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Split a string if character is present else don't split

I have a string like below in python
testing_abc
I want to split string based on _ and extract the 2 element
I have done like below
split_string = string.split('_')[1]
I am getting the correct output as expected
abc
Now I want this to work for below strings
1) xyz
When I use
split_string = string.split('_')[1]
I get below error
list index out of range
expected output I want is xyz
2) testing_abc_bbc
When I use
split_string = string.split('_')[1]
I get abc as output
expected output I want is abc_bbc
Basically What I want is
1) If string contains `_` then print everything after the first `_` as variable
2) If string doesn't contain `_` then print the string as variable
How can I achieve what I want

Set the maxsplit argument of split to 1 and then take the last element of the resulting list.
>>> "testing_abc".split("_", 1)[-1]
'abc'
>>> "xyz".split("_", 1)[-1]
'xyz'
>>> "testing_abc_bbc".split("_", 1)[-1]
'abc_bbc'

You can use list slicing and str.join in case _ is in the string, and you can just get the first element of the split (which is the only element) in the other case:
sp = string.split('_')
result = '_'.join(sp[1:]) if len(sp) > 1 else sp[0]

All of the ways are good but there is a very simple and optimum way for this.
Try:
s = 'aaabbnkkjbg_gghjkk_ttty'
try:
ans = s[s.index('_')+1:]
except:
ans = s

Ok so your error is supposed to happen/expected because you are using '_' as your delimiter and it doesn't contain it.
See How to check a string for specific characters? for character checking.
If you want to only split iff the string contains a '_' and only on the first one,
input_string = "blah_end"
delimiter = '_'
if delimiter in input_string:
result = input_string.split("_", 1)[1] # The ",1" says only split once
else:
# Do whatever here. If you want a space, " " to be a delimiter too you can try that.
result = input_string

this code will solve your problem
txt = "apple_banana_cherry_orange"
# setting the maxsplit parameter to 1, will return a list with 2 elements!
x = txt.split("_", 1)
print(x[-1])

Stripping Hex code from a plain text file in Python [duplicate]

I have a string. How do I remove all text after a certain character? (In this case ...)
The text after will ... change so I that's why I want to remove all characters after a certain one.

Split on your separator at most once, and take the first piece:
sep = '...'
stripped = text.split(sep, 1)[0]
You didn't say what should happen if the separator isn't present. Both this and Alex's solution will return the entire string in that case.

Assuming your separator is '...', but it can be any string.
text = 'some string... this part will be removed.'
head, sep, tail = text.partition('...')
>>> print head
some string
If the separator is not found, head will contain all of the original string.
The partition function was added in Python 2.5.
S.partition(sep) -> (head, sep, tail)
Searches for the separator sep in S, and returns the part before it,
the separator itself, and the part after it. If the separator is not
found, returns S and two empty strings.

If you want to remove everything after the last occurrence of separator in a string I find this works well:
<separator>.join(string_to_split.split(<separator>)[:-1])
For example, if string_to_split is a path like root/location/child/too_far.exe and you only want the folder path, you can split by "/".join(string_to_split.split("/")[:-1]) and you'll get
root/location/child

Without a regular expression (which I assume is what you want):
def remafterellipsis(text):
where_ellipsis = text.find('...')
if where_ellipsis == -1:
return text
return text[:where_ellipsis + 3]
or, with a regular expression:
import re
def remwithre(text, there=re.compile(re.escape('...')+'.*')):
return there.sub('', text)

import re
test = "This is a test...we should not be able to see this"
res = re.sub(r'\.\.\..*',"",test)
print(res)
Output: "This is a test"

The method find will return the character position in a string. Then, if you want remove every thing from the character, do this:
mystring = "123⋯567"
mystring[ 0 : mystring.index("⋯")]
>> '123'
If you want to keep the character, add 1 to the character position.

From a file:
import re
sep = '...'
with open("requirements.txt") as file_in:
lines = []
for line in file_in:
res = line.split(sep, 1)[0]
print(res)

This is in python 3.7 working to me
In my case I need to remove after dot in my string variable fees
fees = 45.05
split_string = fees.split(".", 1)
substring = split_string[0]
print(substring)

Yet another way to remove all characters after the last occurrence of a character in a string (assume that you want to remove all characters after the final '/').
path = 'I/only/want/the/containing/directory/not/the/file.txt'
while path[-1] != '/':
path = path[:-1]

another easy way using re will be
import re, clr
text = 'some string... this part will be removed.'
text= re.search(r'(\A.*)\.\.\..+',url,re.DOTALL|re.IGNORECASE).group(1)
// text = some string

