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pandas.Dataframe.interpolate() does not extrapolate even if it is asked to, depending on interpolation method #31949

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typorian opened this issue Feb 13, 2020 · 16 comments
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API - Consistency Internal Consistency of API/Behavior Docs Missing-data np.nan, pd.NaT, pd.NA, dropna, isnull, interpolate

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@typorian
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typorian commented Feb 13, 2020

Code Sample, a copy-pastable example if possible

import pandas as pd
import numpy as np

a = pd.Series([0, 1, np.nan, 3, 4, np.nan, np.nan, np.nan, np.nan])
a_int=a.interpolate(method='cubic', limit_area=None)

Problem description

Some of the offered methods (it seems all of them that are provided by interp1d) are unable to extrapolate over np.nan. However, the limit_area switch for df.interpolate() indicates you can force extrapolation. A combination of limit_area=None and an incompatible method should raise a warning.

There used to be a similar issue where extrapolation over trailing NaN was done unintentionally, so maybe the fix for that overdid it. #8000

Expected Output

Extrapolation over the NaNs in the array is expected. Using a different method, such as pchip achieves this.

Output of pd.show_versions()

[paste the output of pd.show_versions() here below this line]
INSTALLED VERSIONS

commit : None
python : 3.7.2.final.0
python-bits : 64
OS : Windows
OS-release : 10
machine : AMD64
processor : Intel64 Family 6 Model 63 Stepping 2, GenuineIntel
byteorder : little
LC_ALL : None
LANG : en
LOCALE : None.None

pandas : 0.25.3 (also tested with 1.0.0)
numpy : 1.15.4
pytz : 2018.9
dateutil : 2.7.5
pip : 20.0.2
setuptools : 41.0.1
Cython : 0.29.15
pytest : None
hypothesis : None
sphinx : 1.8.3
blosc : None
feather : None
xlsxwriter : None
lxml.etree : 4.3.3
html5lib : None
pymysql : None
psycopg2 : None
jinja2 : 2.10
IPython : 7.5.0
pandas_datareader: None
bs4 : None
bottleneck : None
fastparquet : None
gcsfs : None
lxml.etree : 4.3.3
matplotlib : 3.0.3
numexpr : None
odfpy : None
openpyxl : 2.5.12
pandas_gbq : None
pyarrow : None
pytables : None
s3fs : None
scipy : 1.2.1
sqlalchemy : None
tables : None
xarray : None
xlrd : 1.2.0
xlwt : None
xlsxwriter : None

@fercook
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fercook commented Apr 19, 2020

I second this.

Also, even when it works, it doesn't. The implied meaning of "extrapolate" is that it will continue on the last available trend. However, the observed result is that the last value is repeated.

In:

a = pd.Series([0, 1, np.nan, 3, 4, np.nan, np.nan, np.nan, np.nan])
a.interpolate(method='linear', limit_area=None)

Out:

0    0.0
1    1.0
2    2.0
3    3.0
4    4.0
5    4.0
6    4.0
7    4.0
8    4.0

@RealJTG
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RealJTG commented Apr 27, 2020

I also stumbled on this bug.

Also examples in current documentation are confusing - extrapolation mentioned there "fill NaNs outside valid values (extrapolate)" https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.interpolate.html:

df = pd.DataFrame([(0.0, np.nan, -1.0, 1.0),
                   (np.nan, 2.0, np.nan, np.nan),
                   (2.0, 3.0, np.nan, 9.0),
                   (np.nan, 4.0, -4.0, 16.0)],
                  columns=list('abcd'))
df
     a    b    c     d
0  0.0  NaN -1.0   1.0
1  NaN  2.0  NaN   NaN
2  2.0  3.0  NaN   9.0
3  NaN  4.0 -4.0  16.0

df.interpolate(method='linear', limit_direction='forward', axis=0)
     a    b    c     d
0  0.0  NaN -1.0   1.0
1  1.0  2.0 -2.0   5.0
2  2.0  3.0 -3.0   9.0
3  2.0  4.0 -4.0  16.0

