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]

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 could define a function containing the loop you want, and call it as a one-liner:

def compare_isclose(df1, df2, atol_list):
    feature_list = []
    for i, col in enumerate(df1.columns):
        feature_list.append(np.isclose(df1[col], df2[col], atol=atol_list[i]))
    feature_df = pd.DataFrame.from_records(feature_list).T
    return feature_df

The function can be used this way:

atol_list = [10,20,30]
feature_df = compare_isclose(df1, df2, atol_list)

It only works under the assumption that the number of elements in atol_list equals the number of columns in the dataframe, and that the column names of the 2 dataframes are identical.

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

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.