Working with missing data
In this section, we will discuss missing (also referred to as NA) values inpandas.
Note
The choice of using NaN
internally to denote missing data was largelyfor simplicity and performance reasons. It differs from the MaskedArrayapproach of, for example, scikits.timeseries
. We are hopeful thatNumPy 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.
Values considered “missing”
As data comes in many shapes and forms, pandas aims to be flexible with regardto handling missing data. While NaN
is the default missing value marker forreasons of computational speed and convenience, we need to be able to easilydetect this value with data of different types: floating point, integer,boolean, and general object. In many cases, however, the Python None
willarise and we wish to also consider that “missing” or “not available” or “NA”.
Note
If you want to consider inf
and -inf
to be “NA” in computations,you can set pandas.options.mode.use_inf_as_na = True
.
- 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
To make detecting missing values easier (and across different array dtypes),pandas provides the isna()
andnotna()
functions, which are also methods onSeries 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.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
- In [10]: df2.isna()
- 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
Warning
One has to be mindful that in Python (and NumPy), the nan's
don’t compare equal, but None's
do.Note that pandas/NumPy uses the fact that np.nan != np.nan
, and treats None
like np.nan
.
- In [11]: None == None # noqa: E711
- Out[11]: True
- In [12]: np.nan == np.nan
- Out[12]: False
So as compared to above, a scalar equality comparison versus a None/np.nan
doesn’t provide useful information.
- 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
Integer dtypes and missing data
Because NaN
is a float, a column of integers with even one missing valuesis cast to floating-point dtype (see Support for integer NA for more). Pandasprovides a nullable integer array, which can be used by explicitly requestingthe dtype:
- In [14]: pd.Series([1, 2, np.nan, 4], dtype=pd.Int64Dtype())
- Out[14]:
- 0 1
- 1 2
- 2 NaN
- 3 4
- dtype: Int64
Alternatively, the string alias dtype='Int64'
(note the capital "I"
) can beused.
See Nullable integer data type for more.
Datetimes
For datetime64[ns] types, NaT
represents missing values. This is a pseudo-nativesentinel value that can be represented by NumPy in a singular dtype (datetime64[ns]).pandas objects provide compatibility between NaT
and NaN
.
- 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
- object 1
- bool 1
- dtype: int64
Inserting missing data
You can insert missing values by simply assigning to containers. Theactual missing value used will be chosen based on the dtype.
For example, numeric containers will always use NaN
regardless ofthe missing value type chosen:
- 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
Likewise, datetime containers will always use NaT
.
For object containers, pandas will use the value given:
- 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
Calculations with missing data
Missing values propagate naturally through arithmetic operations between pandasobjects.
- 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
The descriptive statistics and computational methods discussed in thedata structure overview (and listed here and here) are all written toaccount 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 0.
- Cumulative methods like
cumsum()
andcumprod()
ignore NA values by default, but preserve them in the resulting arrays. To override this behaviour and include NA values, useskipna=False
.
- 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
- 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
Sum/prod of empties/nans
Warning
This behavior is now standard as of v0.22.0 and is consistent with the default in numpy
; previously sum/prod of all-NA or empty Series/DataFrames would return NaN.See v0.22.0 whatsnew for more.
The sum of an empty or all-NA Series or column of a DataFrame is 0.
- In [36]: pd.Series([np.nan]).sum()
- Out[36]: 0.0
- In [37]: pd.Series([]).sum()
- Out[37]: 0.0
The product of an empty or all-NA Series or column of a DataFrame is 1.
- In [38]: pd.Series([np.nan]).prod()
- Out[38]: 1.0
- In [39]: pd.Series([]).prod()
- Out[39]: 1.0
NA values in GroupBy
NA groups in GroupBy are automatically excluded. This behavior is consistentwith R, for example:
- In [40]: df
- 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 NaN -0.706771 -1.039575
- In [41]: df.groupby('one').mean()
- Out[41]:
- 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 dealingwith missing data.
Filling missing values: fillna
fillna()
can “fill in” NA values with non-NA data in a coupleof ways, which we illustrate:
Replace NA with a scalar value
- 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
- In [44]: df2['one'].fillna('missing')
- Out[44]:
- a missing
- c missing
- e 0.119209
- f -2.10457
- h missing
- Name: one, dtype: object
Fill gaps forward or backward
Using the same filling arguments as reindexing, wecan propagate non-NA values forward or backward:
- 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
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 [47]: df
- Out[47]:
- 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 [48]: df.fillna(method='pad', limit=1)
- Out[48]:
- 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:
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 “lastknown value” is available at every time point.
ffill()
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 Seriesmust match the columns of the frame you wish to fill. Theuse case of this is to fill a DataFrame with the mean of that column.
