I think using functions here is unnecessary. It is better and mainly faster to use boolean indexing:

m = (df['Name'] == 'Alisa') & (df['Age'] > 24)
print(m)
0      True
1     False
2     False
3     False
4     False
5     False
6      True
7     False
8     False
9     False
10    False
11    False
dtype: bool

#invert mask by ~
df1 = df[~m]

For more complicated filtering, you could use a function which must return a boolean value:

def filter_fn(row):
    if row['Name'] == 'Alisa' and row['Age'] > 24:
        return False
    else:
        return True

df = pd.DataFrame(d, columns=['Name', 'Age', 'Score'])
m = df.apply(filter_fn, axis=1)
print(m)
0     False
1      True
2      True
3      True
4      True
5      True
6     False
7      True
8      True
9      True
10     True
11     True
dtype: bool

df1 = df[m]

You can find the answer here:

Is it possible to use a custom filter function in pandas?

You can try adding a global counter inside the is_email() function to count how many falses were provided and use .apply() on the email column.

dt2 = dt[dt['email'].apply(is_email)]

You can do this using DataFrame.apply, which applies a function along a given axis,

In [3]: df = pandas.DataFrame(np.random.randn(5, 3), columns=['a', 'b', 'c'])

In [4]: df
Out[4]: 
          a         b         c
0 -0.001968 -1.877945 -1.515674
1 -0.540628  0.793913 -0.983315
2 -1.313574  1.946410  0.826350
3  0.015763 -0.267860 -2.228350
4  0.563111  1.195459  0.343168

In [6]: df[df.apply(lambda x: x['b'] > x['c'], axis=1)]
Out[6]: 
          a         b         c
1 -0.540628  0.793913 -0.983315
2 -1.313574  1.946410  0.826350
3  0.015763 -0.267860 -2.228350
4  0.563111  1.195459  0.343168

I think you can just filter based in 13th symbol of your string:

import pandas as pd

# Enter some data. We want to filter out all rows where the number at pos 13,14 > 9
df = pd.DataFrame({
    'ID': ['ABCD-3Z-A93Z-01A-11R-A37O-07',
           'ABCD-6D-AA2E-11A-11R-A37O-07',
           'ABCD-6D-AA2E-01A-11R-A37O-07',
           'ABCD-A3-3307-01A-01R-0864-07',
           'ABCD-6D-AA2E-01A-11R-A37O-07',
           'ABCD-6D-AA2E-10A-11R-A37O-07',
           'ABCD-6D-AA2E-09A-11R-A37O-07'],
    'year': [2012, 2012, 2013, 2014, 2014, 2017, 2015]
})
# convert to df

df['KeepRow'] = df['ID'].apply(lambda x: x[13] == '0')

or simply:

df[df['ID'].apply(lambda x: x[13] == '0')]