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Python how to get rid of a nested list

Good evening,
I have a python variable like so
myList = ["['Ben'", " 'Dillon'", " 'Rawr'", " 'Mega'", " 'Tote'", " 'Case']"]
I would like it to look like this instead
myList = ['Ben', 'Dillon', 'Rawr', 'Mega', 'Tote', 'Case']
If I do something like this
','.join(myList)
It gives me what I want but the type is a String
I also would like it to keep the type of List. I have tried using the Join method and split method. And I have been debugging use the type() method. It tells me that the type in the original scenario is a list.
I appreciate any and all help on this.

Join the inner list elements, then call ast.literal_eval() to parse it as a list of strings.
import ast
myList = ast.literal_eval(",".join(myList))

Also can be done by truncating Strings, therefore avoiding the import of ast.
myList[5] = (myList[5])[:-1]
for n in range(0, len(myList)):
myList[n] = (myList[n])[2:-1]

Getting data outside of a list in python

I have list with one item in it, then I try to dismantle, & rebuild it.
Not really sure if it is the 'right' way, but for now it will do.
I tried using replace \ substitute, other means of manipulating the list, but it didn't go too far, so this is what I came up with:
This is the list I get : alias_account = ['account-12345']
I then use this code to remove the [' in the front , and '] from the back.
NAME = ('%s' % alias_account).split(',')
for x in NAME:
key = x.split("-")[0]
value = x.split("-")[1]
alias_account = value[:-2]
alias_account1 = key[2:]
alias_account = ('%s-%s') % (alias_account1, alias_account)
This works beautifully when running print alias_account.
The problem starts when I have a list that have ['acc-ount-12345'] or ['account']
So my question is, how to include all of the possibilities?
Should I use try\except with other split options?
or is there more fancy split options ?

To access a single list element, you can index its position in square brackets:
alias_account[0]
To hide the quotes marking the result as a string, you can use print():
print(alias_account[0])

Python: Removing characters from a string and then returning it

For example, given a list of strings prices = ["US$200", "CA$80", "GA$500"],
I am trying to only return ["US", "CA", "GA"].
Here is my code - what am I doing wrong?
def get_country_codes(prices):
prices = ""
list = prices.split()
list.remove("$")
"".join(list)
return list

Since each of the strings in the prices argument has the form '[country_code]$[number]', you can split each of them on '$' and take the first part.
Here's an example of how you can do this:
def get_country_codes(prices):
return [p.split('$')[0] for p in prices]
So get_country_codes(['US$200', 'CA$80', 'GA$500']) returns ['US', 'CA', 'GA'].
Also as a side note, I would recommend against naming a variable list as this will override the built-in value of list, which is the type list itself.

There are multiple problems with your code, and you have to fix all of them to make it work:
def get_country_codes(prices):
prices = ""
Whatever value your caller passed in, you're throwing that away and replacing it with "". You don't want to do that, so just get rid of that last line.
list = prices.split()
You really shouldn't be calling this list list. Also, split with no argument splits on spaces, so what you get may not be what you want:
>>> "US$200, CA$80, GA$500".split()
['US$200,', 'CA$80,', 'GA$500']
I suppose you can get away with having those stray commas, since you're just going to throw them away. But it's better to split with your actual separators, the ', '. So, let's change that line:
prices = prices.split(", ")
list.remove("$")
This removes every value in the list that's equal to the string "$". There are no such values, so it does nothing.
More generally, you don't want to throw away any of the strings in the list. Instead, you want to replace the strings, with strings that are truncated at the $. So, you need a loop:
countries = []
for price in prices:
country, dollar, price = price.partition('$')
countries.append(country)
If you're familiar with list comprehensions, you can rewrite this as a one-liner:
countries = [price.partition('$')[0] for price in prices]
"".join(list)
This just creates a new string and then throws it away. You have to assign it to something if you want to use it, like this:
result = "".join(countries)
But… do you really want to join anything here? It sounds like you want the result to be a list of strings, ['US', 'CA', 'GA'], not one big string 'USCAGA', right? So, just get rid of this line.
return list
Just change the variable name to countries and you're done.

Since your data is structured where the first two characters are the county code you can use simple string slicing.
def get_country_codes(prices):
return [p[:2] for p in prices]

You call the function sending the prices parameter but your first line initialize to an empty string:
prices = ''
I would also suggest using the '$' character as the split character, like:
list = prices.split('$')
try something like this:
def get_country_codes(prices):
list = prices.split('$')
return list[0]

Python Lists -Syntax for '['

I need to declare certain values in List.
Values looks like this:
["compute","controller"], ["compute"] ,["controller"]
I know the List syntax in python is
example = []
I am not sure how I will include square brackets and double quotes in the List.
Could anyone please help.
I tried the following:
cls.node = ["\["compute"\]","\["controller"\]"]
cls.node = ["[\"compute\"]","[\"controller\"]"]
Both did not work.

I think you mean list not dictionary because that is the syntax of a list:
You can simply do it using the following format '"Hello"':
cls.node = ['["compute"]','["controller"]']
cls.node = ['["compute"]','["controller"]']
Demo:
s = ['["hello"]', '["world"]']
for i in s:
print i
[OUTPUT]
["hello"]
["world"]

How to convert a tuple to a string in Python?

After a MySQL select statement, I am left with the following:
set([('1#a.com',), ('2#b.net',), ('3#c.com',), ('4#d.com',), ('5#e.com',), ('6#f.net',), ('7#h.net',), ('8#g.com',)])
What I would like to have is a
emaillist = "\n".join(queryresult)
to in the end, have a string:
1#a.com
2#b.net
3#c.com
etc
What would be the proper way to convert this nested tuple into string?

As long as you're sure you have just one element per tuple:
'\n'.join(elem[0] for elem in queryresult)

A list comprehension along the lines of
[item[0] for item in queryresult]
should help. eg,
emaillist = "\n".join([item[0] for item in queryresult])
.. this makes strong assumptions about the type and structure of queryresult, however.
edit
A generator expression is even better at doing this:
emaillist = "\n".join(item[0] for item in queryresult)
thanks to #delnan for this update

Try this and see
>>> something=set([('1#a.com',), ('2#b.net',), ('3#c.com',), ('4#d.com',), ('5#e.com',), ('6#f.net',), ('7#h.net',), ('8#g.com',)])
>>> print '\n'.join(''.join(s) for s in something)
6#f.net
7#h.net
2#b.net
1#a.com
8#g.com
5#e.com
4#d.com
3#c.com
Note, join can only work on strings, but the items of the set are tuples. To make the join work, you have to iterate over the set to convert each item as string. Finally once done you can join them in the way your heart desires
On a side note. A circumbendibus way of doing it :-)
>>> print "\n".join(re.findall("\'(.*?)\'",pprint.pformat(something)))
1#a.com
2#b.net
3#c.com
4#d.com
5#e.com
6#f.net
7#h.net
8#g.com

new = '\n'.join(x[0] for x in old)