- In [49]: dff = pd.DataFrame(np.random.randn(10, 3), columns=list('ABC'))
- In [50]: dff.iloc[3:5, 0] = np.nan
- In [51]: dff.iloc[4:6, 1] = np.nan
- In [52]: dff.iloc[5:8, 2] = np.nan
- 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
Same result as above, but is aligning the ‘fill’ value which isa Series in this case.
- In [56]: dff.where(pd.notna(dff), dff.mean(), axis='columns')
- Out[56]:
- 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 missingdata. To do this, use dropna()
:
- 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)
An equivalent dropna()
is available for Series.DataFrame.dropna has considerably more options than Series.dropna, which can beexamined in the API.
Interpolation
New in version 0.23.0: The limit_area
keyword argument was added.
Both Series and DataFrame objects have interpolate()
that, by default, performs linear interpolation at missing data points.
- 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 [62]: ts.count()
- Out[62]: 66
- In [63]: ts.plot()
- Out[63]: <matplotlib.axes._subplots.AxesSubplot at 0x7f450959b710>
- 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
- In [65]: ts.interpolate().count()
- Out[65]: 100
- In [66]: ts.interpolate().plot()
- Out[66]: <matplotlib.axes._subplots.AxesSubplot at 0x7f450953b690>
Index aware interpolation is available via the method
keyword:
- 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
For a floating-point index, use method='values'
:
- 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
You can also interpolate with a DataFrame:
- 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
The method
argument gives access to fancier interpolation methods.If you have scipy installed, you can 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, consider
method='akima'
.
Warning
These methods require scipy
.
- In [76]: df.interpolate(method='barycentric')
- Out[76]:
- 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 [77]: df.interpolate(method='pchip')
- Out[77]:
- 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 [78]: df.interpolate(method='akima')
- Out[78]:
- 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 specifythe degree or order of the approximation:
- In [79]: df.interpolate(method='spline', order=2)
- Out[79]:
- 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 [80]: df.interpolate(method='polynomial', order=2)
- Out[80]:
- A B
- 0 1.000000 0.250000
- 1 2.100000 -2.703846
- 2 3.451351 -1.453846
- 3 4.700000 4.000000
- 4 5.600000 12.200000
- 5 6.800000 14.400000
Compare several methods:
- In [81]: np.random.seed(2)
- In [82]: ser = pd.Series(np.arange(1, 10.1, .25) ** 2 + np.random.randn(37))
- In [83]: missing = np.array([4, 13, 14, 15, 16, 17, 18, 20, 29])
- In [84]: ser[missing] = np.nan
- In [85]: methods = ['linear', 'quadratic', 'cubic']
- In [86]: df = pd.DataFrame({m: ser.interpolate(method=m) for m in methods})
- In [87]: df.plot()
- Out[87]: <matplotlib.axes._subplots.AxesSubplot at 0x7f450951ac10>
Another use case is interpolation at new values.Suppose you have 100 observations from some distribution. And let’s supposethat you’re particularly interested in what’s happening around the middle.You can mix pandas’ reindex
and interpolate
methods to interpolateat the new values.
- In [88]: ser = pd.Series(np.sort(np.random.uniform(size=100)))
- # interpolate at new_index
- In [89]: new_index = ser.index | pd.Index([49.25, 49.5, 49.75, 50.25, 50.5, 50.75])
- In [90]: interp_s = ser.reindex(new_index).interpolate(method='pchip')
- In [91]: interp_s[49:51]
- Out[91]:
- 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
keywordargument. Use this argument to limit the number of consecutive NaN
valuesfilled since the last valid observation:
- In [92]: ser = pd.Series([np.nan, np.nan, 5, np.nan, np.nan,
- ....: np.nan, 13, np.nan, np.nan])
- ....:
- In [93]: ser
- Out[93]:
- 0 NaN
- 1 NaN
- 2 5.0
- 3 NaN
- 4 NaN
- 5 NaN
- 6 13.0
- 7 NaN
- 8 NaN
- dtype: float64
- # fill all consecutive values in a forward direction
- In [94]: ser.interpolate()
- Out[94]:
- 0 NaN
- 1 NaN
- 2 5.0
- 3 7.0
- 4 9.0
- 5 11.0
- 6 13.0
- 7 13.0
- 8 13.0
- dtype: float64
- # fill one consecutive value in a forward direction
- 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
By default, NaN
values are filled in a forward
direction. Uselimit_direction
parameter to fill backward
or from both
directions.
