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how to replace a comma in python, which is pressed to the letter [duplicate]

I'm trying to remove specific characters from a string using Python. This is the code I'm using right now. Unfortunately it appears to do nothing to the string.
for char in line:
if char in " ?.!/;:":
line.replace(char,'')
How do I do this properly?

Strings in Python are immutable (can't be changed). Because of this, the effect of line.replace(...) is just to create a new string, rather than changing the old one. You need to rebind (assign) it to line in order to have that variable take the new value, with those characters removed.
Also, the way you are doing it is going to be kind of slow, relatively. It's also likely to be a bit confusing to experienced pythonators, who will see a doubly-nested structure and think for a moment that something more complicated is going on.
Starting in Python 2.6 and newer Python 2.x versions *, you can instead use str.translate, (see Python 3 answer below):
line = line.translate(None, '!##$')
or regular expression replacement with re.sub
import re
line = re.sub('[!##$]', '', line)
The characters enclosed in brackets constitute a character class. Any characters in line which are in that class are replaced with the second parameter to sub: an empty string.
Python 3 answer
In Python 3, strings are Unicode. You'll have to translate a little differently. kevpie mentions this in a comment on one of the answers, and it's noted in the documentation for str.translate.
When calling the translate method of a Unicode string, you cannot pass the second parameter that we used above. You also can't pass None as the first parameter. Instead, you pass a translation table (usually a dictionary) as the only parameter. This table maps the ordinal values of characters (i.e. the result of calling ord on them) to the ordinal values of the characters which should replace them, or—usefully to us—None to indicate that they should be deleted.
So to do the above dance with a Unicode string you would call something like
translation_table = dict.fromkeys(map(ord, '!##$'), None)
unicode_line = unicode_line.translate(translation_table)
Here dict.fromkeys and map are used to succinctly generate a dictionary containing
{ord('!'): None, ord('#'): None, ...}
Even simpler, as another answer puts it, create the translation table in place:
unicode_line = unicode_line.translate({ord(c): None for c in '!##$'})
Or, as brought up by Joseph Lee, create the same translation table with str.maketrans:
unicode_line = unicode_line.translate(str.maketrans('', '', '!##$'))
* for compatibility with earlier Pythons, you can create a "null" translation table to pass in place of None:
import string
line = line.translate(string.maketrans('', ''), '!##$')
Here string.maketrans is used to create a translation table, which is just a string containing the characters with ordinal values 0 to 255.

Am I missing the point here, or is it just the following:
string = "ab1cd1ef"
string = string.replace("1", "")
print(string)
# result: "abcdef"
Put it in a loop:
a = "a!b#c#d$"
b = "!##$"
for char in b:
a = a.replace(char, "")
print(a)
# result: "abcd"

>>> line = "abc##!?efg12;:?"
>>> ''.join( c for c in line if c not in '?:!/;' )
'abc##efg12'

With re.sub regular expression
Since Python 3.5, substitution using regular expressions re.sub became available:
import re
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
Example
import re
line = 'Q: Do I write ;/.??? No!!!'
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
'QDoIwriteNo'
Explanation
In regular expressions (regex), | is a logical OR and \ escapes spaces and special characters that might be actual regex commands. Whereas sub stands for substitution, in this case with the empty string ''.

The asker almost had it. Like most things in Python, the answer is simpler than you think.
>>> line = "H E?.LL!/;O:: "
>>> for char in ' ?.!/;:':
... line = line.replace(char,'')
...
>>> print line
HELLO
You don't have to do the nested if/for loop thing, but you DO need to check each character individually.

For the inverse requirement of only allowing certain characters in a string, you can use regular expressions with a set complement operator [^ABCabc]. For example, to remove everything except ascii letters, digits, and the hyphen:
>>> import string
>>> import re
>>>
>>> phrase = ' There were "nine" (9) chick-peas in my pocket!!! '
>>> allow = string.letters + string.digits + '-'
>>> re.sub('[^%s]' % allow, '', phrase)
'Therewerenine9chick-peasinmypocket'
From the python regular expression documentation:
Characters that are not within a range can be matched by complementing
the set. If the first character of the set is '^', all the characters
that are not in the set will be matched. For example, [^5] will match
any character except '5', and [^^] will match any character except
'^'. ^ has no special meaning if it’s not the first character in the
set.

line = line.translate(None, " ?.!/;:")

>>> s = 'a1b2c3'
>>> ''.join(c for c in s if c not in '123')
'abc'

Strings are immutable in Python. The replace method returns a new string after the replacement. Try:
for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
This is identical to your original code, with the addition of an assignment to line inside the loop.
Note that the string replace() method replaces all of the occurrences of the character in the string, so you can do better by using replace() for each character you want to remove, instead of looping over each character in your string.