... but this is not a linear extrapolation

     a
0  0.0
1  NaN
2  2.0
3  NaN

     a
0  0.0
1  1.0
2  2.0
3  2.0

@Dr-Irv Dr-Irv added Bug Docs Missing-data np.nan, pd.NaT, pd.NA, dropna, isnull, interpolate labels Oct 15, 2020
@Dr-Irv
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Dr-Irv commented Oct 15, 2020

Based on discussions in #8000 it seems we need an argument to specify that extrapolation at the beginning and end of the series can be specified. Alternatively, the docs could reflect that such extrapolation is not provided by interpolate

@Dr-Irv Dr-Irv added API - Consistency Internal Consistency of API/Behavior and removed Bug labels Oct 15, 2020
@flxai
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flxai commented Mar 1, 2021

@Dr-Irv Is this still on the project's roadmap?

@Dr-Irv
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Dr-Irv commented Mar 1, 2021

@Dr-Irv Is this still on the project's roadmap?

We are always open to a PR that would address this issue. For issues like this, we don't put them on a roadmap. We are just open to the community addressing them!

@flxai
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flxai commented Mar 1, 2021

@Dr-Irv Thanks for the explanation. Unfortunately I am not familiar enough with Pandas to see myself making a real contribution. But here is the solution I am currently working with:

def extrapolate_linear(s):
    s = s.copy()
    # Indices of not-nan values
    idx_nn = s.index[~s.isna()]
    
    # At least two data points needed for trend analysis
    assert len(idx_nn) >= 2
    
    # Outermost indices
    idx_l = idx_nn[0]
    idx_r = idx_nn[-1]
    
    # Indices left and right of outermost values
    idx_ll = s.index[s.index < idx_l]
    idx_rr = s.index[s.index > idx_r]
    
    # Derivative of not-nan indices / values
    v = s[idx_nn].diff()
    
    # Left- and right-most derivative values
    v_l = v[1]
    v_r = v[-1]
    f_l = idx_l - idx_nn[1]
    f_r = idx_nn[-2] - idx_r
    
    # Set values left / right of boundaries
    l_l = lambda idx: (idx_l - idx) / f_l * v_l + s[idx_l]
    l_r = lambda idx: (idx_r - idx) / f_r * v_r + s[idx_r]
    x_l = pd.Series(idx_ll).apply(l_l)
    x_l.index = idx_ll
    x_r = pd.Series(idx_rr).apply(l_r)
    x_r.index = idx_rr
    s[idx_ll] = x_l
    s[idx_rr] = x_r
    
    return s

Exemplary usage is documented in this notebook. Hopefully this is inspiration enough for more competent people to make a PR.

EDIT: I've vectorized the function.

@khaeru
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khaeru commented Jul 13, 2021

Based on discussions in #8000 it seems we need an argument to specify that extrapolation at the beginning and end of the series can be specified. Alternatively, the docs could reflect that such extrapolation is not provided by interpolate.

I don't think that is a correct interpretation. The pandas docs specify that **kwargs are "Keyword arguments to pass on to the interpolating function." They also link directly to the docs for one such interpolating function, scipy.interpolate.interp1d. These in turn specify that one of these keyword arguments is fill_value and:

  • If “extrapolate”, then points outside the data range will be extrapolated.
    New in version 0.17.0.

So the following should work, with the current method signature and no new argument:

import numpy as np
import pandas as pd

# A 1-D Series with missing external values
x = [0.5, 1, 2, 3, 20]
y = [np.NaN, 1, 4, 9, np.NaN]
s = pd.Series(y, index=x)

# Expected usage
kw = dict(method="quadratic", fill_value="extrapolate")
s.interpolate(**kw)

But this fails. A value is extrapolated on the 'forward' end of s, but no value is extrapolated for x=0.5 on the 'backward' end:

0.5       NaN
1.0       1.0
2.0       4.0
3.0       9.0
20.0    400.0
dtype: float64

The code simply does not do what is advertised by the docs, so this is clearly (IMO) a bug. (Also a surprising one, since scipy 0.17 was 5 years ago, and one would think that pandas' use of basic features in numpy and scipy was stable and tested.)