- # fill one consecutive value backwards
- In [96]: ser.interpolate(limit=1, limit_direction='backward')
- Out[96]:
- 0 NaN
- 1 5.0
- 2 5.0
- 3 NaN
- 4 NaN
- 5 11.0
- 6 13.0
- 7 NaN
- 8 NaN
- dtype: float64
- # fill one consecutive value in both directions
- In [97]: ser.interpolate(limit=1, limit_direction='both')
- Out[97]:
- 0 NaN
- 1 5.0
- 2 5.0
- 3 7.0
- 4 NaN
- 5 11.0
- 6 13.0
- 7 13.0
- 8 NaN
- dtype: float64
- # fill all consecutive values in both directions
- In [98]: ser.interpolate(limit_direction='both')
- Out[98]:
- 0 5.0
- 1 5.0
- 2 5.0
- 3 7.0
- 4 9.0
- 5 11.0
- 6 13.0
- 7 13.0
- 8 13.0
- dtype: float64
By default, NaN
values are filled whether they are inside (surrounded by)existing valid values, or outside existing valid values. Introduced in v0.23the limit_area
parameter restricts filling to either inside or outside values.
- # fill one consecutive inside value in both directions
- In [99]: ser.interpolate(limit_direction='both', limit_area='inside', limit=1)
- Out[99]:
- 0 NaN
- 1 NaN
- 2 5.0
- 3 7.0
- 4 NaN
- 5 11.0
- 6 13.0
- 7 NaN
- 8 NaN
- dtype: float64
- # fill all consecutive outside values backward
- In [100]: ser.interpolate(limit_direction='backward', limit_area='outside')
- Out[100]:
- 0 5.0
- 1 5.0
- 2 5.0
- 3 NaN
- 4 NaN
- 5 NaN
- 6 13.0
- 7 NaN
- 8 NaN
- dtype: float64
- # fill all consecutive outside values in both directions
- In [101]: ser.interpolate(limit_direction='both', limit_area='outside')
- Out[101]:
- 0 5.0
- 1 5.0
- 2 5.0
- 3 NaN
- 4 NaN
- 5 NaN
- 6 13.0
- 7 13.0
- 8 13.0
- dtype: float64
Replacing generic values
Often times we want to replace arbitrary values with other values.
replace()
in Series and replace()
in DataFrame provides an efficient yetflexible way to perform such replacements.
For a Series, you can replace a single value or a list of values by anothervalue:
- 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
You can replace a list of values by a list of other values:
- 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 also specify a mapping dict:
- In [105]: ser.replace({0: 10, 1: 100})
- Out[105]:
- 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 [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
Instead of replacing with specified values, you can treat all given values asmissing and interpolate over them:
- 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
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 regardingbackslashes than strings without this prefix. Backslashes in raw stringswill be interpreted as an escaped backslash, e.g., r'\' == '\'
. Youshould read about themif this is unclear.
Replace the ‘.’ with NaN
(str -> str):
- In [109]: d = {'a': list(range(4)), 'b': list('ab..'), 'c': ['a', 'b', np.nan, 'd']}
- In [110]: df = pd.DataFrame(d)
- In [111]: df.replace('.', np.nan)
- Out[111]:
- 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 [112]: df.replace(r'\s*\.\s*', np.nan, regex=True)
- Out[112]:
- 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 [113]: df.replace(['a', '.'], ['b', np.nan])
- Out[113]:
- 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 [114]: df.replace([r'\.', r'(a)'], ['dot', r'\1stuff'], regex=True)
- Out[114]:
- a b c
- 0 0 astuff astuff
- 1 1 b b
- 2 2 dot NaN
- 3 3 dot d
Only search in column 'b'
(dict -> dict):
- In [115]: df.replace({'b': '.'}, {'b': np.nan})
- Out[115]:
- 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 forsearching instead (dict of regex -> dict):
- In [116]: df.replace({'b': r'\s*\.\s*'}, {'b': np.nan}, regex=True)
- Out[116]:
- 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 [117]: df.replace({'b': {'b': r''}}, regex=True)
- Out[117]:
- a b c
- 0 0 a a
- 1 1 b
- 2 2 . NaN
- 3 3 . d
Alternatively, you can pass the nested dictionary like so:
- In [118]: df.replace(regex={'b': {r'\s*\.\s*': np.nan}})
- Out[118]:
- 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 (dictof regex -> dict of regex), this works for lists as well.
- In [119]: df.replace({'b': r'\s*(\.)\s*'}, {'b': r'\1ty'}, regex=True)
- Out[119]:
- 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 matchwill be replaced with a scalar (list of regex -> regex).
- In [120]: df.replace([r'\s*\.\s*', r'a|b'], np.nan, regex=True)
- Out[120]:
- 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 theto_replace
argument as the regex
argument. In this case the value
argument must be passed explicitly by name or regex
must be a nesteddictionary. The previous example, in this case, would then be:
- In [121]: df.replace(regex=[r'\s*\.\s*', r'a|b'], value=np.nan)
- Out[121]:
- 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 youwant to use a regular expression.
Note
Anywhere in the above replace
examples that you see a regular expressiona compiled regular expression is valid as well.