Parsing a MAC address with python

How can I convert a hex value "0000.0012.13a4" into "00:00:00:12:13:A4"?

text = '0000.0012.13a4'
text = text.replace('.', '').upper() # a little pre-processing
# chunk into groups of 2 and re-join
out = ':'.join([text[i : i + 2] for i in range(0, len(text), 2)])
print(out)
00:00:00:12:13:A4

import re
old_string = "0000.0012.13a4"
new_string = ':'.join(s for s in re.split(r"(\w{2})", old_string.upper()) if s.isalnum())
print(new_string)
OUTPUT
> python3 test.py
00:00:00:12:13:A4
>
Without modification, this approach can handle some other MAC formats that you might run into like, "00-00-00-12-13-a4"

Try following code
import re
hx = '0000.0012.13a4'.replace('.','')
print(':'.join(re.findall('..', hx)))
Output: 00:00:00:12:13:a4

There is a pretty simple three step solution:
First we strip those pesky periods.
step1 = hexStrBad.replace('.','')
Then, if the formatting is consistent:
step2 = step1[0:2] + ':' + step1[2:4] + ':' + step1[4:6] + ':' + step1[6:8] + ':' + step1[8:10] + ':' + step1[10:12]
step3 = step2.upper()
It's not the prettiest, but it will do what you need!

It's unclear what you're asking exactly, but if all you want is to make a string all uppercase, use .upper()
Try to clarify your question somewhat, because if you're asking about converting some weirdly formatted string into what looks like a MAC address, we need to know that to answer your question.

Adding "/" to strings if they are not already there

How do I add a "/" to the beginning and ending of a string in python? For example:
Input: test
Output: /test/
Input: /test
Output: /test/
Input: test/
Ouput: /test/
What's a neat way to do this? The only way I could think of was writing different if statements for each case. Surely, there's a better way to do this?

Strip off any slashes from the existing string, then put two new ones around that.
text = "/%s/" % text.strip("/")

Remove the old '/'s, if any, and then add them back:
'/' + s.strip('/') + '/'

There are several ways to do this, but the fastest (for the computer) is to simply test and augment each end. Using str as the string:
if str[0] != '/':
str = '/' + str
... and repeat for the other end.
Another way is to strip any existing slashes, and then add them to both ends.
If you know that there are no double-slashes within the string, you can add them and then replace doubles (in case you created them at the end):
str = '/' + str + '/'
str.replace('//', '/')

You can strip the string first and then add '/' to beginning and end like this
'/' + <your_string>.strip('/') + '/'

Well the following code solves the purpose of your question. You can simply use an or instead of separate if or elif statements.
def test(x):
if x=="test" or x == "/test" or x=="test/":
print("/",x.strip("/"),"/")

How to delete everything after a certain character in a string?

How would I delete everything after a certain character of a string in python? For example I have a string containing a file path and some extra characters. How would I delete everything after .zip? I've tried rsplit and split , but neither included the .zip when deleting extra characters.
Any suggestions?

Just take the first portion of the split, and add '.zip' back:
s = 'test.zip.zyz'
s = s.split('.zip', 1)[0] + '.zip'
Alternatively you could use slicing, here is a solution where you don't need to add '.zip' back to the result (the 4 comes from len('.zip')):
s = s[:s.index('.zip')+4]
Or another alternative with regular expressions:
import re
s = re.match(r'^.*?\.zip', s).group(0)

str.partition:
>>> s='abc.zip.blech'
>>> ''.join(s.partition('.zip')[0:2])
'abc.zip'
>>> s='abc.zip'
>>> ''.join(s.partition('.zip')[0:2])
'abc.zip'
>>> s='abc.py'
>>> ''.join(s.partition('.zip')[0:2])
'abc.py'

Use slices:
s = 'test.zip.xyz'
s[:s.index('.zip') + len('.zip')]
=> 'test.zip'
And it's easy to pack the above in a little helper function:
def removeAfter(string, suffix):
return string[:string.index(suffix) + len(suffix)]
removeAfter('test.zip.xyz', '.zip')
=> 'test.zip'

I think it's easy to create a simple lambda function for this.
mystrip = lambda s, ss: s[:s.index(ss) + len(ss)]
Can be used like this:
mystr = "this should stay.zipand this should be removed."
mystrip(mystr, ".zip") # 'this should stay.zip'

You can use the re module:
import re
re.sub('\.zip.*','.zip','test.zip.blah')