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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])

How to partial split and take the first portion of string in Python?

Have a scenario where I wanted to split a string partially and pick up the 1st portion of the string.
Say String could be like aloha_maui_d0_b0 or new_york_d9_b10. Note: After d its numerical and it could be any size.
I wanted to partially strip any string before _d* i.e. wanted only _d0_b0 or _d9_b10.
Tried below code, but obviously it removes the split term as well.
print(("aloha_maui_d0_b0").split("_d"))
#Output is : ['aloha_maui', '0_b0']
#But Wanted : _d0_b0
Is there any other way to get the partial portion? Do I need to try out in regexp?

How about just
stArr = "aloha_maui_d0_b0".split("_d")
st2 = '_d' + stArr[1]
This should do the trick if the string always has a '_d' in it

You can use index() to split in 2 parts:
s = 'aloha_maui_d0_b0'
idx = s.index('_d')
l = [s[:idx], s[idx:]]
# l = ['aloha_maui', '_d0_b0']
Edit: You can also use this if you have multiple _d in your string:
s = 'aloha_maui_d0_b0_d1_b1_d2_b2'
idxs = [n for n in range(len(s)) if n == 0 or s.find('_d', n) == n]
parts = [s[i:j] for i,j in zip(idxs, idxs[1:]+[None])]
# parts = ['aloha_maui', '_d0_b0', '_d1_b1', '_d2_b2']

I have two suggestions.
partition()
Use the method partition() to get a tuple containing the delimiter as one of the elements and use the + operator to get the String you want:
teste1 = 'aloha_maui_d0_b0'
partitiontest = teste1.partition('_d')
print(partitiontest)
print(partitiontest[1] + partitiontest[2])
Output:
('aloha_maui', '_d', '0_b0')
_d0_b0
The partition() methods returns a tuple with the first element being what is before the delimiter, the second being the delimiter itself and the third being what is after the delimiter.
The method does that to the first case of the delimiter it finds on the String, so you can't use it to split in more than 3 without extra work on the code. For that my second suggestion would be better.
replace()
Use the method replace() to insert an extra character (or characters) right before your delimiter (_d) and use these as the delimiter on the split() method.
teste2 = 'new_york_d9_b10'
replacetest = teste2.replace('_d', '|_d')
print(replacetest)
splitlist = replacetest.split('|')
print(splitlist)
Output:
new_york|_d9_b10
['new_york', '_d9_b10']
Since it replaces all cases of _d on the String for |_d there is no problem on using it to split in more than 2.
Problem?
A situation to which you may need to be careful would be for unwanted splits because of _d being present in more places than anticipated.
Following the apparent logic of your examples with city names and numericals, you might have something like this:
teste3 = 'rio_de_janeiro_d3_b32'
replacetest = teste3.replace('_d', '|_d')
print(replacetest)
splitlist = replacetest.split('|')
print(splitlist)
Output:
rio|_de_janeiro|_d3_b32
['rio', '_de_janeiro', '_d3_b32']
Assuming you always have the numerical on the end of the String and _d won't happen inside the numerical, rpartition() could be a solution:
rpartitiontest = teste3.rpartition('_d')
print(rpartitiontest)
print(rpartitiontest[1] + rpartitiontest[2])
Output:
('rio_de_janeiro', '_d', '3_b32')
_d3_b32
Since rpartition() starts the search on the String's end and only takes the first match to separate the terms into a tuple, you won't have to worry about the first term (city's name?) causing unexpected splits.

Use regex's split and keep delimiters capability:
import re
patre = re.compile(r"(_d\d)")
#👆 👆
#note the surrounding parenthesises - they're what drives "keep"
for line in """aloha_maui_d0_b0 new_york_d9_b10""".split():
parts = patre.split(line)
print("\n", line)
print(parts)
p1, p2 = parts[0], "".join(parts[1:])
print(p1, p2)
output:
aloha_maui_d0_b0
['aloha_maui', '_d0', '_b0']
aloha_maui _d0_b0
new_york_d9_b10
['new_york', '_d9', '_b10']
new_york _d9_b10
credit due: https://stackoverflow.com/a/15668433

how to get second last and last value in a string after separator in python

In Python, how do you get the last and second last element in string ?
string "client_user_username_type_1234567"
expected output : "type_1234567"

Try this :
>>> s = "client_user_username_type_1234567"
>>> '_'.join(s.split('_')[-2:])
'type_1234567'

