Pandas DataFrames with NaNs equality comparison

问题:

In the context of unit testing some functions, I'm trying to establish the equality of 2 DataFrames using python pandas:

ipdb> expect
                            1   2
2012-01-01 00:00:00+00:00 NaN   3
2013-05-14 12:00:00+00:00   3 NaN

ipdb> df
identifier                  1   2
timestamp
2012-01-01 00:00:00+00:00 NaN   3
2013-05-14 12:00:00+00:00   3 NaN

ipdb> df[1][0]
nan

ipdb> df[1][0], expect[1][0]
(nan, nan)

ipdb> df[1][0] == expect[1][0]
False

ipdb> df[1][1] == expect[1][1]
True

ipdb> type(df[1][0])
<type 'numpy.float64'>

ipdb> type(expect[1][0])
<type 'numpy.float64'>

ipdb> (list(df[1]), list(expect[1]))
([nan, 3.0], [nan, 3.0])

ipdb> df1, df2 = (list(df[1]), list(expect[1])) ;; df1 == df2
False

Given that I'm trying to test the entire of expect against the entire of df, including NaN positions, what am I doing wrong?

What is the simplest way to compare equality of Series/DataFrames including NaNs?

回答1:

You can use assert_frame_equals with check_names=False (so as not to check the index/columns names), which will raise if they are not equal:

In [11]: from pandas.util.testing import assert_frame_equal

In [12]: assert_frame_equal(df, expected, check_names=False)

You can wrap this in a function with something like:

try:
    assert_frame_equal(df, expected, check_names=False)
    return True
except AssertionError:
    return False

---

In more recent pandas this functionality has been added as .equals:

df.equals(expected)

回答2:

One of the properties of NaN is that NaN != NaN is True.

Check out this answer for a nice way to do this using numexpr.

(a == b) | ((a != a) & (b != b))

says this (in pseudocode):

a == b or (isnan(a) and isnan(b))

So, either a equals b, or both a and b are NaN.

If you have small frames then assert_frame_equal will be okay. However, for large frames (10M rows) assert_frame_equal is pretty much useless. I had to interrupt it, it was taking so long.

In [1]: df = DataFrame(rand(1e7, 15))

In [2]: df = df[df > 0.5]

In [3]: df2 = df.copy()

In [4]: df
Out[4]:
<class 'pandas.core.frame.DataFrame'>
Int64Index: 10000000 entries, 0 to 9999999
Columns: 15 entries, 0 to 14
dtypes: float64(15)

In [5]: timeit (df == df2) | ((df != df) & (df2 != df2))
1 loops, best of 3: 598 ms per loop

timeit of the (presumably) desired single bool indicating whether the two DataFrames are equal:

In [9]: timeit ((df == df2) | ((df != df) & (df2 != df2))).values.all()
1 loops, best of 3: 687 ms per loop

回答3:

Like @PhillipCloud answer, but more written out

In [26]: df1 = DataFrame([[np.nan,1],[2,np.nan]])

In [27]: df2 = df1.copy()

They really are equivalent

In [28]: result = df1 == df2

In [29]: result[pd.isnull(df1) == pd.isnull(df2)] = True

In [30]: result
Out[30]: 
      0     1
0  True  True
1  True  True

A nan in df2 that doesn't exist in df1

In [31]: df2 = DataFrame([[np.nan,1],[np.nan,np.nan]])

In [32]: result = df1 == df2

In [33]: result[pd.isnull(df1) == pd.isnull(df2)] = True

In [34]: result
Out[34]: 
       0     1
0   True  True
1  False  True

You can also fill with a value you know not to be in the frame

In [38]: df1.fillna(-999) == df1.fillna(-999)
Out[38]: 
      0     1
0  True  True
1  True  True

回答4:

df.fillna(0) == df2.fillna(0)

You can use fillna(). Documenation here.

from pandas import DataFrame

# create a dataframe with NaNs
df = DataFrame([{'a': 1, 'b': 2}, {'a': 5, 'b': 10, 'c': 20}])
df2 = df

# comparison fails!
print df == df2

# all is well 
print df.fillna(0) == df2.fillna(0)

回答5:

Any equality comparison using == with np.NaN is False, even np.NaN == np.NaN is False.

Simply, df1.fillna('NULL') == df2.fillna('NULL'), if 'NULL' is not a value in the original data.

To be safe, do the following:

Example a) Compare two dataframes with NaN values

bools = (df1 == df2)
bools[pd.isnull(df1) & pd.isnull(df2)] = True
assert bools.all().all()

Example b) Filter rows in df1 that do not match with df2

bools = (df1 != df2)
bools[pd.isnull(df1) & pd.isnull(df2)] = False
df_outlier = df1[bools.all(axis=1)]

(Note: this is wrong - bools[pd.isnull(df1) == pd.isnull(df2)] = False)