How do I find index of last second integer in this string? I'm not sure to if I can use rindex() and if so then how.
The index of 7 in the above string is 6.
I can't wrap my head around writing something like stre.rindex(isnumeric())
This will find the penultimate occurrence of a sequence of digits in a string and print its offset.
import re
a = "hEY3 a7yY5"
offsets = [m.start() for m in re.finditer(r'\d+', a)]
print(-1 if len(offsets) < 2 else offsets[-2])
If you are averse to re then you can do it like this:
offsets = []
for i, c in enumerate(a):
if c.isdigit():
if offsets:
if offsets[-1] != i - 1:
offsets.append(i)
else:
offsets = [i]
print(-1 if len(offsets) < 2 else offsets[-2])
Output:
6
Note that the same result would be printed if the string is "hEY7 a75yY5" - i.e., this also handles sequences of more than one digit
You may try something like this
a = "hEY3 a7yy5"
out = [i for i,j in enumerate(a) if j.isnumeric()][-2]
print(out)
An option to get the index of the second last digit using a pattern, and use the start() method of the Match object:
s = "hEY3 a7yY5"
pattern = r"\d(?=[^\d\n]*\d[^\d\n]*\Z)"
m = re.search(pattern, s)
if m:
print(m.start())
Output
6
The pattern matches:
\d Match a single digit(?= Positive lookahead, assert what is to the right is
[^\d\n]* Optionally match any char except a digit or a newline\d Match a single digit[^\d\n]* Optionally match any char except a digit or a newline\Z End of string) Close the lookahead