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.

Due to imprecise float comparison you can or your comparison with np.isclose, isclose takes a relative and absolute tolerance param so the following should work:

Copydf['result'] = df['actual_credit'].ge(df['min_required_credit']) | np.isclose(df['actual_credit'], df['min_required_credit'])

You also need to be careful to create a copy of the DataFrame, otherwise the csvdata_old will be updated with csvdata (since it points to the same object):

csvdata_old = csvdata.copy()

To check whether they are equal, you can use assert_frame_equal as in this answer:

from pandas.util.testing import assert_frame_equal
assert_frame_equal(csvdata, csvdata_old)

You can wrap this in a function with something like:

try:
    assert_frame_equal(csvdata, csvdata_old)
    return True
except:  # appeantly AssertionError doesn't catch all
    return False

There was discussion of a better way...

If I got it correctly, you just want to find the closest value from second dataframe.

masses = mass_q_sequence['Mass']
mass_pos = df_seq2['mass_pos']
# using broadcasting and finding indices of closest mass for each mass_pos:
closest_mass_indices = np.argmin(np.abs(masses.reshape(1, -1) - mass_pos.reshape(-1, 1)), axis=1) 
df['closest_mass'] = masses[closest_mass_indices]

math.isclose is not a function that can cooperate with pandas and numpy arrays.

You need numpy.isclose:

temp_df = orig_df[
                (np.isclose(orig_df["Alpha"], num))
                & (orig_df["Method"] == method_type)
            ]

np is numpy in this case.

I believe you need repeat columns values by numpy.broadcast_to before using numpy.isclose:

np.random.seed(142)

df = pd.DataFrame({'column':np.random.rand(10)})
print (df)
     column
0  0.902062
1  0.557808
2  0.655985
3  0.832471
4  0.199884
5  0.127254
6  0.771439
7  0.432289
8  0.385282
9  0.783643

A = [0.432, 0.783, 0.902]

#repeat by length of number of list A
len_A = len(A)
a = np.broadcast_to(df['column'].values[:, None], (len(df),len_A))
print (a)
[[0.90206152 0.90206152 0.90206152]
 [0.55780754 0.55780754 0.55780754]
 [0.65598471 0.65598471 0.65598471]
 [0.83247141 0.83247141 0.83247141]
 [0.19988419 0.19988419 0.19988419]
 [0.12725426 0.12725426 0.12725426]
 [0.77143911 0.77143911 0.77143911]
 [0.43228855 0.43228855 0.43228855]
 [0.38528223 0.38528223 0.38528223]
 [0.78364337 0.78364337 0.78364337]]

#pandas solution
m = pd.concat([df['column']] * len_A, axis=1)
print (m)
     column    column    column
0  0.902062  0.902062  0.902062
1  0.557808  0.557808  0.557808
2  0.655985  0.655985  0.655985
3  0.832471  0.832471  0.832471
4  0.199884  0.199884  0.199884
5  0.127254  0.127254  0.127254
6  0.771439  0.771439  0.771439
7  0.432289  0.432289  0.432289
8  0.385282  0.385282  0.385282
9  0.783643  0.783643  0.783643

m = np.isclose(a, b=A, atol=0.004)
print (m)
[[False False  True]
 [False False False]
 [False False False]
 [False False False]
 [False False False]
 [False False False]
 [False False False]
 [ True False False]
 [False False False]
 [False  True False]]

Last get all values with True per rows by any:

print (m.any(axis=1))
[ True False False False False False False  True False  True]

And last filter by boolean indexing:

print (df[m.any(axis=1)])
     column
0  0.902062
7  0.432289
9  0.783643