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Split string into list by separate delimiters, but only by certain instances of said delimiters [duplicate]

This question already has answers here:
Split string with multiple delimiters in Python [duplicate]
(5 answers)
Closed 2 years ago.
I think what I'm trying to achieve is fairly common but I can't find reference for it on the internet; either that or I'm misphrasing what I'm trying to do.
This is the string I would like to split:
array_1:target:radec, 0:00:00.00, -90:00:00.0
I would like to split it by the first two colons (':'), and by the first comma & space (', '), such that I get
['array_1', 'target', 'radec', '0:00:00.00, -90:00:00.0']
I've tried to run split() with arguments twice on the original string, and it fails on the second split() because I'm trying to split something that's already a list. All the other answers I can find seem to focus on splitting the string by all instances of a delimiter, but I want the last field in the list 0:00:00.00, -90:00:00.0 to remain like it is.

First split it by the first ", " (using maxsplit=1), then the first element of the resulting list split by ":":
s = "array_1:target:radec, 0:00:00.00, -90:00:00.0"
temp = s.split(", ", maxsplit=1)
temp[0] = temp[0].split(":")
result = temp[0] + [temp[1]]
The result:
['array_1', 'target', 'radec', '0:00:00.00, -90:00:00.0']

How about
l1 = s.split()
l2 = l1[0].split(':') + l1[1:]
This will first split by whitespace separator, then split the first element (only) by a colon separator, and then join the lists. Result:
['array_1', 'target', 'radec,', '0:00:00.00,', '-90:00:00.0']

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.

Python: What is the Best way to split a string of 9 characters into 3 characters each and join them using delimiters? [duplicate]

This question already has answers here:
How to iterate over a list in chunks
(39 answers)
Closed 8 years ago.
I have a string "111222333" inside a CSV file. I would like to convert this into something like "\111\222\333"
Currently my python code is :
refcode = "111222333"
returnstring = "\\" + refcode[:3] + "\\" + refcode[3:6] + "\\" + refcode[-3:] + "\\"
I know there must be a better way to do this. May I know what are the better ways to do the same thing. Please help.

You could use re for that:
import re
refcode = "111222333"
returnstring = '\\'.join(re.match('()(\S{3})(\S{3})(\S{3})()', refcode).groups())
Explanation:
You have a string of 9 characters (let's say they are not any kind of whitespace chatacters, so we could represent it with \S).
We create a matching regexp using it, so (\S{3}) is a group of three sequential non-space characters (like letters, numbers, exclamation marks etc.).
(\S{3})(\S{3})(\S{3}) are three groups with 3 characters in each one.
If we call .groups() on it, we'll have a tuple of the matched groups, just like that:
In [1]: re.match('(\S{3})(\S{3})(\S{3})', refcode).groups()
Out[1]: ('111', '222', '333')
If we join it using a \ string, we'll get a:
In [29]: print "\\".join(re.match('(\S{3})(\S{3})(\S{3})', refcode).groups())
111\222\333
But you want to add the backslashes on the both sides of the string as well!
So we could create an empty group - () - on the each side of the regular expression.

How to split strings by only the first special character that's within the string? [duplicate]

This question already has answers here:
Splitting on first occurrence
(5 answers)
Closed 8 years ago.
I'd like to split the below string by only the first equal sign in that string
String:
s= 'ButtonParams=U3ViamVjdCxFbWFpbA=='
Desired_String:
s= ['ButtonParams','U3ViamVjdCxFbWFpbA==']
When I do s.split("="), I get the following, which is what I do not want:
s.split("=")
['ButtonParams', 'U3ViamVjdCxFbWFpbA', '', '']
I need to run this function across a list of strings where this is the case, so scalability is important here.

split accepts an optional "maxsplit" parameter: if you set it to 1 it will split on the first = character it finds and return the remainder of the string:
>>> s.split('=', 1)
['ButtonParams', 'U3ViamVjdCxFbWFpbA==']

s.split("=", maxsplit=1) is the best but
import re
print (re.split('=',s,1))
The output is
['ButtonParams', 'U3ViamVjdCxFbWFpbA==']
As you have tagged regex
A little deviation from the post
If the expected output was ['ButtonParams', 'U3ViamVjdCxFbWFpbA'] then you can have the following liscos (list comprehensions)
[i for i in s.split('=') if i is not '']
[i for i in s.split('=') if i ] (Contributed by Adam Smith)

str.split accepts an optional argument maxsplit which determines how many splits (maximally) to make, e.g.:
"ButtonParams=U3ViamVjdCxFbWFpbA==".split("=", maxsplit=1)
# ['ButtonParams', 'U3ViamVjdCxFbWFpbA==']

Select last chars of string until whitespace in Python [duplicate]

This question already has answers here:
Python: Cut off the last word of a sentence?
(10 answers)
Closed 8 years ago.
Is there any efficient way to select the last characters of a string until there's a whitespace in Python?
For example I have the following string:
str = 'Hello my name is John'
I want to return 'John'. But if the str was:
str = 'Hello my name is Sally'
I want to retrun 'Sally'

Just split the string on whitespace, and get the last element of the array. Or use rsplit() to start splitting from end:
>>> st = 'Hello my name is John'
>>> st.rsplit(' ', 1)
['Hello my name is', 'John']
>>>
>>> st.rsplit(' ', 1)[1]
'John'
The 2nd argument specifies the number of split to do. Since you just want last element, we just need to split once.
As specified in comments, you can just pass None as 1st argument, in which case the default delimiter which is whitespace will be used:
>>> st.rsplit(None, 1)[-1]
'John'
Using -1 as index is safe, in case there is no whitespace in your string.

It really depends what you mean by efficient, but the simplest (efficient use of programmer time) way I can think of is:
str.split()[-1]
This fails for empty strings, so you'll want to check that.

I think this is what you want:
str[str.rfind(' ')+1:]
this creates a substring from str starting at the character after the right-most-found-space, and up until the last character.
This works for all strings - empty or otherwise (unless it's not a string object, e.g. a None object would throw an error)