Pandas DataFrames with NaNs equality comparison
Question:
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?
Answers:
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
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.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)
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
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)
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)
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?
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
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.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)
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
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)
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)