I was surprised that no one had yet recommended using the builtin filter function.
import operator
import string # only for the example you could use a custom string
s = "1212edjaq"
Say we want to filter out everything that isn't a number. Using the filter builtin method "...is equivalent to the generator expression (item for item in iterable if function(item))" [Python 3 Builtins: Filter]
sList = list(s)
intsList = list(string.digits)
obj = filter(lambda x: operator.contains(intsList, x), sList)))
In Python 3 this returns
>> <filter object # hex>
To get a printed string,
nums = "".join(list(obj))
print(nums)
>> "1212"
I am not sure how filter ranks in terms of efficiency but it is a good thing to know how to use when doing list comprehensions and such.
UPDATE
Logically, since filter works you could also use list comprehension and from what I have read it is supposed to be more efficient because lambdas are the wall street hedge fund managers of the programming function world. Another plus is that it is a one-liner that doesnt require any imports. For example, using the same string 's' defined above,
num = "".join([i for i in s if i.isdigit()])
That's it. The return will be a string of all the characters that are digits in the original string.
If you have a specific list of acceptable/unacceptable characters you need only adjust the 'if' part of the list comprehension.
target_chars = "".join([i for i in s if i in some_list])
or alternatively,
target_chars = "".join([i for i in s if i not in some_list])

Using filter, you'd just need one line
line = filter(lambda char: char not in " ?.!/;:", line)
This treats the string as an iterable and checks every character if the lambda returns True:
>>> help(filter)
Help on built-in function filter in module __builtin__:
filter(...)
filter(function or None, sequence) -> list, tuple, or string
Return those items of sequence for which function(item) is true. If
function is None, return the items that are true. If sequence is a tuple
or string, return the same type, else return a list.

Try this one:
def rm_char(original_str, need2rm):
''' Remove charecters in "need2rm" from "original_str" '''
return original_str.translate(str.maketrans('','',need2rm))
This method works well in Python 3

Here's some possible ways to achieve this task:
def attempt1(string):
return "".join([v for v in string if v not in ("a", "e", "i", "o", "u")])
def attempt2(string):
for v in ("a", "e", "i", "o", "u"):
string = string.replace(v, "")
return string
def attempt3(string):
import re
for v in ("a", "e", "i", "o", "u"):
string = re.sub(v, "", string)
return string
def attempt4(string):
return string.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "")
for attempt in [attempt1, attempt2, attempt3, attempt4]:
print(attempt("murcielago"))
PS: Instead using " ?.!/;:" the examples use the vowels... and yeah, "murcielago" is the Spanish word to say bat... funny word as it contains all the vowels :)
PS2: If you're interested on performance you could measure these attempts with a simple code like:
import timeit
K = 1000000
for i in range(1,5):
t = timeit.Timer(
f"attempt{i}('murcielago')",
setup=f"from __main__ import attempt{i}"
).repeat(1, K)
print(f"attempt{i}",min(t))
In my box you'd get:
attempt1 2.2334518376057244
attempt2 1.8806643818474513
attempt3 7.214925774955572
attempt4 1.7271184513757465
So it seems attempt4 is the fastest one for this particular input.

