I need to find the string in between two other strings. How could I do this?
string = "vHELLOv"
Finding "HELLO" which is in between of two lowercase "v"'s
Another example could be:
string = "/World/"
Finding "World" in the middle of the two "/"s
Try this. Use regex (regular expression) library for pattern matching.
import re
# 1st pattern: vHELLOv
re.findall(r"v(.*?)v", "vHELLOv")
# 2nd pattern: /HELLO/
re.findall(r"/(.*?)/", "/HELLO/")
## NOTE: For regex in python you do not have to escape / with a \.
# But, if you want to use a similar regex universally elsewhere,
# consider escaping each / with a \.
#
# Example: r"/(.*?)/" ==> r"\/(.*?)\/"
EDIT: Added lazy quantifier .*? instead of .* inside the () as suggested by Jan.
pattern: r"v(.*)v" Greedy --> Example on regex101.com
pattern: r"v(.*?)v" Lazy --> Example on regex101.com BETTER
If you are talking about two identical strings surrounding another, you can use the split method:
string = "vHELLOv"
surrounder = "v"
x = string.split(surrounder)
print(x[1])
To remove punctuation, you could use the following regex expression in Python:
import re
s = "/World/"
new_s = re.sub(r'\W', "", s)
You can do it in various ways as
import re
text = "vHELLOv" #or txt = "/World/"
re.match(r"(v|\/)(\w*)(v|\/)", text).group(2)
or
re.search(r"(v|\/)(\w*)(v|\/)", text).group(2)
This is the regex methods You can do it also with string replace as
txt.replace(txt[0], "") if txt[0] == txt[-1] else txt
"Another example could be" - Just those words made me think you are not only interested in v or / but any type of character that surrounds a string of interest. Therefor you could try a pattern like:
^(.)(.+)\1$
Or if you want to define the possibilities:
^([\/v])(.+)\1$
This way you would capture the first character in a capture group that you can refer to later down the line. You don't need to distinquish between v or / or any character perse. Now grab group 2 from a search result, e.g.:
import re
s = 'vHELLOv'
print(re.search(r'^(.)(.+)\1$', s).group(2))
returns:
HELLO
remodule)"vHELLOv"[1:-1]