A slightly more compact workaround than @flxai's:

from scipy.interpolate import interp1d

# A mask indicating where `s` is not null
m = s.notna()

# Construct an interpolator from the non-null values
# NB 'kind' instead of 'method'!
kw = dict(kind="quadratic", fill_value="extrapolate")
f = interp1d(s[m].index, s[m], **kw) 

# Apply this to the indices of the nulls; reconstruct a series
s2 = pd.Series(f(s[~m].index), index=s[~m].index)

# Fill `s` using the values from `s2`
result = s.fillna(s2)

# Previous 3 statements combined:
result = s.fillna(
    pd.Series(
        interp1d(s[m].index, s[m], **kw)(s[~m].index),
        index=s[~m].index,
    )
)

This gives the expected result:

0.5       0.25
1.0       1.00
2.0       4.00
3.0       9.00
20.0    400.00
dtype: float64

@lyndonchan
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lyndonchan commented Jul 14, 2021

@khaeru thanks for your code - I think I found a more elegant solution. As you mentioned, the extrapolation only works in the forward direction, so you can just flip the series, apply another extrapolation, and flip it back again.

Using your example (just one extrapolation):

import numpy as np
import pandas as pd

# A 1-D Series with missing external values
x = [0.5, 1, 2, 3, 20]
y = [np.NaN, 1, 4, 9, np.NaN]
s = pd.Series(y, index=x)

# Expected usage
kw = dict(method="quadratic", fill_value="extrapolate")
s.interpolate(**kw)

gives us this:

0.5       NaN
1.0       1.0
2.0       4.0
3.0       9.0
20.0    400.0
dtype: float64

But with my solution:

s.interpolate(**kw).iloc[::-1].interpolate(**kw).iloc[::-1]

we get this:

0.5       0.25
1.0       1.00
2.0       4.00
3.0       9.00
20.0    400.00
dtype: float64

@khaeru
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khaeru commented Jul 14, 2021

Sure, that also works! I don't know pandas' internals, so I can't guess whether your workaround, mine, or some other would perform best on large(r) series. People should consider their particular use-cases, and check/test.

Since this is a bug, these are only workarounds. The real ‘solution’ will be a PR by someone who knows the internals well enough to make one.

@zhihua-zheng
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@khaeru @lyndonchan To extrapolate in both directions, use limit_direction="both", which is not obvious at all.

import numpy as np
import pandas as pd

# A 1-D Series with missing external values
x = [0.5, 1, 2, 3, 20]
y = [np.NaN, 1, 4, 9, np.NaN]
s = pd.Series(y, index=x)

# Expected usage
kw = dict(method="quadratic", fill_value="extrapolate", limit_direction="both")
s.interpolate(**kw)

This gives:

0.5       0.25
1.0       1.00
2.0       4.00
3.0       9.00
20.0    400.00
dtype: float64

@valschmidt
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The method @zhihua-zheng provides, unfortunately does not work properly for method="time". Specifying the 'time' method will result in the first non-NaN value repeated to the beginning and the last non-NaN value repeated to the end, similar to what was shown by @RealJTG.

Here's an example:

import datetime
import pandas as pd

start=datetime.datetime.now()
dt = datetime.timedelta(1)

x = [start,
     start+dt,
     start+2*dt,
     start+3*dt,
     start+4*dt]
y = [np.NaN,2.0,np.NaN,3.0,np.NaN]

s = pd.Series(y,index=x)
kw = dict(method='time',limit_direction='both',fill_value="extrapolate")
s2 = s.interpolate(**kw)
print(s2)

2023-07-30 03:10:46.806929    2.0
2023-07-31 03:10:46.806929    2.0
2023-08-01 03:10:46.806929    2.5
2023-08-02 03:10:46.806929    3.0
2023-08-03 03:10:46.806929    3.0
dtype: float64

@kmuehlbauer
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kmuehlbauer commented Sep 15, 2023

Late to the party here, but we've experienced similar issues over at xarray.