Here's my Python 2/3 compatible version. Since the translate api has changed.
def remove(str_, chars):
"""Removes each char in `chars` from `str_`.
Args:
str_: String to remove characters from
chars: String of to-be removed characters
Returns:
A copy of str_ with `chars` removed
Example:
remove("What?!?: darn;", " ?.!:;") => 'Whatdarn'
"""
try:
# Python2.x
return str_.translate(None, chars)
except TypeError:
# Python 3.x
table = {ord(char): None for char in chars}
return str_.translate(table)

#!/usr/bin/python
import re
strs = "how^ much for{} the maple syrup? $20.99? That's[] ricidulous!!!"
print strs
nstr = re.sub(r'[?|$|.|!|a|b]',r' ',strs)#i have taken special character to remove but any #character can be added here
print nstr
nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr)#for removing special character
print nestr

You can also use a function in order to substitute different kind of regular expression or other pattern with the use of a list. With that, you can mixed regular expression, character class, and really basic text pattern. It's really useful when you need to substitute a lot of elements like HTML ones.
*NB: works with Python 3.x
import re # Regular expression library
def string_cleanup(x, notwanted):
for item in notwanted:
x = re.sub(item, '', x)
return x
line = "<title>My example: <strong>A text %very% $clean!!</strong></title>"
print("Uncleaned: ", line)
# Get rid of html elements
html_elements = ["<title>", "</title>", "<strong>", "</strong>"]
line = string_cleanup(line, html_elements)
print("1st clean: ", line)
# Get rid of special characters
special_chars = ["[!##$]", "%"]
line = string_cleanup(line, special_chars)
print("2nd clean: ", line)
In the function string_cleanup, it takes your string x and your list notwanted as arguments. For each item in that list of elements or pattern, if a substitute is needed it will be done.
The output:
Uncleaned: <title>My example: <strong>A text %very% $clean!!</strong></title>
1st clean: My example: A text %very% $clean!!
2nd clean: My example: A text very clean

My method I'd use probably wouldn't work as efficiently, but it is massively simple. I can remove multiple characters at different positions all at once, using slicing and formatting.
Here's an example:
words = "things"
removed = "%s%s" % (words[:3], words[-1:])
This will result in 'removed' holding the word 'this'.
Formatting can be very helpful for printing variables midway through a print string. It can insert any data type using a % followed by the variable's data type; all data types can use %s, and floats (aka decimals) and integers can use %d.
Slicing can be used for intricate control over strings. When I put words[:3], it allows me to select all the characters in the string from the beginning (the colon is before the number, this will mean 'from the beginning to') to the 4th character (it includes the 4th character). The reason 3 equals till the 4th position is because Python starts at 0. Then, when I put word[-1:], it means the 2nd last character to the end (the colon is behind the number). Putting -1 will make Python count from the last character, rather than the first. Again, Python will start at 0. So, word[-1:] basically means 'from the second last character to the end of the string.
So, by cutting off the characters before the character I want to remove and the characters after and sandwiching them together, I can remove the unwanted character. Think of it like a sausage. In the middle it's dirty, so I want to get rid of it. I simply cut off the two ends I want then put them together without the unwanted part in the middle.
If I want to remove multiple consecutive characters, I simply shift the numbers around in the [] (slicing part). Or if I want to remove multiple characters from different positions, I can simply sandwich together multiple slices at once.
Examples:
words = "control"
removed = "%s%s" % (words[:2], words[-2:])
removed equals 'cool'.
words = "impacts"
removed = "%s%s%s" % (words[1], words[3:5], words[-1])
removed equals 'macs'.
In this case, [3:5] means character at position 3 through character at position 5 (excluding the character at the final position).
Remember, Python starts counting at 0, so you will need to as well.

In Python 3.5
e.g.,
os.rename(file_name, file_name.translate({ord(c): None for c in '0123456789'}))
To remove all the number from the string

How about this:
def text_cleanup(text):
new = ""
for i in text:
if i not in " ?.!/;:":
new += i
return new

Below one.. with out using regular expression concept..
ipstring ="text with symbols!##$^&*( ends here"
opstring=''
for i in ipstring:
if i.isalnum()==1 or i==' ':
opstring+=i
pass
print opstring

Recursive split:
s=string ; chars=chars to remove
def strip(s,chars):
if len(s)==1:
return "" if s in chars else s
return strip(s[0:int(len(s)/2)],chars) + strip(s[int(len(s)/2):len(s)],chars)
example:
print(strip("Hello!","lo")) #He!

