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Andy McDonald5,016 Points
Trouble negating numbers from a string
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:
- Why is the last number in the set not negated?
- 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.
- 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?
import re string = '1234567890' def good_numbers(stri): return re.findall(r'''[\d][^5678]''', stri, re.X|re.I)
Chris FreemanTreehouse Moderator 67,639 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(...
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,
- 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!!!