Slicing the second last character

If you just want to delete the second last character, your can do like this,

s = "[2,3,1,1,]"

s[:-2] + s[-1]
# '[2,3,1,1]'

s[:-2] -> Will slice the string from 0 to -2 index location (without -2 index)

s[-1] -> Will fetch the last element

s[:-2] + s[-1] -> concatenation of the strigs

If you're sure you have that string, slice both characters and add the ] back on!

source_string = "[2,3,1,1,]"
if source_string.endswith(",]"):
    source_string = source_string[:-2] + "]"

However, often lists stored as strings are not very useful - you may really want to convert the whole thing to a collection of numbers (perhaps manually removing the "[]" and splitting by ,, or using ast.literal_eval()), potentially converting it back to a string to display later

>>> source_string = "[2,3,1,1,]"
>>> import ast
>>> my_list = ast.literal_eval(source_string)
>>> my_list
[2, 3, 1, 1]
>>> str(my_list)
'[2, 3, 1, 1]'

You can use this for this one case

txt = "[2,3,1,1,]"
print(f"{txt[:-2]}{txt[-1]}")

Even tho storing lists as string is a bit weird

txt[:-2] will get the characters from 0 index to -2 index

txt[-1] will get the last character which is "]"

then I concatenate both with an f"" string

You can use this if you don't wanna use an f string

print(txt[:-2], txt[-1], sep="")

the "sep" argument is so there won't be space between the two prints

Using built-in functions such as str.rstrip

l = '[2,3,1,1,]'

l_new = f"{l.rstrip('],')}]" # it doesn't matter the order!
print(l_new)

or str.rpartition

a, _, b = l.rpartition(',')
l_new = a + b
print(l_new)

you can avoid an explicit slicing. See doc for details.

The 1st approach is universal, hence doesn't produce any side-effects. The 2nd may give rise to side-effects if the final last character is not a ,. If necessary use a check, i.e. str.endswith, to fix it.