You could use the re module's regular expression replacement. Using the ^ expression allows you to pick exactly what you want from your string.
import re
text = "This is absurd!"
text = re.sub("[^a-zA-Z]","",text) # Keeps only Alphabets
print(text)
Output to this would be "Thisisabsurd". Only things specified after the ^ symbol will appear.

# for each file on a directory, rename filename
file_list = os.listdir (r"D:\Dev\Python")
for file_name in file_list:
os.rename(file_name, re.sub(r'\d+','',file_name))

Even the below approach works
line = "a,b,c,d,e"
alpha = list(line)
while ',' in alpha:
alpha.remove(',')
finalString = ''.join(alpha)
print(finalString)
output: abcde

The string method replace does not modify the original string. It leaves the original alone and returns a modified copy.
What you want is something like: line = line.replace(char,'')
def replace_all(line, )for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
return line
However, creating a new string each and every time that a character is removed is very inefficient. I recommend the following instead:
def replace_all(line, baddies, *):
"""
The following is documentation on how to use the class,
without reference to the implementation details:
For implementation notes, please see comments begining with `#`
in the source file.
[*crickets chirp*]
"""
is_bad = lambda ch, baddies=baddies: return ch in baddies
filter_baddies = lambda ch, *, is_bad=is_bad: "" if is_bad(ch) else ch
mahp = replace_all.map(filter_baddies, line)
return replace_all.join('', join(mahp))
# -------------------------------------------------
# WHY `baddies=baddies`?!?
# `is_bad=is_bad`
# -------------------------------------------------
# Default arguments to a lambda function are evaluated
# at the same time as when a lambda function is
# **defined**.
#
# global variables of a lambda function
# are evaluated when the lambda function is
# **called**
#
# The following prints "as yellow as snow"
#
# fleece_color = "white"
# little_lamb = lambda end: return "as " + fleece_color + end
#
# # sometime later...
#
# fleece_color = "yellow"
# print(little_lamb(" as snow"))
# --------------------------------------------------
replace_all.map = map
replace_all.join = str.join

If you want your string to be just allowed characters by using ASCII codes, you can use this piece of code:
for char in s:
if ord(char) < 96 or ord(char) > 123:
s = s.replace(char, "")
It will remove all the characters beyond a....z even upper cases.

How to extract a substring from a string in Python 3

I am trying to pull a substring out of a function result, but I'm having trouble figuring out the best way to strip the necessary string out using Python.
Output Example:
[<THIS STRING-STRING-STRING THAT THESE THOSE>]
In this example, I would like to grab "STRING-STRING-STRING" and throw away all the rest of the output. In this example, "[<THIS " &" THAT THESE THOSE>]" are static.

Many many ways to solve this. Here are two examples:
First one is a simple replacement of your unwanted characters.
targetstring = '[<THIS STRING-STRING-STRING THAT THESE THOSE>]'
#ALTERNATIVE 1
newstring = targetstring.replace(r" THAT THESE THOSE>]", '').replace(r"[<THIS ", '')
print(newstring)
and this drops everything except your target pattern:
#ALTERNATIVE 2
match = "STRING-STRING-STRING"
start = targetstring.find(match)
stop = len(match)
targetstring[start:start+stop]
These can be shortened but thought it might be useful for OP to have them written out.
I found this extremely useful, might be of help to you as well: https://www.computerhope.com/issues/ch001721.htm

If by '"[<THIS " &" THAT THESE THOSE>]" are static' you mean that they are always the exact same string, then:
s = "[<THIS STRING-STRING-STRING THAT THESE THOSE>]"
before = len("[<THIS ")
after = len(" THAT THESE THOSE>]")
s[before:-after]
# 'STRING-STRING-STRING'

Like so (as long as the postition of the characters in the string doesn't change):
myString = "[<THIS STRING-STRING-STRING THAT THESE THOSE>]"
myString = myString[7:27]

Another alternative method;
import re
my_str = "[<THIS STRING-STRING-STRING THAT THESE THOSE>]"
string_pos = [(s.start(), s.end()) for s in list(re.finditer('STRING-STRING-STRING', my_str))]
start, end = string_pos[0]
print(my_str[start: end + 1])
STRING-STRING-STRING
If the STRING-STRING-STRING occurs multiple times in the string, start and end indexes of the each occurrences will be given as tuples in string_pos.

