 -0.374279
    h       NaN -1.157886 -0.551865

    [5 rows x 3 columns]

    In [18]: a + b
    Out[18]:
            one  three       two
    a       NaN    NaN  2.276938
    c       NaN    NaN  0.051306
    e  1.727874    NaN  0.504923
    f  2.106405    NaN -4.677190
    h       NaN    NaN -2.315772

    [5 rows x 3 columns]

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 [19]: df
    Out[19]:
            one       two     three
    a       NaN  1.138469 -2.400634
    c       NaN  0.025653 -1.386071
    e  0.863937  0.252462  1.500571
    f  1.053202 -2.338595 -0.374279
    h       NaN -1.157886 -0.551865

    [5 rows x 3 columns]

    In [20]: df['one'].sum()
    Out[20]: 1.917139050150438

    In [21]: df.mean(1)
    Out[21]:
    a   -0.631082
    c   -0.680209
    e    0.872323
    f   -0.553224
    h   -0.854876
    dtype: float64

    In [22]: df.cumsum()
    Out[22]:
            one       two     three
    a       NaN  1.138469 -2.400634
    c       NaN  1.164122 -3.786705
    e  0.863937  1.416584 -2.286134
    f  1.917139 -0.922011 -2.660413
    h       NaN -2.079897 -3.212278

    [5 rows x 3 columns]

### NA values in GroupBy

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

## 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 [23]: df2
    Out[23]:
            one       two     three four   five  timestamp
    a       NaN  1.138469 -2.400634  bar   True        NaT
    c       NaN  0.025653 -1.386071  bar  False        NaT
    e  0.863937  0.252462  1.500571  bar   True 2012-01-01
    f  1.053202 -2.338595 -0.374279  bar   True 2012-01-01
    h       NaN -1.157886 -0.551865  bar  False        NaT

    [5 rows x 6 columns]

    In [24]: df2.fillna(0)
    Out[24]:
            one       two     three four   five  timestamp
    a  0.000000  1.138469 -2.400634  bar   True 1970-01-01
    c  0.000000  0.025653 -1.386071  bar  False 1970-01-01
    e  0.863937  0.252462  1.500571  bar   True 2012-01-01
    f  1.053202 -2.338595 -0.374279  bar   True 2012-01-01
    h  0.000000 -1.157886 -0.551865  bar  False 1970-01-01

    [5 rows x 6 columns]

    In [25]: df2['four'].fillna('missing')
    Out[25]:
    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 [26]: df
    Out[26]:
            one       two     three
    a       NaN  1.138469 -2.400634
    c       NaN  0.025653 -1.386071
    e  0.863937  0.252462  1.500571
    f  1.053202 -2.338595 -0.374279
    h       NaN -1.157886 -0.551865

    [5 rows x 3 columns]

    In [27]: df.fillna(method='pad')
    Out[27]:
            one       two     three
    a       NaN  1.138469 -2.400634
    c       NaN  0.025653 -1.386071
    e  0.863937  0.252462  1.500571
    f  1.053202 -2.338595 -0.374279
    h  1.053202 -1.157886 -0.551865

    [5 rows x 3 columns]

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 [28]: df
    Out[28]:
       one       two     three
    a  NaN  1.138469 -2.400634
    c  NaN  0.025653 -1.386071
    e  NaN       NaN       NaN
    f  NaN       NaN       NaN
    h  NaN -1.157886 -0.551865

    [5 rows x 3 columns]

    In [29]: df.fillna(method='pad', limit=1)
    Out[29]:
       one       two     three
    a  NaN  1.138469 -2.400634
    c  NaN  0.025653 -1.386071
    e  NaN  0.025653 -1.386071
    f  NaN       NaN       NaN
    h  NaN -1.157886 -0.551865

    [5 rows x 3 columns]

To remind you, these are the available filling methods:

Method | Action
--- | ---
pad / ffill | Fill values forward
bfill / backfill | Fill values backward

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

New in version 0.12.

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

    In [31]: dff.iloc[3:5,0] = np.nan

    In [32]: dff.iloc[4:6,1] = np.nan

    In [33]: dff.iloc[5:8,2] = np.nan

    In [34]: dff
    Out[34]:
              A         B         C
    0  1.592673  1.559318  1.562443
    1  0.763264  0.162027 -0.902704
    2  1.106010 -0.199234  0.458265
    3       NaN  0.128594  1.147862
    4       NaN       NaN -2.417312
    5  0.972827       NaN       NaN
    6  0.086926 -0.445645       NaN
    7 -1.420361 -0.015601       NaN
    8 -0.798334 -0.557697  0.381353
    9  1.337122 -1.531095  1.331458

    [10 rows x 3 columns]

