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How to get value of "string"?

Given strings like:
"hello"
'hello'
I want to remove only first and last char if:
They are the same
They are " or '
I.e., given 'hello' I'm expecting hello. Given 'hello" I'm not expecting it to change.
I was able to do this by reading first char and last char, validating they are the same + validating they are equal to ' or " and validating it's not the the same index for char (because I don't want this: ' to end up as the empty string). With all edge cases checking I ended with 10s of lines.
What's your approach to solve this?
In simple words, Given a string in Python format I want to return its data and if it's not valid to keep it as is.

Sounds like a job for regular expressions with groups:
import re
re.sub(r'^([\'"])(.*)(\1)$', r'\2', s)
Which reads as:
^ - match the beginning of the string
(['"]) - either single or double quote (group 1)
(.*) any (possibly, empty) sequence of characters in between (group 2)
(\1) - the same character as in group 1
$ - end of the string
If the string matches the pattern above, replace it with the content of the group 2.
For example:
>>> s = re.sub(r'^([\'"])(.*)(\1)$', r'\2', "'hello'")
>>> print(s)
hello
An alternative way could be with ast.literal_eval(), but it won't handle non-matching quotes.

I would use str.endswith and str.startswith, although it still gets a bit long:
def readstring(string):
if len(string)>1 and (string.startswith('"') and string.endswith('"') or string.startswith("'") and string.endswith("'")):
return string[1:-1]
return string

Split Python String by letters and keep deliminators

Using regex, how can i split a string and keep it's deliminators in the returned results? I'm trying to split a string containing numbers and strings by a set of letters followed by any numerical value including '.' however it's not appearing to work correctly.
Below is my test string, im using python 2.7 and it's not producing what id expect.
s = 'M160.394,83.962L121.5,52L86.31,73.378L58,104.917L89.75,C136.667L158.542,136.667L185,110.208L160.394,83.962Z'
parts = filter(None, re.split('([MLHVCSQTAZ][^MLHVCSQTAZ]+)', s, re.IGNORECASE))
print len(parts), parts
>>> 3 ['M160.394,83.962', 'L121.5,52', 'L86.31,73.378L58,104.917L89.75,C136.667L158.542,136.667L185,110.208L160.394,83.962Z']
I would expect it to give me this
>>> 10 ['M160.394,83.962', 'L121.5,52', 'L86.31,73.378', 'L58,104.917', 'L89.75,', 'C136.667', 'L158.542,136.667', 'L185,110.208', 'L160.394,83.962', 'Z']
It should output a list of strings where each string starts with a letter, found in the original regex MLHVCSQTAZ

In your code you are passing re.IGNORECASE as 3rd argument to re.split but 3rd argument of re.split is maxsplit not flags.
re.IGNORECASE equals to 2 hence your input is split only two times.
You may use:
>>> list(filter(None, re.split(r'([MLHVCSQTAZ][^MLHVCSQTAZ]+)', s, 0, re.I)))
['M160.394,83.962', 'L121.5,52', 'L86.31,73.378', 'L58,104.917', 'L89.75,', 'C136.667', 'L158.542,136.667', 'L185,110.208', 'L160.394,83.962', 'Z']
Or use inline mode for ignore case:
re.split(r'(?i)([MLHVCSQTAZ][^MLHVCSQTAZ]+)', s)
I suggest using this simple re.findall code that uses almost identical regex:
parts = re.findall('(?i)[MLHVCSQTAZ][^MLHVCSQTAZ]*', s)
Reference: SRE_FLAG_IGNORECASE = 2 in lib/python2.7/sre_constants.py (thanks to comment from #vks)

