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Start your free trialdylan hamm
3,667 PointsI don't really know how to do this one
I have attempted to find all numbers and ignore only 5, 6, and 7 however the code finds all numbers and includes 5,6, and 7
import re
string = '1234567890'
good_numbers = re.findall(r'\d*[^567]', string)
2 Answers
Stuart Wright
41,120 PointsYour regular expression reads:
"Match any number of digits, followed by one character that is not 5, 6 or 7."
This should do what the challenge is asking, which is match any number of characters except 5, 6 or 7 :
import re
string = '1234567890'
good_numbers = re.findall(r'[^567]*', string)
Tri Pham
18,671 PointsI didn't like the way the question was phrased. I thought they should have added the expected result. The expected result is actually ['1', '2', '3', '4', '8', '9', '0']. Yours got ['1234567890']. Can you solve it now?
Tri Pham
18,671 Pointsactually now I go no clue what they wanted :(. Stuart's answer works too and produces ['1234', '', '', '', '890', ''].... my solution was good_numbers = re.findall(r'[^567]', string) and produced ['1', '2', '3', '4', '8', '9', '0'].
Stuart Wright
41,120 PointsYes good point, it's not very clear what they want. Reading the question again I think yours is the solution they're after, but it's odd that mine gets accepted too.
dylan hamm
3,667 Pointsdylan hamm
3,667 PointsI figured it out shortly after posting the question but I appreciate the help