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Python Regular Expressions in Python Introduction to Regular Expressions Negated Numbers

Andy McDonald
Andy McDonald
Python Development Techdegree Student 8,745 Points

Trouble negating numbers from a string

import re

string = '1234567890'

def good_numbers(stri): return re.findall(r'''[\d][^5678]''', stri, re.X|re.I)


I need a function that returns all numbers that arent 5, 6, and 7. I got to this function from realizing I'm utterly out of ideas on what to do. I noticed that the last number in the negated set still shows up in the string... So my question is:

  1. Why is the last number in the set not negated?
  2. When doing a findall on \d alone. It returns each individual number. But as soon as a add the negated set. It returns them in double digits.... ie 12, 34,ect instead of 1, 2, 3, 4. I'm really curious as to why that is happening.
  3. And finally, what it is that I'm doing wrong??? What it is it about this function that isn't right? Why cant I get 1, 2, 3, 4, 8, 9, 0?

Trying to coomplete quiz: https://teamtreehouse.com/library/regular-expressions-in-python/introduction-to-regular-expressions/negated-numbers

import re

string = '1234567890'

def good_numbers(stri):
    return re.findall(r'''[\d][^5678]''', stri, re.X|re.I)

1 Answer

Chris Freeman
Chris Freeman
Treehouse Moderator 68,063 Points

Hey Andy McDonald, you’re on the right path but the structure is not correct.

  • the challenge asks to define a variable, not define a function. Use the form good_numbers = re.findall(...
  • with r‘[\d][^5678]', you are matching two characters at a time. This regex says “match any digit that isn’t followed by a 5, 6, 7, or 8”.

By looking at pairs of numbers, the pattern “consumes” both numbers, so only five matching attempts are done in a string of ten characters:

  • “12”, passes since 2 is not in “5678”
  • “34”, passes since 4 is not in “5678”
  • “56”, fails since 6 is in “5678”
  • “78”, fails since 8 is in “5678”
  • “90”, passes since 0 is not in “5678”

What you want instead is “any single character that is not 5, 6, 7”, that is, r’[^567]’


  • you can use instead of ’’’, since you don’t a multiple line pattern
  • neither of the re options are required since you are not using verbose mode and do not need to ignore case

Post back if you need more help. Good luck!!!