You can use re.findall:
import re
s = 'M160.394,83.962L121.5,52L86.31,73.378L58,104.917L89.75,C136.667L158.542,136.667L185,110.208L160.394,83.962Z'
result = re.findall('[A-Z][\.\d,]+|[A-Z]', s)
Output:
['M160.394,83.962', 'L121.5,52', 'L86.31,73.378', 'L58,104.917', 'L89.75,', 'C136.667', 'L158.542,136.667', 'L185,110.208', 'L160.394,83.962', 'Z']

parts = filter(None, re.split('([MLHVCSQTAZ][^MLHVCSQTAZ]+)', s, flags=re.IGNORECASE))
You need to use flags.Check re.split function definition.
Default re does not support 0 width assertion split.So you can also use regex module for that.
import regex
s = 'M160.394,83.962L121.5,52L86.31,73.378L58,104.917L89.75,C136.667L158.542,136.667L185,110.208L160.394,83.962Z'
print regex.split('(?=[MLHVCSQTAZ][^MLHVCSQTAZ])', s, flags=regex.IGNORECASE|regex.VERSION1)

remove all characters aside from the decimal numbers immediately followed by the £ sign

I have text with values like:
this is a value £28.99 (0.28/ml)
I want to remove everything to return the price only so it returns:
£28.99
there could be any number of digits between the £ and .
I think
r"£[0-9]*\.[0-9]{2}"
matches the pattern I want to keep but i'm unsure on how to remove everything else and keep the pattern instead of replacing the pattern like in usual re.sub() cases.

I want to remove everything to return the price only so it returns:
Why not trying to extract the proper information instead?
import re
s = "this is a value £28.99 (0.28/ml)"
m = re.search("£\d*(\.\d+)?",s)
if m:
print(m.group(0))
to find several occurrences use findall or finditer instead of search

You don't care how many digits are before the decimal, so using the zero-or-more matcher was correct. However, you could just rely on the digit class (\d) to provide that more succinctly.
The same is true of after the decimal. You only need two so your limiting the matches to 2 is correct.
The issue then comes in with how you actually capture the value. You can use a capturing group to be sure that you only ever get the value you care about.
Complete regex:
(£\d*.\d{2})
Sample code:
import re
r = re.compile("(£\d*.\d{2})")
match = r.findall("this is a value £28.99 (0.28/ml)")
if match: # may bring back an empty list; check for that here
print(match[0]) # uses the first group, and will print £28.99

If it's a string, you can do something like this:
x = "this is a value £28.99 (0.28/ml)"
x_list = x.split()
for i in x_list:
if "£" in i: #or if i.startswith("£") Credit – Jean-François Fabre
value=i
print(value)
>>>£28.99

You can try:
import re
t = "this is a value £28.99 (0.28/ml)"
r = re.sub(".*(£[\d.]+).*", r"\1", t)
print(r)
Output:
£28.99
Python Demo

python regular expression to match strings

I want to parse a string, such as:
package: name='jp.tjkapp.droid1lwp' versionCode='2' versionName='1.1'
uses-permission:'android.permission.WRITE_APN_SETTINGS'
uses-permission:'android.permission.RECEIVE_BOOT_COMPLETED'
uses-permission:'android.permission.ACCESS_NETWORK_STATE'
I want to get:
string1: jp.tjkapp.droidllwp`
string2: 1.1
Because there are multiple uses-permission, I want to get permission as a list, contains:
WRITE_APN_SETTINGS, RECEIVE_BOOT_COMPLETED and ACCESS_NETWORK_STATE.
Could you help me write the python regular expression to get the strings I want?
Thanks.

Assuming the code block you provided is one long string, here stored in a variable called input_string:
name = re.search(r"(?<=name\=\')[\w\.]+?(?=\')", input_string).group(0)
versionName = re.search(r"(?<=versionName\=\')\d+?\.\d+?(?=\')", input_string).group(0)
permissions = re.findall(r'(?<=android\.permission\.)[A-Z_]+(?=\')', input_string)
Explanation:
name
(?<=name\=\'): check ahead of the main string in order to return only strings that are preceded by name='. The \ in front of = and ' serve to escape them so that the regex knows we're talking about the = string and not a regex command. name=' is not also returned when we get the result, we just know that the results we get are all preceded by it.
[\w\.]+?: This is the main string we're searching for. \w means any alphanumeric character and underscore. \. is an escaped period, so the regex knows we mean . and not the regex command represented by an unescaped period. Putting these in [] means we're okay with anything we've stuck in brackets, so we're saying that we'll accept any alphanumeric character, _, or .. + afterwords means at least one of the previous thing, meaning at least one (but possibly more) of [\w\.]. Finally, the ? means don't be greedy--we're telling the regex to get the smallest possible group that meets these specifications, since + could go on for an unlimited number of repeats of anything matched by [\w\.].
(?=\'): check behind the main string in order to return only strings that are followed by '. The \ is also an escape, since otherwise regex or Python's string execution might misinterpret '. This final ' is not returned with our results, we just know that in the original string, it followed any result we do end up getting.

