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Python how to remove equal sign "=" in strings?

Python how to remove = in strings?
a = 'bbb=ccc'
a.rstrip('=')
# returns 'bbb=ccc'
a.rstrip('\=')
# alse returns 'bbb=ccc'
how to match = ?

You can replace it with an empty string:
a.replace("=", "")
For reference:
https://docs.python.org/3/library/stdtypes.html#str.replace

You can use the replace method (easiest):
a = 'bbb=ccc'
a.replace('=', '')
or the translate method (probably faster on large amounts of data):
a = 'bbb=ccc'
a.translate(None, '=')
or the re.sub method (most powerful, i.e. can do much more):
import re
re.sub('=', '', 'aaa=bbb')

strip removes characters from the beginning and from the end of the string!
From the documentation:
str.strip([chars])
Return a copy of the string with leading and trailing characters removed.
Since you "=" is neither at the beggining nor at the end of your string, you can't use strip for your purpose. You need to use replace.
a.replace("=", "")

Strip removing more characters than expected

Can anyone explain what's going on here:
s = 'REFPROP-MIX:METHANOL&WATER'
s.lstrip('REFPROP-MIX') # this returns ':METHANOL&WATER' as expected
s.lstrip('REFPROP-MIX:') # returns 'THANOL&WATER'
What happened to that 'ME'? Is a colon a special character for lstrip? This is particularly confusing because this works as expected:
s = 'abc-def:ghi'
s.lstrip('abc-def') # returns ':ghi'
s.lstrip('abd-def:') # returns 'ghi'

str.lstrip removes all the characters in its argument from the string, starting at the left. Since all the characters in the left prefix "REFPROP-MIX:ME" are in the argument "REFPROP-MIX:", all those characters are removed. Likewise:
>>> s = 'abcadef'
>>> s.lstrip('abc')
'def'
>>> s.lstrip('cba')
'def'
>>> s.lstrip('bacabacabacabaca')
'def'
str.lstrip does not remove whole strings (of length greater than 1) from the left. If you want to do that, use a regular expression with an anchor ^ at the beginning:
>>> import re
>>> s = 'REFPROP-MIX:METHANOL&WATER'
>>> re.sub(r'^REFPROP-MIX:', '', s)
'METHANOL&WATER'

The method mentioned by #PadraicCunningham is a good workaround for the particular problem as stated.
Just split by the separating character and select the last value:
s = 'REFPROP-MIX:METHANOL&WATER'
res = s.split(':', 1)[-1] # 'METHANOL&WATER'

Python, can't replace generator object

I need to change replace a string's punctuation marks with space.
The problem is that I need to do it in one line.
for example: there's a string: 'H,+-=/e^##%ll-!!..o'
the result should be : 'H-----e----ll-----o'
where '-' symbolizes ' ' (space)
when I do
replace((c for c in string.punctuation),' ')
I get the error:
TypeError: Can't convert 'generator' object to str implicitly
I tried to put it in a list, in a set even in a dict.
but this error keeps on coming back.
how can I surpass this?

str.replace() doesn't take a list or generator, it'd only take a string, and even then won't do what you want. The method replaces one whole sequence of characters with another, so even x.replace(string.puntuation, '-') would only replace whole occurrences of the string.punctuation string in x with one dash.
Use string.maketrans() and str.translate() instead:
import string
translationmap = string.maketrans(string.punctuation, '-' * len(string.punctuation))
x = x.translate(translationmap)
Demo:
>>> import string
>>> x = 'H,+-=/e^##%ll-!!..o'
>>> import string
>>> translationmap = string.maketrans(string.punctuation, '-' * len(string.punctuation))
>>> x.translate(translationmap)
'H-----e----ll-----o'
str.translate() is hands-down the fastest method to map characters to other characters, or delete characters from a string.
On Python 3, str.translate() (or in Python 2, unicode.translate()) takes a mapping instead:
translationmap = {ord(c): '-' for c in string.punctuation}
x.translate(translationmap)

Try following
import string
''.join(map(lambda x : '-' if x in string.punctuation else x,
'H,+-=/e^##%ll-!!..o'))

You could also use re.sub for this:
>>> from re import sub
>>> sub("\W", "-", "H,+-=/e^##%ll-!!..o")
'H-----e----ll-----o'
>>>
\W captures all non-word characters.
Note that the above code will keep underscores. If you don't want them, replace \W with [\W_].

