Python, regex to exclude matches of numbers

Question

To use regex to extract any numbers of length greater than 2, in a string, but also exclude "2016", here is what I have:

import re

string = "Employee ID DF856, Year 2016, Department Finance, Team 2, Location 112 "

print re.findall(r'\d{3,}', string)

output:

['856', '2016', '112']

I tried to change it to below to exclude "2016" but all failed.

print re.findall(r'\d{3,}/^(!2016)/', string)
print re.findall(r"\d{3,}/?!2016/", string)
print re.findall(r"\d{3,}!'2016'", string)

What is the right way to do it? Thank you.

the question was extended, please see the final comment made by Wiktor Stribiżew for the update.

Answer 1

You may use

import re
s = "Employee ID DF856, Year 2016, Department Finance, Team 2, Location 112 20161 12016 120162"
print(re.findall(r'(?<!\d)(?!2016(?!\d))\d{3,}', s))

See the Python demo and a regex demo.

Details

• (?<!\d) - no digit allowed iommediately to the left of the current location
• (?!2016(?!\d)) - no 2016 not followed with another digit is allowed immediately to the right of the current location
• \d{3,} - 3 or more digits.

An alternative solution with some code:

import re
s = "Employee ID DF856, Year 2016, Department Finance, Team 2, Location 112 20161 12016 120162"
print([x for x in re.findall(r'\d{3,}', s) if x != "2016"])

Here, we extract any chunks of 3 or more digits (re.findall(r'\d{3,}', s)) and then filter out those equal to 2016.

Answer 2

You want to use a negative lookahead. The correct syntax is:

\D(?!2016)(\d{3,})\b

Results in:

In [24]: re.findall(r'\D(?!2016)(\d{3,})\b', string)
Out[24]: ['856', '112']

Or using a negative lookbehind:

In [26]: re.findall(r'\D(\d{3,})(?<!2016)\b', string)
Out[26]: ['856', '112']