Getting a string that comes after a '%' symbol and should end before other characters (no numbers and characters).
for example:
string = 'Hi %how are %YOU786$ex doing'
it should return as a list.
['how', 'you']
I tried
string = text.split()
sample = []
for i in string:
if '%' in i:
sample.append(i[1:index].lower())
return sample
but it I don't know how to get rid of 'you786$ex'.
EDIT: I don't want to import re
You can use a regular expression.
>>> import re
>>>
>>> s = 'Hi %how are %YOU786$ex doing'
>>> re.findall('%([a-z]+)', s.lower())
>>> ['how', 'you']
This can be most easily done with re.findall():
import re
re.findall(r'%([a-z]+)', string.lower())
This returns:
['how', 'you']
Or you can use str.split() and iterate over the characters:
sample = []
for token in string.lower().split('%')[1:]:
word = ''
for char in token:
if char.isalpha():
word += char
else:
break
sample.append(word)
sample would become:
['how', 'you']
Use Regex (Regular Expressions).
First, create a Regex pattern for your task. You could use online tools to test it. See regex for your task: https://regex101.com/r/PMSvtK/1
Then just use this regex in Python:
import re
def parse_string(string):
return re.findall("\%([a-zA-Z]+)", string)
print(parse_string('Hi %how are %YOU786$ex doing'))
Output:
['how', 'YOU']