You can also use re.findall:
import re
s = "client_user_username_type_1234567"
result = re.findall('[a-zA-Z]+_\d+$', s)[0]
Output:
'type_1234567'

There's no set function that will do this for you, you have to use what Python gives you and for that I present:
split slice and join
"_".join("one_two_three".split("_")[-2:])
In steps:
Split the string by the common separator, "_"
s.split("_")
Slice the list so that you get the last two elements by using a negative index
s.split("_")[-2:]
Now you have a list composed of the last two elements, now you have to merge that list again so it's like the original string, with separator "_".
"_".join("one_two_three".split("_")[-2:])
That's pretty much it. Another way to investigate is through regex.

python string, delete character, count from right

I have some strings I created with elements coming from many sources, number of elements will vary each time the program is run; I created a sample string that my program creates now.
I want to count in [:-3] for the following string and delete the last comma:
'{"SEignjExQfumwZRacPNHvq8UcsBjKWPERB":1.00000000,"SCaWymicaunRLAxNSTTRhVxLMAB9PaKBDK":2.80000000,"SGFHTxuRLttUShUjZyFMzs8NgC1JopSUK6":1.20000000,}'
So my string looks like:
'{"SEignjExQfumwZRacPNHvq8UcsBjKWPERB":1.00000000,"SCaWymicaunRLAxNSTTRhVxLMAB9PaKBDK":2.80000000,"SGFHTxuRLttUShUjZyFMzs8NgC1JopSUK6":1.20000000}'
I just cant quite get there, help appreciated.

To remove the third last character from the string you can use:
string[:-3] + string[-2:]
>>> string = "hellothere"
>>> string[:-3] + string[-2:]
'hellothre'

I would use rsplit to split on the right most occurrence of a substring (limiting to two results) and then join them with an empty string
''.join(s.rsplit(',', 2))

a = '{"SEignjExQfumwZRacPNHvq8UcsBjKWPERB":1.00000000,"SCaWymicaunRLAxNSTTRhVxLMAB9PaKBDK":2.80000000,"SGFHTxuRLttUShUjZyFMzs8NgC1JopSUK6":1.20000000,}'
a[:len(a) - 2] + a[len(a) - 1:]
You could obviously use different expressions in the brackets, I just wanted to show that you could use any expressions you wanted.

you can try with rfind to find the last comma
s = '{"SEignjExQfumwZRacPNHvq8UcsBjKWPERB":1.00000000,"SCaWymicaunRLAxNSTTRhVxLMAB9PaKBDK":2.80000000,"SGFHTxuRLttUShUjZyFMzs8NgC1JopSUK6":1.20000000,}'
idx = s.rfind(",")
s[:idx]+s[idx+1:]
you get,
'{"SEignjExQfumwZRacPNHvq8UcsBjKWPERB":1.00000000,"SCaWymicaunRLAxNSTTRhVxLMAB9PaKBDK":2.80000000,"SGFHTxuRLttUShUjZyFMzs8NgC1JopSUK6":1.20000000}'

Using regex:
>>> print re.sub(r ",(?=[^.]*$)", r '', s)
{"SEignjExQfumwZRacPNHvq8UcsBjKWPERB":1.00000000,"SCaWymicaunRLAxNSTTRhVxLMAB9PaKBDK":2.80000000,"SGFHTxuRLttUShUjZyFMzs8NgC1JopSUK6":1.20000000}
This will match a ',' all before a any potential NOT ','. It matches the last ',' right before the end of a string.

how to get the same required string with better and shorter way

s = 'myName.Country.myHeight'
required = s.split('.')[0]+'.'+s.split('.')[1]
print required
myName.Country
How can I get the same 'required' string with better and shorter way?

Use str.rpartition like this
s = 'myName.Country.myHeight'
print s.rpartition(".")[0]
# myName.Country
rpartition returns a three element tuple,
1st element being the string before the separator
then the separator itself
and the the string after the separator
So, in our case,
s = 'myName.Country.myHeight'
print s.rpartition(".")
# ('myName.Country', '.', 'myHeight')
And we have picked only the first element.
Note: If you want to do it from the left, instead of doing it from the right, we have a sister function called str.partition.

You have a few options.
1
print s.rsplit('.',1)[0]
2
print s[:s.rfind('.')]
3
print s.rpartition('.')[0]

Well, that seems just fine to me... But here are a few other ways I can think of :
required = ".".join(s.split(".")[0:2]) // only one split
// using regular expressions
import re
required = re.sub(r"\.[^\.]$", "", s)
The regex only works if there are no dots in the last part you want to split off.