>>> x = 'Pear.good'
>>> y = x.replace('.good','')
>>> y
'Pear'
>>> x
'Pear.good'

.replace doesn't change the string, it returns a copy of the string with the replacement. You can't change the string directly because strings are immutable.

You need to take the return values from x.replace and put them in a new set.

In Python 3.9+ you could remove the suffix using str.removesuffix('mysuffix'). From the docs:

If the string ends with the suffix string and that suffix is not empty, return string[:-len(suffix)]. Otherwise, return a copy of the original string

So you can either create a new empty set and add each element without the suffix to it:

set1  = {'Apple.good', 'Orange.good', 'Pear.bad', 'Pear.good', 'Banana.bad', 'Potato.bad'}

set2 = set()
for s in set1:
   set2.add(s.removesuffix(".good").removesuffix(".bad"))

Or create the new set using a set comprehension:

set2 = {s.removesuffix(".good").removesuffix(".bad") for s in set1}
   
print(set2)

Output:

{'Orange', 'Pear', 'Apple', 'Banana', 'Potato'}

All you need is a bit of black magic!

>>> a = ["cherry.bad","pear.good", "apple.good"]
>>> a = list(map(lambda x: x.replace('.good','').replace('.bad',''),a))
>>> a
['cherry', 'pear', 'apple']

When there are multiple substrings to remove, one simple and effective option is to use re.sub with a compiled pattern that involves joining all the substrings-to-remove using the regex OR (|) pipe.

import re

to_remove = ['.good', '.bad']
strings = ['Apple.good','Orange.good','Pear.bad']

p = re.compile('|'.join(map(re.escape, to_remove))) # escape to handle metachars
[p.sub('', s) for s in strings]
# ['Apple', 'Orange', 'Pear']

You could do this:

import re
import string
set1={'Apple.good','Orange.good','Pear.bad','Pear.good','Banana.bad','Potato.bad'}

for x in set1:
    x.replace('.good',' ')
    x.replace('.bad',' ')
    x = re.sub('\.good$', '', x)
    x = re.sub('\.bad$', '', x)
    print(x)

# practices 2
str = "Amin Is A Good Programmer"
new_set = str.replace('Good', '')
print(new_set)

 

print : Amin Is A  Programmer

I did the test (but it is not your example) and the data does not return them orderly or complete

>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> newind = {x.replace('p','') for x in ind}
>>> newind
{'1', '2', '8', '5', '4'}

I proved that this works:

>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> newind = [x.replace('p','') for x in ind]
>>> newind
['5', '1', '8', '4', '2', '8']

or

>>> newind = []
>>> ind = ['p5','p1','p8','p4','p2','p8']
>>> for x in ind:
...     newind.append(x.replace('p',''))
>>> newind
['5', '1', '8', '4', '2', '8']

If list

I was doing something for a list which is a set of strings and you want to remove all lines that have a certain substring you can do this

import re
def RemoveInList(sub,LinSplitUnOr):
    indices = [i for i, x in enumerate(LinSplitUnOr) if re.search(sub, x)]
    A = [i for j, i in enumerate(LinSplitUnOr) if j not in indices]
    return A

where sub is a patter that you do not wish to have in a list of lines LinSplitUnOr

for example

A=['Apple.good','Orange.good','Pear.bad','Pear.good','Banana.bad','Potato.bad']
sub = 'good'
A=RemoveInList(sub,A)

Then A will be