One very obvious problem is that fill_value is broken for all interpolation schemes which use numpy.interp:

if method in NP_METHODS:
# np.interp requires sorted X values, #21037
indexer = np.argsort(indices[valid])
yvalues[invalid] = np.interp(
indices[invalid], indices[valid][indexer], yvalues[valid][indexer]
)

Although numpy.interp doesn't have a notion of fill_value, it has similar usable kwargs left and right. Unfortunately no kwargs are transported to numpy.interp.

One solution to fix this inconsistent behaviour is to either allow left/right kwargs to be transported to numpy.interp in the above code or use fill_value to set left/right (xarray is doing something along these lines).

The second issue is that a proper use of left/right would be limited to either left or right depending on limit_direction (default forward) which can't be deactivated. With limit_direction="both" this would then work as expected.

As the numpy.interp issue is not directly connected to this issue here, I've opened a new bug report #55144.

@joooeey
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joooeey commented Sep 26, 2024

I second what others have said, the current behaviour is very surprising:

import numpy as np
import pandas as pd

# A 1-D Series with missing external values
x = [0.5, 1, 1.5, 2, 2.5, 3, 20]
y = [np.NaN, 1, np.NaN, 4, np.NaN, 9, np.NaN]
s = pd.Series(y, index=x)

# Expected usage
print(s.interpolate(method="index", limit_direction="both"))

This yields an utter mess:

0.5     1.0
1.0     1.0
1.5     2.5
2.0     4.0
2.5     6.5
3.0     9.0
20.0    9.0
dtype: float64

How is this a mess? Well it uses the linear interpolation that I specified for interpolation but for extrapolation it uses bfill and ffill instead of linear extrapolation.

I can work around this by using the following instead:

print(s.interpolate(method="slinear", limit_direction="both", fill_value="extrapolate"))

with output:

0.5     -0.5
1.0      1.0
1.5      2.5
2.0      4.0
2.5      6.5
3.0      9.0
20.0    94.0
dtype: float64

but this workaround was hard to find.

Either s.interpolate(method="index", limit_direction="both") or s.interpolate(method="slinear", limit_direction="both") should be enough to get the correct output (the dataframe just above).

@joooeey
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joooeey commented Sep 26, 2024

Furthermore, the default to extrapolate forward (limit_direction= kwarg) strikes me as totally arbitrary. IMHO no extrapolation should be the default as that is the safest (except for method="ffill" and method="bfill" which imply an extrapolation direction but we drop those in 3.0.0 as far as I can see which is a great idea since ffill is already accessible via Series.ffill and Series.interpolate(method="zero")).

@joooeey
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joooeey commented Sep 26, 2024

Is there a chance to get the changes I suggest (do the right thing for extrapolation and don't extrapolate by default) into 3.0.0? A major release would be required to clean up the API.

Is a PR welcome? Has someone already been working on parts of my suggestions?

@joooeey
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joooeey commented Sep 26, 2024

I notice that the workaround I mentioned above doesn't work if there are duplicate x-values:

import numpy as np
import pandas as pd

# A 1-D Series with missing external values
x = [0.5, 2, 2, 2.5, 3, 20]
y = [np.NaN, 1, 4, np.NaN, 9, np.NaN]
s = pd.Series(y, index=x)

# Expected usage
print(s.interpolate(method="index", limit_direction="both", fill_value="extrapolate"))

interpolates just fine but doesn't extrapolate:

0.5     1.0
2.0     1.0
2.0     4.0
2.5     6.5
3.0     9.0
20.0    9.0
dtype: float64

but delegating to scipy:

print(s.interpolate(method="slinear", limit_direction="both", fill_value="extrapolate"))

raises an error:

[...]
  File [censored]/python3.12/site-packages/scipy/interpolate/_bsplines.py:1385 in make_interp_spline
    raise ValueError("Expect x to not have duplicates")

ValueError: Expect x to not have duplicates

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