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angel moreta2,912 Points
i need a hint =D
why is not excluding 567 ?
output:['1234567890'] i know im pretty close so guys please a good explanation would be great help , something else: since in my string var are only numbers i didnt bother using \w and i used \d instead so i dont know if that might cause not removing 567 :/
import re string = '1234567890' def good_numbers(string): return re.findall(r'[\d]+[^567]',string)
Brandon Spangler8,756 Points
Hint one: The challenge says to find a variable, it doesn't say to write a function!
Hint two: So, what it looks like findall() does is search through the string from left to right searching for the pattern specified. You have asked it to find a pattern that is possibly endless set of numbers with one number on the end of the pattern that is not 5,6, or 7. So it's returning ['1234567890'] which is the pattern you have specified. The [\d]+ says to python "look for a set of one or more numbers that ends with one number that is not 5,6,or 7"; so 1,2,3,4,5,6,7,8, and 9 are being "eaten" by [\d]+ and the [^567] returns the 0 at the end. If the string ended in, say, 5, this new string would not have met your specifications. This might sound confusing and I don't know how to say it exactly, but if you tell python to look for the pattern "one number that is not 5,6, or 7", findall() will look through the string and return each number (in a string) that is not 5,6, or 7. So, if you define your pattern correctly, python will look at each number individually, decide if it is 5,6,or 7 and it will add the number, as a string, to a list.
I hope this sort of helps lol. I'm sure I made no sense though. I used https://www.debuggex.com/cheatsheet/regex/python and https://docs.python.org/3/howto/regex.html to help me solve this particular challenge. Good luck!