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.

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

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

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"

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.

#!/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

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

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

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

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.

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)

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])

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

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

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 ''.

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

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!

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.

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:DevPython")

   for file_name in file_list:

       os.rename(file_name, re.sub(r'd+','',file_name))

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.