Numeric replacement
replace()
is similar to fillna()
.
- In [122]: df = pd.DataFrame(np.random.randn(10, 2))
- In [123]: df[np.random.rand(df.shape[0]) > 0.5] = 1.5
- In [124]: df.replace(1.5, np.nan)
- Out[124]:
- 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 is possible by passing a list.
- In [125]: df00 = df.iloc[0, 0]
- In [126]: df.replace([1.5, df00], [np.nan, 'a'])
- Out[126]:
- 0 1
- 0 a -1.02141
- 1 0.432396 -0.32358
- 2 0.423825 0.79918
- 3 1.26261 0.751965
- 4 NaN NaN
- 5 NaN NaN
- 6 -0.498174 -1.0608
- 7 0.591667 -0.183257
- 8 1.01985 -1.48247
- 9 NaN NaN
- In [127]: df[1].dtype
- Out[127]: dtype('float64')
You can also operate on the DataFrame in place:
- In [128]: df.replace(1.5, np.nan, inplace=True)
Warning
When replacing multiple bool
or datetime64
objects, the firstargument to replace
(to_replace
) must match the type of the valuebeing replaced. For example,
- >>> s = pd.Series([True, False, True])
- >>> s.replace({'a string': 'new value', True: False}) # raises
- TypeError: Cannot compare types 'ndarray(dtype=bool)' and 'str'
will raise a TypeError
because one of the dict
keys is not of thecorrect type for replacement.
However, when replacing a single object such as,
- In [129]: s = pd.Series([True, False, True])
- In [130]: s.replace('a string', 'another string')
- Out[130]:
- 0 True
- 1 False
- 2 True
- dtype: bool
the original NDFrame
object will be returned untouched. We’re working onunifying this API, but for backwards compatibility reasons we cannot breakthe latter behavior. See GH6354 for more details.
Missing data casting rules and indexing
While pandas supports storing arrays of integer and boolean type, these typesare not capable of storing missing data. Until we can switch to using a nativeNA type in NumPy, we’ve established some “casting rules”. When a reindexingoperation introduces missing data, the Series will be cast according to therules introduced in the table below.
data type | Cast to |
---|---|
integer | float |
boolean | object |
float | no cast |
object | no cast |
For example:
- In [131]: s = pd.Series(np.random.randn(5), index=[0, 2, 4, 6, 7])
- In [132]: s > 0
- Out[132]:
- 0 True
- 2 True
- 4 True
- 6 True
- 7 True
- dtype: bool
- In [133]: (s > 0).dtype
- Out[133]: dtype('bool')
- In [134]: crit = (s > 0).reindex(list(range(8)))
- In [135]: crit
- Out[135]:
- 0 True
- 1 NaN
- 2 True
- 3 NaN
- 4 True
- 5 NaN
- 6 True
- 7 True
- dtype: object
- In [136]: crit.dtype
- Out[136]: dtype('O')
Ordinarily NumPy will complain if you try to use an object array (even if itcontains boolean values) instead of a boolean array to get or set values froman ndarray (e.g. selecting values based on some criteria). If a boolean vectorcontains NAs, an exception will be generated:
- In [137]: reindexed = s.reindex(list(range(8))).fillna(0)
- In [138]: reindexed[crit]
- ---------------------------------------------------------------------------
- ValueError Traceback (most recent call last)
- <ipython-input-138-0dac417a4890> in <module>
- ----> 1 reindexed[crit]
- /pandas/pandas/core/series.py in __getitem__(self, key)
- 1108 key = list(key)
- 1109
- -> 1110 if com.is_bool_indexer(key):
- 1111 key = check_bool_indexer(self.index, key)
- 1112
- /pandas/pandas/core/common.py in is_bool_indexer(key)
- 128 if not lib.is_bool_array(key):
- 129 if isna(key).any():
- --> 130 raise ValueError(na_msg)
- 131 return False
- 132 return True
- ValueError: cannot index with vector containing NA / NaN values
However, these can be filled in using fillna()
and it will work fine:
- In [139]: reindexed[crit.fillna(False)]
- Out[139]:
- 0 0.126504
- 2 0.696198
- 4 0.697416
- 6 0.601516
- 7 0.003659
- dtype: float64
- In [140]: reindexed[crit.fillna(True)]
- Out[140]:
- 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
Pandas provides a nullable integer dtype, but you must explicitly request itwhen creating the series or column. Notice that we use a capital “I” inthe dtype="Int64"
.
- In [141]: s = pd.Series([0, 1, np.nan, 3, 4], dtype="Int64")
- In [142]: s
- Out[142]:
- 0 0
- 1 1
- 2 NaN
- 3 3
- 4 4
- dtype: Int64
See Nullable integer data type for more.