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Rowwise min() and max() fails for column with NaNs

I am trying to take the rowwise max (and min) of two columns containing dates

from datetime import date
import pandas as pd
import numpy as np    

df = pd.DataFrame({'date_a' : [date(2015, 1, 1), date(2012, 6, 1),
                               date(2013, 1, 1), date(2016, 6, 1)],
                   'date_b' : [date(2012, 7, 1), date(2013, 1, 1), 
                               date(2014, 3, 1), date(2013, 4, 1)]})

df[['date_a', 'date_b']].max(axis=1)
Out[46]: 
0    2015-01-01
1    2013-01-01
2    2014-03-01
3    2016-06-01

as expected. However, if the dataframe contains a single NaN value, the whole operation fails

df_nan = pd.DataFrame({'date_a' : [date(2015, 1, 1), date(2012, 6, 1),
                                   np.NaN, date(2016, 6, 1)],
                       'date_b' : [date(2012, 7, 1), date(2013, 1, 1), 
                                   date(2014, 3, 1), date(2013, 4, 1)]})

df_nan[['date_a', 'date_b']].max(axis=1)
Out[49]: 
0   NaN 
1   NaN
2   NaN
3   NaN
dtype: float64

What is going on here? I was expecting this result

0    2015-01-01
1    2013-01-01
2    NaN
3    2016-06-01

How can this be achieved?

like image 383
mortysporty Avatar asked Aug 26 '17 20:08

mortysporty


3 Answers

I would say the best solution is to use the appropriate dtype. Pandas provides a very well integrated datetime dtype. So note, you are using object dtypes...

>>> df
       date_a      date_b
0  2015-01-01  2012-07-01
1  2012-06-01  2013-01-01
2         NaN  2014-03-01
3  2016-06-01  2013-04-01
>>> df.dtypes
date_a    object
date_b    object
dtype: object

But note, the problem disappears when you use

>>> df2 = df.apply(pd.to_datetime)
>>> df2
      date_a     date_b
0 2015-01-01 2012-07-01
1 2012-06-01 2013-01-01
2        NaT 2014-03-01
3 2016-06-01 2013-04-01
>>> df2.min(axis=1)
0   2012-07-01
1   2012-06-01
2   2014-03-01
3   2013-04-01
dtype: datetime64[ns]
like image 87
juanpa.arrivillaga Avatar answered Oct 19 '22 12:10

juanpa.arrivillaga


This appears to happen when date objects are mixed with floats (such as NaN) in columns. By default, the numeric_only flag is set because of the single float value. For example, replace your df_nan with this:

df_float = pd.DataFrame({'date_a' : [date(2015, 1, 1), date(2012, 6, 1),
                                    1.023, date(2016, 6, 1)],
                        'date_b' : [date(2012, 7, 1), 3.14, 
                                    date(2014, 3, 1), date(2013, 4, 1)]})

print(df_float.max(1))

0   NaN
1   NaN
2   NaN
3   NaN
dtype: float64

If the flag is manually set to false, this would rightly throw a TypeError because:

print(date(2015, 1, 1) < 1.0)

TypeError                                 Traceback (most recent call last)
<ipython-input-362-ccbf44ddb40a> in <module>()
      1 
----> 2 print(date(2015, 1, 1) < 1.0)

TypeError: unorderable types: datetime.date() < float()

However, pandas seems to coerce everything to NaN. As a workaround, converting to str using df.astype appears to do it:

out = df_nan.astype(str).max(1)
print(out) 
0    2015-01-01
1    2013-01-01
2           nan
3    2016-06-01
dtype: object

In this case, sorting lexicographically yields the same solution as before.

Otherwise, as juan suggests, you can cast to datetime using pd.to_datetime:

out = df_nan.apply(pd.to_datetime, errors='coerce').max(1)
print(out)

0   2015-01-01
1   2013-01-01
2   2014-03-01
3   2016-06-01
dtype: datetime64[ns]
like image 6
cs95 Avatar answered Oct 19 '22 13:10

cs95


The following should work:

>>> df_nan.where(df_nan.T.notnull().all()).max(axis=1)
Out[1]:
0    2015-01-01
1    2013-01-01
2          None
3    2016-06-01
dtype: object

Where:

  1. df_nan.T.notnull().all() computes a mask of row containing no np.nan
  2. df_nan.where() applies the former mask to the dataframe
  3. .max(axis=1) gets the row-wise maximum

This works because the maximum of an array where all values are np.nan is None. It allows to keep track of rows where a value is missing by not showing a maximum.

But this decision is up to you, otherwise the solution of @juanpa.arrivillaga that converts NaN to NaT is what you want.

like image 1
FabienP Avatar answered Oct 19 '22 12:10

FabienP