Add space between double quotes

p = '"fred", "bert", "blah"'

p.replace('\"'," ")

' fred , bert , blah '

This doesn't work as sep requires several parameters in print to have an effect.

You have many ways.

split and unpacking:

s = 'Hi I am Mike'
print(*s.split(), sep="...")

split and str.join:

s = 'Hi I am Mike'
print('...'.join(s.split()))

str.replace:

s = 'Hi I am Mike'
print(s.replace(' ', '...'))

Output: Hi...I...am...Mike

To replace all occurrences of t and T with a space in the string input, use the following:

input = input.replace('t', ' ').replace('T', ' ')

Or with regular expressions:

import re
input = re.sub('[tT]', ' ', input)

If you want to avoid replace, a faster method would just split and join. This is faster simply because .split and .join are fast:

"20%".join(string.split(" "))

For a more thorough review, I'll point out that your functions aren't equivalent. The first strips whitespace and the second doesn't. One of them must be wrong!

In the second case:

def replace_token_inplace(s, token=" "):
    for index, char in enumerate(s):
        if ord(char) == ord(token):
            s[index] = '20%'
    return s

you are doing several non-idiomatic things. For one, you are mutating and returning a list. It's better to just not return it if you mutate:

def replace_token_inplace(s, token=" "):
    for index, char in enumerate(s):
        if ord(char) == ord(token):
            s[index] = '20%'

Secondly, it'll probably be faster to do a copying transform:

def replace_token_inplace(s, token=" "):
    for char in s:
        if ord(char) == ord(token):
            yield '20%'
        else:
            yield char

which can also be written

def replace_token_inplace(s, token=" "):
    for char in s:
        yield '20%' if ord(char) == ord(token) else char

or even

def replace_token_inplace(s, token=" "):
    return ('20%' if ord(char) == ord(token) else char for char in s)

If you want to return a list, use square brackets instead of round ones.

import re
s1=re.sub(r"[^0-9 ]"," ",s)

You can use re here.

To prevent . of floating numbers use

(?!(?<=\d)\.(?=\d))[^0-9 ]

You don't need regular expressions. Python has a built-in string method that does what you need:

mystring.replace(" ", "_")

The piece of code:

a = ''.join([ch for ch in text if ch not in exclude])

is equivalent to

string_without_punctuation = ''
exclude = set(string.punctuation) # =set('!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~')
for character in text:
    if character not in exclude:
        string_without_punctuation += character

You could simply do this to replace the punctuation with spaces:

string_without_punctuation = ''
exclude = set(string.punctuation) # =set('!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~')
for character in text:
    if character not in exclude:
        string_without_punctuation += character
    else:
        string_without_punctuation += ' '

You may use a (?<=\d) to require a digit before the whitespace:

release = re.sub(r"(?<=\d)\s+", ",", release)

See the regex demo

Details

• (?<=\d) - a positive lookbehind that requires a digit to appear immediately to the left of the current location
• \s+ - 1 or more whitespace chars.

You can use replace()

text.replace(' ', 'x')