You can do this without regex by reading the file content line by line.
>>> def split_string(s):
... if s.startswith('package'):
... return [i.split('=')[1] for i in s.split() if "=" in i]
... elif s.startswith('uses-permission'):
... return s.split('.')[-1]
...
>>> split_string("package: name='jp.tjkapp.droid1lwp' versionCode='2' versionName='1.1'")
["'jp.tjkapp.droid1lwp'", "'2'", "'1.1'"]
>>> split_string("uses-permission:'android.permission.WRITE_APN_SETTINGS'")
"WRITE_APN_SETTINGS'"
>>> split_string("uses-permission:'android.permission.RECEIVE_BOOT_COMPLETED'")
"RECEIVE_BOOT_COMPLETED'"
>>> split_string("uses-permission:'android.permission.ACCESS_NETWORK_STATE'")
"ACCESS_NETWORK_STATE'"
>>>

Here is one example code
#!/usr/bin/env python
inputFile = open("test.txt", "r").readlines()
for line in inputFile:
if line.startswith("package"):
words = line.split()
string1 = words[1].split("=")[1].replace("'","")
string2 = words[3].split("=")[1].replace("'","")
test.txt file contains input data you mentioned earlier..

python regex for repeating string

I am wanting to verify and then parse this string (in quotes):
string = "start: c12354, c3456, 34526; other stuff that I don't care about"
//Note that some codes begin with 'c'
I would like to verify that the string starts with 'start:' and ends with ';'
Afterward, I would like to have a regex parse out the strings. I tried the following python re code:
regx = r"start: (c?[0-9]+,?)+;"
reg = re.compile(regx)
matched = reg.search(string)
print ' matched.groups()', matched.groups()
I have tried different variations but I can either get the first or the last code but not a list of all three.
Or should I abandon using a regex?
EDIT: updated to reflect part of the problem space I neglected and fixed string difference.
Thanks for all the suggestions - in such a short time.

In Python, this isn’t possible with a single regular expression: each capture of a group overrides the last capture of that same group (in .NET, this would actually be possible since the engine distinguishes between captures and groups).
Your easiest solution is to first extract the part between start: and ; and then using a regular expression to return all matches, not just a single match, using re.findall('c?[0-9]+', text).

You could use the standard string tools, which are pretty much always more readable.
s = "start: c12354, c3456, 34526;"
s.startswith("start:") # returns a boolean if it starts with this string
s.endswith(";") # returns a boolean if it ends with this string
s[6:-1].split(', ') # will give you a list of tokens separated by the string ", "

This can be done (pretty elegantly) with a tool like Pyparsing:
from pyparsing import Group, Literal, Optional, Word
import string
code = Group(Optional(Literal("c"), default='') + Word(string.digits) + Optional(Literal(","), default=''))
parser = Literal("start:") + OneOrMore(code) + Literal(";")
# Read lines from file:
with open('lines.txt', 'r') as f:
for line in f:
try:
result = parser.parseString(line)
codes = [c[1] for c in result[1:-1]]
# Do something with teh codez...
except ParseException exc:
# Oh noes: string doesn't match!
continue
Cleaner than a regular expression, returns a list of codes (no need to string.split), and ignores any extra characters in the line, just like your example.

import re
sstr = re.compile(r'start:([^;]*);')
slst = re.compile(r'(?:c?)(\d+)')
mystr = "start: c12354, c3456, 34526; other stuff that I don't care about"
match = re.match(sstr, mystr)
if match:
res = re.findall(slst, match.group(0))
results in
['12354', '3456', '34526']