Remove variable parts of a string that start and end the same

I have a string as the following:
'1:CH,AG,ME,GS,AP,CH,HE,AC,AC,AG,CA,HE,AT,AT,AC,AT,OG,NE,AG,AC,CS,OD\n&:TA,EB,PA,AC,BR,TH,PO,AC,2I,AC,TH,PE,TH,AZ,AZ,ZE,CS,OD,CH,EO,ZE,OG\n&:TH,ZE,ZE,HE,HE,HP,HP,OG,HP,ZE\n2:ZE,FD,FD,AG,EO,OG,AG,NE,RU,GS,HP,ZE,ZE,HM,HM,PC,PC,AS,AS,TY,TY,AG\n&:AG,GS,NO,EU,ZF,HE,AT,AT,OD,OD,EB,OD,GS,TR,OD,AC,TR,GS,OD,TR,OD,AT,GS\n&:CA,GS,NE,GS,AG,PS,HL,AG,NE,ID,AJ,AX,DI,OD,ME,AT,GS,MU,HO,PB,LT,9Z,PT,9Y\n&:9W,9X,AR,9V,9U,9T,AX,9S,9R,AT,AJ,DI,ST,EA,AG,ME,NE,MU,9Q,9P,9O,9N,9M,9L\n&:9K,ID,MG,OD,FY,AU,AU,HR,HR,9J,TL,9I,9H,9G,9F,AC,BR,AC,9E,9D,9C,9B,99\n'
As you can see, I would like to get the '\n(number or & here):' replaced by ','
Since they all start with '\n' and end with ':' I believe that there should be a way to replace them all at once.
The output would be as the sort:
'CH,AG,ME,GS,AP,CH,HE,AC,AC,AG,CA,HE,AT,AT,AC,AT,OG,NE,AG,AC,CS,OD,TA,EB,PA,AC,BR,TH,PO,AC,2I,AC,TH,PE,TH,AZ,AZ,ZE,CS,OD,CH,EO,ZE,OG,TH,ZE,ZE,HE,HE,HP,HP,OG,HP,ZE,ZE,FD,FD,AG,EO,OG,AG,NE,RU,GS,HP,ZE,ZE,HM,HM,PC,PC,AS,AS,TY,TY,AG,AG,GS,NO,EU,ZF,HE,AT,AT,OD,OD,EB,OD,GS,TR,OD,AC,TR,GS,OD,TR,OD,AT,GS,CA,GS,NE,GS,AG,PS,HL,AG,NE,ID,AJ,AX,DI,OD,ME,AT,GS,MU,HO,PB,LT,9Z,PT,9Y,9W,9X,AR,9V,9U,9T,AX,9S,9R,AT,AJ,DI,ST,EA,AG,ME,NE,MU,9Q,9P,9O,9N,9M,9L,9K,ID,MG,OD,FY,AU,AU,HR,HR,9J,TL,9I,9H,9G,9F,AC,BR,AC,9E,9D,9C,9B,99'
What could work was making a for lop for numbers and &.
string.replace('\n&:',',')
for i in range(1,20):
string.replace('\ni:',',')
But I believe there must be a better way.

