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# 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)

print(good_numbers(string))

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?
negate.py
```import re

string = '1234567890'

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

MOD

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]’`