Ruby's "next" implementation in Python [duplicate] - python

This question already has answers here:
How can I get the next string, in alphanumeric ordering, in Python?
(4 answers)
Closed 6 years ago.
If Python has an implementation of Ruby's next method? I mean something what works exactly the same as in Ruby, so if I type e.g. "z".next it will return "aa" (instead of just next sign in ascii table), "az".next will return "ba" and so on.

I don't believe there is a built-in method for this in Python. A similar question was asked on How can I get the next string, in alphanumeric ordering, in Python? and the accepted answer gives a solution.

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re.match('z', 'az') returns None [duplicate]

This question already has answers here:
What is the difference between re.search and re.match?
(9 answers)
Closed 1 year ago.
See title. The behavior here is counterintuitive. I think maybe I am missing some flags or something? Why does the regex z not match the string az?
The reason is that match only matches the beginning of a string. Must use search to do the thing that match would do in all other programming languages.
Sorry to throw shade at you Python, but you're the odd one out here.

Example of Match.expand usage [duplicate]

This question already has answers here:
Use Python Match object in string with backreferences
(1 answer)
Regex in python: is it possible to get the match, replacement, and final string?
(2 answers)
Closed 3 years ago.
What is an example usage of Match.expand ? It doesn't give any examples in the python docs (which is the first I've heard of the method), and only states:
Match.expand(template)
Return the string obtained by doing backslash substitution on the template string template, as done by the sub() method.
How would this actually be used and how could it be useful?
you can find more explanations and examples here:
match.expand
In general:
it allows you to expand the match you have found and modify its' prefix

getting error while using count in string "BANANA" to find "ANA" as substring using str.count() inbuild function of python [duplicate]

This question already has answers here:
Python string count not working properly? [duplicate]
(2 answers)
Closed 4 years ago.
when executing the command in python
str="BANANA"
print(str.count("ANA"))
give answer as 1.
but real answer would be 2. how to solve this
It's true that this might be misleading most tutorial sites say that count returns the number of occurences in the string, it should actually say it returns the nnumber of non-overlapping occurences in the string

Detect what a python string begins with [duplicate]

This question already has answers here:
Checking whether a string starts with XXXX
(5 answers)
Closed 5 years ago.
To detect whether a python string ends with a particular substring, say ".txt", there is a convenient built-in python string method;
if file_string.endswith(".txt"):
I would like to to detect whether a python string begins with a particular substring, say "begin_like_this". I am not able find a convenient method that I can use like this;
if file_string.beginswith("begin_like_this"):
I am using Python 3.6.
You're looking for str.startswith
if file_string.startswith("begin_like_this"):

How to find the position of every instance of a character in a list [duplicate]

This question already has answers here:
How do i find the position of MORE THAN ONE substring in a string (Python 3.4.3 shell)
(3 answers)
Closed 6 years ago.
I have a program where I need to identify the location of every instance of the letter A in a quote. Something like I would do with quote.index("A"), but I need every instance of A, not just the first.
I know this question has been asked before but I'm very, very new to Python and I'm having trouble understanding the answers to those questions.
If anyone could give me a dumbed down explanation of how to do this, I'd be incredibly thankful because I'm utterly lost.
If i understand correctly, you have a string and you want to keep all A's locations in e different array.
Then you can try something like that.
quote = "some quote"
locs = []
for i in range(len(quote)) :
if quote[i] == 'A' :
locs.append(i)
print(locs)

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