You can use regex to get the job done:
Input:
import re
text = '1:CH,AG,ME,GS,AP,CH,HE,AC,AC,AG,CA,HE,AT,AT,AC,AT,OG,NE,AG,AC,CS,OD\n&:TA,EB,PA,AC,BR,TH,PO,AC,2I,AC,TH,PE,TH,AZ,AZ,ZE,CS,OD,CH,EO,ZE,OG\n&:TH,ZE,ZE,HE,HE,HP,HP,OG,HP,ZE\n2:ZE,FD,FD,AG,EO,OG,AG,NE,RU,GS,HP,ZE,ZE,HM,HM,PC,PC,AS,AS,TY,TY,AG\n&:AG,GS,NO,EU,ZF,HE,AT,AT,OD,OD,EB,OD,GS,TR,OD,AC,TR,GS,OD,TR,OD,AT,GS\n&:CA,GS,NE,GS,AG,PS,HL,AG,NE,ID,AJ,AX,DI,OD,ME,AT,GS,MU,HO,PB,LT,9Z,PT,9Y\n&:9W,9X,AR,9V,9U,9T,AX,9S,9R,AT,AJ,DI,ST,EA,AG,ME,NE,MU,9Q,9P,9O,9N,9M,9L\n&:9K,ID,MG,OD,FY,AU,AU,HR,HR,9J,TL,9I,9H,9G,9F,AC,BR,AC,9E,9D,9C,9B,99\n'
text = re.sub(r'\n&*(\d*:)*',',', text[2:]).rstrip(',')
Output:
'CH,AG,ME,GS,AP,CH,HE,AC,AC,AG,CA,HE,AT,AT,AC,AT,OG,NE,AG,AC,CS,OD,TA,EB,PA,AC,BR,TH,PO,AC,2I,AC,TH,PE,TH,AZ,AZ,ZE,CS,OD,CH,EO,ZE,OG,TH,ZE,ZE,HE,HE,HP,HP,OG,HP,ZE,ZE,FD,FD,AG,EO,OG,AG,NE,RU,GS,HP,ZE,ZE,HM,HM,PC,PC,AS,AS,TY,TY,AG,AG,GS,NO,EU,ZF,HE,AT,AT,OD,OD,EB,OD,GS,TR,OD,AC,TR,GS,OD,TR,OD,AT,GS,CA,GS,NE,GS,AG,PS,HL,AG,NE,ID,AJ,AX,DI,OD,ME,AT,GS,MU,HO,PB,LT,9Z,PT,9Y,9W,9X,AR,9V,9U,9T,AX,9S,9R,AT,AJ,DI,ST,EA,AG,ME,NE,MU,9Q,9P,9O,9N,9M,9L,9K,ID,MG,OD,FY,AU,AU,HR,HR,9J,TL,9I,9H,9G,9F,AC,BR,AC,9E,9D,9C,9B,99'

You can use a regular expression replace:
s = '1:CH,AG,ME,GS,AP,CH,HE,AC,AC,AG,CA,HE,AT,AT,AC,AT,OG,NE,AG,AC,CS,OD\n&:TA,EB,PA,AC,BR,TH,PO,AC,2I,AC,TH,PE,TH,AZ,AZ,ZE,CS,OD,CH,EO,ZE,OG\n&:TH,ZE,ZE,HE,HE,HP,HP,OG,HP,ZE\n2:ZE,FD,FD,AG,EO,OG,AG,NE,RU,GS,HP,ZE,ZE,HM,HM,PC,PC,AS,AS,TY,TY,AG\n&:AG,GS,NO,EU,ZF,HE,AT,AT,OD,OD,EB,OD,GS,TR,OD,AC,TR,GS,OD,TR,OD,AT,GS\n&:CA,GS,NE,GS,AG,PS,HL,AG,NE,ID,AJ,AX,DI,OD,ME,AT,GS,MU,HO,PB,LT,9Z,PT,9Y\n&:9W,9X,AR,9V,9U,9T,AX,9S,9R,AT,AJ,DI,ST,EA,AG,ME,NE,MU,9Q,9P,9O,9N,9M,9L\n&:9K,ID,MG,OD,FY,AU,AU,HR,HR,9J,TL,9I,9H,9G,9F,AC,BR,AC,9E,9D,9C,9B,99\n'
s = re.sub(r"(\n\d*?:)|(\n&:)", ",", s).strip() # replaces the middle bits with commas and strips trailing \n
s = re.sub(r"^(\d*?:)|(&:)", "", s) # removes the initial 1: or similar

Excluding a specific string of characters in a str()-function

A small issue I've encountered during coding.
I'm looking to print out the name of a .txt file.
For example, the file is named: verdata_florida.txt, or verdata_newyork.txt
How can I exclude .txt and verdata_, but keep the string between? It must work for any number of characters, but .txt and verdata_ must be excluded.
This is where I am so far, I've already defined filename to be input()
print("Average TAM at", str(filename[8:**????**]), "is higher than ")

