Mar-15-2021, 04:22 PM
There is also str.count method:
>>> help(str.count) Help on method_descriptor: count(...) S.count(sub[, start[, end]]) -> int Return the number of non-overlapping occurrences of substring sub in string S[start:end]. Optional arguments start and end are interpreted as in slice notation. >>> 'abrakadabra'.count('abra') 2 >>> 'abrakadabra'.count('some') 0
I'm not 'in'-sane. Indeed, I am so far 'out' of sane that you appear a tiny blip on the distant coast of sanity. Bucky Katt, Get Fuzzy
Da Bishop: There's a dead bishop on the landing. I don't know who keeps bringing them in here. ....but society is to blame.
Da Bishop: There's a dead bishop on the landing. I don't know who keeps bringing them in here. ....but society is to blame.