How to delete some characters from a string by matching certain character in python

i am trying to delete certain portion of a string if a match found in the string as below
string = 'Newyork, NY'
I want to delete all the characters after the comma from the string including comma, if comma is present in the string
Can anyone let me now how to do this .

Use .split():
string = string.split(',', 1)[0]
We split the string on the comma once, to save python the work of splitting on more commas.
Alternatively, you can use .partition():
string = string.partition(',')[0]
Demo:
>>> 'Newyork, NY'.split(',', 1)[0]
'Newyork'
>>> 'Newyork, NY'.partition(',')[0]
'Newyork'
.partition() is the faster method:
>>> import timeit
>>> timeit.timeit("'one, two'.split(',', 1)[0]")
0.52929401397705078
>>> timeit.timeit("'one, two'.partition(',')[0]")
0.26499605178833008

You can split the string with the delimiter ",":
string.split(",")[0]
Example:
'Newyork, NY'.split(",") # ['Newyork', ' NY']
'Newyork, NY'.split(",")[0] # 'Newyork'

Try this :
s = "this, is"
m = s.index(',')
l = s[:m]

A fwe options:
string[:string.index(",")]
This will raise a ValueError if , cannot be found in the string. Here, we find the position of the character with .index then use slicing.
string.split(",")[0]
The split function will give you a list of the substrings that were separated by ,, and you just take the first element of the list. This will work even if , is not present in the string (as there'd be nothing to split in that case, we'd have string.split(...) == [string])

Why does str.lstrip strip an extra character? [duplicate]

This question already has answers here:
How do I remove a substring from the end of a string?
(24 answers)
Closed 4 years ago.
>>> path = "/Volumes/Users"
>>> path.lstrip('/Volume')
's/Users'
>>> path.lstrip('/Volumes')
'Users'
>>>
I expected the output of path.lstrip('/Volumes') to be '/Users'

lstrip is character-based, it removes all characters from the left end that are in that string.
To verify this, try this:
"/Volumes/Users".lstrip("semuloV/") # also returns "Users"
Since / is part of the string, it is removed.
You need to use slicing instead:
if s.startswith("/Volumes"):
s = s[8:]
Or, on Python 3.9+ you can use removeprefix:
s = s.removeprefix("/Volumes")

Strip is character-based. If you are trying to do path manipulation you should have a look at os.path
>>> os.path.split("/Volumes/Users")
('/Volumes', 'Users')

The argument passed to lstrip is taken as a set of characters!
>>> ' spacious '.lstrip()
'spacious '
>>> 'www.example.com'.lstrip('cmowz.')
'example.com'
See also the documentation
You might want to use str.replace()
str.replace(old, new[, count])
# e.g.
'/Volumes/Home'.replace('/Volumes', '' ,1)
Return a copy of the string with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced.
For paths, you may want to use os.path.split(). It returns a list of the paths elements.
>>> os.path.split('/home/user')
('/home', '/user')
To your problem:
>>> path = "/vol/volume"
>>> path.lstrip('/vol')
'ume'
The example above shows, how lstrip() works. It removes '/vol' starting form left. Then, is starts again...
So, in your example, it fully removed '/Volumes' and started removing '/'. It only removed the '/' as there was no 'V' following this slash.
HTH

lstrip doc says:
Return a copy of the string S with leading whitespace removed.
If chars is given and not None, remove characters in chars instead.
If chars is unicode, S will be converted to unicode before stripping
So you are removing every character that is contained in the given string, including both 's' and '/' characters.

Here is a primitive version of lstrip (that I wrote) that might help clear things up for you:
def lstrip(s, chars):
for i in range len(s):
char = s[i]
if not char in chars:
return s[i:]
else:
return lstrip(s[i:], chars)
Thus, you can see that every occurrence of a character in chars is is removed until a character that is not in chars is encountered. Once that happens, the deletion stops and the rest of the string is simply returned