3 ways of doing it:
using str.split twice:
>>> "verdata_florida.txt".split("_")[1].split(".")[0]
'florida'
using str.partition twice (you won't get an exception if the format doesn't match, and probably faster too):
>>> "verdata_florida.txt".partition("_")[2].partition(".")[0]
'florida'
using re, keeping only center part:
>>> import re
>>> re.sub(".*_(.*)\..*",r"\1","verdata_florida.txt")
'florida'
all those above must be tuned if _ and . appear multiple times (must we keep the longest or the shortest string)
EDIT: In your case, though, prefixes & suffixes seem fixed. In that case, just use str.replace twice:
>>> "verdata_florida.txt".replace("verdata_","").replace(".txt","")
'florida'

Assuming you want it to split on the first _ and the last . you can use slicing and the index and rindex functions to get this done. These functions will search for the first occurrence of the substring in the parenthesis and return the index number. If no substring is found, they will throw a ValueError. If the search is desired, but not the ValueError, you can also use find and rfind, which do the same thing but always return -1 if no match is found.
s = 'verdata_new_hampshire.txt'
s_trunc = s[s.index('_') + 1: s.rindex('.')] # or s[s.find('_') + 1: s.rfind('.')]
print(s_trunc) # new_hampshire
Of course, if you are always going to exclude verdata_ and .txt you could always hardcode the slice as well.
print(s[8:-4]) # new_hampshire

You can leverage str.split() on strings. For example:
s = 'verdata_newyork.txt'
s.split('verdata_')
# ['', 'florida.txt']
s.split('verdata_')[1]
# 'florida.txt'
s.split('verdata_')[1].split('.txt')
['florida', '']
s.split('verdata_')[1].split('.txt')[0]
# 'florida'

You can just split string by dot and underscore like this:
string filename = "verdata_prague.txt";
string name = filename.split("."); //verdata_prague
name = name[0].split("_")[1]; //prague
or by replace function:
string filename = "verdata_prague.txt";
string name = filename.replace(".txt",""); //verdata_prague
name = name[0].replace("verdata_","")[1]; //prague

Strip in Python

I have a question regarding strip() in Python. I am trying to strip a semi-colon from a string, I know how to do this when the semi-colon is at the end of the string, but how would I do it if it is not the last element, but say the second to last element.
eg:
1;2;3;4;\n
I would like to strip that last semi-colon.

Strip the other characters as well.
>>> '1;2;3;4;\n'.strip('\n;')
'1;2;3;4'

>>> "".join("1;2;3;4;\n".rpartition(";")[::2])
'1;2;3;4\n'

how about replace?
string1='1;2;3;4;\n'
string2=string1.replace(";\n","\n")

>>> string = "1;2;3;4;\n"
>>> string.strip().strip(";")
"1;2;3;4"
This will first strip any leading or trailing white space, and then remove any leading or trailing semicolon.

Try this:
def remove_last(string):
index = string.rfind(';')
if index == -1:
# Semi-colon doesn't exist
return string
return string[:index] + string[index+1:]
This should be able to remove the last semicolon of the line, regardless of what characters come after it.
>>> remove_last('Test')
'Test'
>>> remove_last('Test;abc')
'Testabc'
>>> remove_last(';test;abc;foobar;\n')
';test;abc;foobar\n'
>>> remove_last(';asdf;asdf;asdf;asdf')
';asdf;asdf;asdfasdf'
The other answers provided are probably faster since they're tailored to your specific example, but this one is a bit more flexible.

You could split the string with semi colon and then join the non-empty parts back again using ; as separator
parts = '1;2;3;4;\n'.split(';')
non_empty_parts = []
for s in parts:
if s.strip() != "": non_empty_parts.append(s.strip())
print "".join(non_empty_parts, ';')

If you only want to use the strip function this is one method:
Using slice notation, you can limit the strip() function's scope to one part of the string and append the "\n" on at the end:
# create a var for later
str = "1;2;3;4;\n"
# format and assign to newstr
newstr = str[:8].strip(';') + str[8:]
Using the rfind() method(similar to Micheal0x2a's solution) you can make the statement applicable to many strings:
# create a var for later
str = "1;2;3;4;\n"
# format and assign to newstr
newstr = str[:str.rfind(';') + 1 ].strip(';') + str[str.rfind(';') + 1:]

re.sub(r';(\W*$)', r'\1', '1;2;3;4;\n') -> '1;2;3;4\n'