    In [35]: dff.fillna(dff.mean())
    Out[35]:
              A         B         C
    0  1.592673  1.559318  1.562443
    1  0.763264  0.162027 -0.902704
    2  1.106010 -0.199234  0.458265
    3  0.455016  0.128594  1.147862
    4  0.455016 -0.112417 -2.417312
    5  0.972827 -0.112417  0.223052
    6  0.086926 -0.445645  0.223052
    7 -1.420361 -0.015601  0.223052
    8 -0.798334 -0.557697  0.381353
    9  1.337122 -1.531095  1.331458

    [10 rows x 3 columns]

    In [36]: dff.fillna(dff.mean()['B':'C'])
    Out[36]:
              A         B         C
    0  1.592673  1.559318  1.562443
    1  0.763264  0.162027 -0.902704
    2  1.106010 -0.199234  0.458265
    3       NaN  0.128594  1.147862
    4       NaN -0.112417 -2.417312
    5  0.972827 -0.112417  0.223052
    6  0.086926 -0.445645  0.223052
    7 -1.420361 -0.015601  0.223052
    8 -0.798334 -0.557697  0.381353
    9  1.337122 -1.531095  1.331458

    [10 rows x 3 columns]

New in version 0.13.

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

    In [37]: dff.where(notnull(dff),dff.mean(),axis='columns')
    Out[37]:
              A         B         C
    0  1.592673  1.559318  1.562443
    1  0.763264  0.162027 -0.902704
    2  1.106010 -0.199234  0.458265
    3  0.455016  0.128594  1.147862
    4  0.455016 -0.112417 -2.417312
    5  0.972827 -0.112417  0.223052
    6  0.086926 -0.445645  0.223052
    7 -1.420361 -0.015601  0.223052
    8 -0.798334 -0.557697  0.381353
    9  1.337122 -1.531095  1.331458

    [10 rows x 3 columns]

### 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 [38]: df
    Out[38]:
       one       two     three
    a  NaN  1.138469 -2.400634
    c  NaN  0.025653 -1.386071
    e  NaN  0.000000  0.000000
    f  NaN  0.000000  0.000000
    h  NaN -1.157886 -0.551865

    [5 rows x 3 columns]

    In [39]: df.dropna(axis=0)
    Out[39]:
    Empty DataFrame
    Columns: [one, two, three]
    Index: []

    [0 rows x 3 columns]

    In [40]: df.dropna(axis=1)
    Out[40]:
            two     three
    a  1.138469 -2.400634
    c  0.025653 -1.386071
    e  0.000000  0.000000
    f  0.000000  0.000000
    h -1.157886 -0.551865

    [5 rows x 2 columns]

    In [41]: df['one'].dropna()
    Out[41]: Series([], name: one, dtype: float64)

dropna is presently only implemented for Series and DataFrame, but will be eventually added to Panel. Series.dropna is a simpler method as it only has one axis to consider. DataFrame.dropna has considerably more options, which can be examined in the API.

### Interpolation

New in version 0.13.0.

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

    In [42]: ts
    Out[42]:
    2000-01-31    0.469112
    2000-02-29         NaN
    2000-03-31         NaN
    2000-04-28         NaN
    2000-05-31         NaN
    ...
    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, Length: 100

    In [43]: ts.count()
    Out[43]: 61

    In [44]: ts.interpolate().count()
    Out[44]: 100

    In [45]: plt.figure()
    Out[45]: <matplotlib.figure.Figure at 0xd562bd0>

    In [46]: ts.interpolate().plot()
    Out[46]: <matplotlib.axes.AxesSubplot at 0xee7e850>

Index aware interpolation is available via the `method` keyword:

    In [47]: ts2
    Out[47]:
    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 [48]: ts2.interpolate()
    Out[48]:
    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 [49]: ts2.interpolate(method='time')
    Out[49]:
    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 [50]: ser
    Out[50]:
    0      0
    1    NaN
    10    10
    dtype: float64

    In [51]: ser.interpolate()
    Out[51]:
    0      0
    1      5
    10    10
    dtype: int64

    In [52]: ser.interpolate(method='values')
    Out[52]:
    0      0
    1      1
    10    10
    dtype: int64

You can also interpolate with a DataFrame:

    In [53]: df = 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 [54]: df
    Out[54]:
         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

    [6 rows x 2 columns]

    In [55]: df.interpolate()
    Out[55]:
         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

    [6 rows x 2 columns]

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. For example, 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.

Warning

These methods require `scipy`.

