Comparing two pandas series for floating point near-equality?

You can use numpy.allclose:

numpy.allclose(a, b, rtol=1e-05, atol=1e-08, equal_nan=False)

Returns True if two arrays are element-wise equal within a tolerance.

The tolerance values are positive, typically very small numbers. The relative difference (rtol * abs(b)) and the absolute difference atol are added together to compare against the absolute difference between a and b.

numpy works well with pandas.Series objects, so if you have two of them - s1 and s2, you can simply do:

np.allclose(s1, s2, atol=...) 

Where atol is your tolerance value.

If you want to avoid numpy, there is another way, use assert_series_equal

import pandas as pd
s1 = pd.Series([1.333333, 1.666666])
s2 = pd.Series([1.333, 1.666])

from pandas.testing import assert_series_equal
assert_series_equal(s1,s2)  

raises an AssertionError. So use the check_less_precise flag

assert_series_equal(s1,s2, check_less_precise= True)  # No assertion error

This doesn't raise an AssertionError as check_less_precise only compares 3 digits after decimal.

See the docs here

Not good to use asserts but if you want to avoid numpy, this is a way.

Note: I'm posting this mostly because I came to this thread via a Google search of something similar and it seemed too long for a comment. Not necessarily the best solution nor strictly "ε of precision"-based, but an alternative using scaling and rounding if you want to do this for vectors (i.e. rows) rather than scalars for a DataFrame (rather than Series) without looping through explicitly:

import numpy as np
import pandas as pd
from sklearn.preprocessing import MinMaxScaler

Xcomb = pd.concat((X, X2), axis=0, ignore_index=True)
# scale
scaler = MinMaxScaler()
scaler.fit(Xcomb)
Xscl = scaler.transform(Xcomb)
# round
df_scl = pd.DataFrame(np.round(Xscl, decimals=8), columns=X.columns)
# post-processing
n_uniq = df_scl.drop_duplicates().shape[0]
n_dup = df.shape[0] + df2.shape[0] - n_uniq
print(f"Number of shared rows: {n_dup}")