    In [56]: df.interpolate(method='barycentric')
    Out[56]:
          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

    [6 rows x 2 columns]

    In [57]: df.interpolate(method='pchip')
    Out[57]:
              A          B
    0  1.000000   0.250000
    1  2.100000   1.130135
    2  3.429309   2.337586
    3  4.700000   4.000000
    4  5.600000  12.200000
    5  6.800000  14.400000

    [6 rows x 2 columns]

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

    In [58]: df.interpolate(method='spline', order=2)
    Out[58]:
              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

    [6 rows x 2 columns]

    In [59]: df.interpolate(method='polynomial', order=2)
    Out[59]:
              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

    [6 rows x 2 columns]

Compare several methods:

    In [60]: np.random.seed(2)

    In [61]: ser = Series(np.arange(1, 10.1, .25)**2 + np.random.randn(37))

    In [62]: bad = np.array([4, 13, 14, 15, 16, 17, 18, 20, 29])

    In [63]: ser[bad] = np.nan

    In [64]: methods = ['linear', 'quadratic', 'cubic']

    In [65]: df = DataFrame({m: ser.interpolate(method=m) for m in methods})

    In [66]: plt.figure()
    Out[66]: <matplotlib.figure.Figure at 0xad08d10>

    In [67]: df.plot()
    Out[67]: <matplotlib.axes.AxesSubplot at 0xae1a210>

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 [68]: ser = Series(np.sort(np.random.uniform(size=100)))

    # interpolate at new_index
    In [69]: new_index = ser.index + Index([49.25, 49.5, 49.75, 50.25, 50.5, 50.75])

    In [70]: interp_s = ser.reindex(new_index).interpolate(method='pchip')

    In [71]: interp_s[49:51]
    Out[71]:
    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

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

    In [72]: ser = Series([1, 3, np.nan, np.nan, np.nan, 11])

    In [73]: ser.interpolate(limit=2)
    Out[73]:
    0     1
    1     3
    2     5
    3     7
    4   NaN
    5    11
    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 [74]: ser = Series([0., 1., 2., 3., 4.])

    In [75]: ser.replace(0, 5)
    Out[75]:
    0    5
    1    1
    2    2
    3    3
    4    4
    dtype: float64

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

    In [76]: ser.replace([0, 1, 2, 3, 4], [4, 3, 2, 1, 0])
    Out[76]:
    0    4
    1    3
    2    2
    3    1
    4    0
    dtype: float64

You can also specify a mapping dict:

    In [77]: ser.replace({0: 10, 1: 100})
    Out[77]:
    0     10
    1    100
    2      2
    3      3
    4      4
    dtype: float64

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

    In [78]: df = DataFrame({'a': [0, 1, 2, 3, 4], 'b': [5, 6, 7, 8, 9]})

    In [79]: df.replace({'a': 0, 'b': 5}, 100)
    Out[79]:
         a    b
    0  100  100
    1    1    6
    2    2    7
    3    3    8
    4    4    9

    [5 rows x 2 columns]

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

    In [80]: ser.replace([1, 2, 3], method='pad')
    Out[80]:
    0    0
    1    0
    2    0
    3    0
    4    4
    dtype: float64

### String/Regular Expression Replacement

Note

Python strings prefixed with the `r` character such as `r'hello world'` are so-called “raw” strings. They have different semantics regarding backslashes than strings without this prefix. Backslashes in raw strings will be interpreted as an escaped backslash, e.g., `r'\' == '\\'`. You should read about them if this is unclear.

Replace the ‘.’ with `nan` (str -> str)

    In [81]: d = {'a': list(range(4)), 'b': list('ab..'), 'c': ['a', 'b', nan, 'd']}

    In [82]: df = DataFrame(d)

    In [83]: df.replace('.', nan)
    Out[83]:
       a    b    c
    0  0    a    a
    1  1    b    b
    2  2  NaN  NaN
    3  3  NaN    d

    [4 rows x 3 columns]

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

    In [84]: df.replace(r'\s*\.\s*', nan, regex=True)
    Out[84]:
       a    b    c
    0  0    a    a
    1  1    b    b
    2  2  NaN  NaN
    3  3  NaN    d

    [4 rows x 3 columns]

Replace a few different values (list -> list)

    In [85]: df.replace(['a', '.'], ['b', nan])
    Out[85]:
       a    b    c
    0  0    b    b
    1  1    b    b
    2  2  NaN  NaN
    3  3  NaN    d

    [4 rows x 3 columns]

list of regex -> list of regex

    In [86]: df.replace([r'\.', r'(a)'], ['dot', '\1stuff'], regex=True)
    Out[86]:
       a       b       c
    0  0  stuff  stuff
    1  1       b       b
    2  2     dot     NaN
    3  3     dot       d

    [4 rows x 3 columns]

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

    In [87]: df.replace({'b': '.'}, {'b': nan})
    Out[87]:
       a    b    c
    0  0    a    a
    1  1    b    b
    2  2  NaN  NaN
    3  3  NaN    d

    [4 rows x 3 columns]

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

    In [88]: df.replace({'b': r'\s*\.\s*'}, {'b': nan}, regex=True)
    Out[88]:
       a    b    c
    0  0    a    a
    1  1    b    b
    2  2  NaN  NaN
    3  3  NaN    d

    [4 rows x 3 columns]

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

    In [89]: df.replace({'b': {'b': r''}}, regex=True)
    Out[89]:
       a  b    c
    0  0  a    a
    1  1       b
    2  2  .  NaN
    3  3  .    d

    [4 rows x 3 columns]

or you can pass the nested dictionary like so

    In [90]: df.replace(regex={'b': {r'\s*\.\s*': nan}})
    Out[90]:
       a    b    c
    0  0    a    a
    1  1    b    b
    2  2  NaN  NaN
    3  3  NaN    d

    [4 rows x 3 columns]

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 [91]: df.replace({'b': r'\s*(\.)\s*'}, {'b': r'\1ty'}, regex=True)
    Out[91]:
       a    b    c
    0  0    a    a
    1  1    b    b
    2  2  .ty  NaN
    3  3  .ty    d

    [4 rows x 3 columns]

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

    In [92]: df.replace([r'\s*\.\s*', r'a|b'], nan, regex=True)
    Out[92]:
       a   b    c
    0  0 NaN  NaN
    1  1 NaN  NaN
    2  2 NaN  NaN
    3  3 NaN    d

    [4 rows x 3 columns]

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 explicity by name or `regex` must be a nested dictionary. The previous example, in this case, would then be

    In [93]: df.replace(regex=[r'\s*\.\s*', r'a|b'], value=nan)
    Out[93]:
       a   b    c
    0  0 NaN  NaN
    1  1 NaN  NaN
    2  2 NaN  NaN
    3  3 NaN    d

    [4 rows x 3 columns]

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

Note

Anywhere in the above `replace` examples that you see a regular expression a compiled regular expression is valid as well.

### Numeric Replacement

Similiar to `DataFrame.fillna`

    In [94]: df = DataFrame(randn(10, 2))

    In [95]: df[rand(df.shape[0]) > 0.5] = 1.5

    In [96]: df.replace(1.5, nan)
    Out[96]:
              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

    [10 rows x 2 columns]

Replacing more than one value via lists works as well

    In [97]: df00 = df.values[0, 0]

    In [98]: df.replace([1.5, df00], [nan, 'a'])
    Out[98]:
               0         1
    0          a -1.021415
    1  0.4323957 -0.323580
    2  0.4238247  0.799180
    3   1.262614  0.751965
    4        NaN       NaN
    5        NaN       NaN
    6 -0.4981742 -1.060799
    7  0.5916665 -0.183257
    8   1.019855 -1.482465
    9        NaN       NaN

    [10 rows x 2 columns]

    In [99]: df[1].dtype
    Out[99]: dtype('float64')

You can also operate on the DataFrame in place

    In [100]: df.replace(1.5, nan, inplace=True)

## 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:

data type | Cast to
--- | ---
integer | float
boolean | object
float | no cast
object | no cast

For example:

    In [101]: s = Series(randn(5), index=[0, 2, 4, 6, 7])

    In [102]: s > 0
    Out[102]:
    0    True
    2    True
    4    True
    6    True
    7    True
    dtype: bool

    In [103]: (s > 0).dtype
    Out[103]: dtype('bool')

    In [104]: crit = (s > 0).reindex(list(range(8)))

    In [105]: crit
    Out[105]:
    0    True
    1     NaN
    2    True
    3     NaN
    4    True
    5     NaN
    6    True
    7    True
    dtype: object

    In [106]: crit.dtype
    Out[106]: 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 [107]: reindexed = s.reindex(list(range(8))).fillna(0)

    In [108]: reindexed[crit]
    ---------------------------------------------------------------------------
    ValueError                                Traceback (most recent call last)
    <ipython-input-108-2da204ed1ac7> in <module>()
    ----> 1 reindexed[crit]

    /home/user1/src/pandas/pandas/core/series.pyc in __getitem__(self, key)
        517             key = list(key)
        518
    --> 519         if _is_bool_indexer(key):
        520             key = _check_bool_indexer(self.index, key)
        521

    /home/user1/src/pandas/pandas/core/common.pyc in _is_bool_indexer(key)
       1747             if not lib.is_bool_array(key):
       1748                 if isnull(key).any():
    -> 1749                     raise ValueError('cannot index with vector containing '
       1750                                      'NA / NaN values')
       1751                 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 [109]: reindexed[crit.fillna(False)]
    Out[109]:
    0    0.126504
    2    0.696198
    4    0.697416
    6    0.601516
    7    0.003659
    dtype: float64

    In [110]: reindexed[crit.fillna(True)]
    Out[110]:
    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

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