I would use ",".join() with a generator expression to do this.

But with your setup you can simply have an if statement to see if you are on your final value for i (20 in this case) and opt to not print your comma then.

You could do something really abusive also, like multiplying the "," by i-19 to concat the comma using string multiplication (i.e. "a"*8 == "aaaaaaaa"). I think that "a"*(i-19) == "" for i =(1,19), and "a"*(i-19) == "a" for i = (20,39) (please don't do that though, haha). needs major refactoring i had a brainfart.

Just use rstrip().

result = your_string.rstrip(',')

You could build a list of strings in your for loop and print afterword using join:

strings = []

for ...:
   # some work to generate string
   strings.append(sting)

print(', '.join(strings))

alternatively, if your something has a well-defined length (i.e you can len(something)), you can select the string terminator differently in the end case:

for i, x in enumerate(something):
   #some operation to generate string

   if i < len(something) - 1:
      print(string, end=', ')
   else:
      print(string)

UPDATE based on real example code:

Taking this piece of your code:

value = input("")
string = ""
for unit_value in value.split(", "):
    if unit_value.split(' ', 1)[0] == "negative":
        neg_value = unit_value.split(' ', 1)[1]
        string = "-" + str(challenge1(neg_value.lower()))
    else:
        string = str(challenge1(unit_value.lower()))

    print(string, end=", ")

and following the first suggestion above, I get:

value = input("")
string = ""
strings = []
for unit_value in value.split(", "):
    if unit_value.split(' ', 1)[0] == "negative":
        neg_value = unit_value.split(' ', 1)[1]
        string = "-" + str(challenge1(neg_value.lower()))
    else:
        string = str(challenge1(unit_value.lower()))

    strings.append(string)

print(', '.join(strings))

If you want to do it your way:

for i in range(1, 21):
    print(i, end="," if i!=20 else "" )

Output:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20

But a better way of doing this would be:

print(*range(1, 21), sep=",")

Notice that in any case: the if or else, first of all you print n. So you can always start by printing it. Now we can switch the comma to the start of the printing:

n = 15
k = 5
def pattern(n):
    # Write your recursive function here
    print(n, end='')
    if n > 0:
        print(', ', end='') 
        pattern(n-k)
        print(', ' + str(n), end='')
pattern(n)

Gives:

15, 10, 5, 0, 5, 10, 15

You can use the sep parameter of print

numbers=[numbex(x) for x in range(1,11)]
print(*numbers, sep=",")

You want to replace it, not strip it:

s = s.replace(',', '')

You could use the join method:

x = int(input())
y = int(input())
z = int(input())
n = int(input())

print("[", end="")
join_list = []
for i in range(0, x + 1):
    for j in range(0, y + 1):
        for k in range(0, z + 1):
            if (i + j + k) != n:
                join_list.append(f"[{i},{j},{k}]")
print(",".join(join_list), end="")
print("]", end="")

Or in combination with comprehesions:

x = int(input())
y = int(input())
z = int(input())
n = int(input())

print(
    "["
    + ",".join(
        f"[{i},{j},{k}]"
        for i in range(x + 1) for j in range(y + 1) for k in range(z + 1)
        if i + j + k != n
    )
    + "]",
    end=""
)

Or use itertools.product instead of the nested comprehensions:

from itertools import product

x = int(input())
y = int(input())
z = int(input())
n = int(input())

print(
    "["
    + ",".join(
        "[{},{},{}]".format(*p)
        for p in product(range(x + 1), range(y + 1), range(z + 1))
        if sum(p) != n
    )
    + "]",
    end=""
)

Or build the list first, then convert it to string and replace the blanks:

from itertools import product

x = int(input())
y = int(input())
z = int(input())
n = int(input())

result = [
    list(p)
    for p in product(range(x + 1), range(y + 1), range(z + 1))
    if sum(p) != n
]
print(str(result).replace(" ", ""), end="")

This is not too good in terms of performance if your string is very long, but it should do

"\n".join(x[:-1] for x in output.splitlines())

Rather than printing each returned answer as you get it, you should add the answer to a list. Then you can print the entire list with join:

def distance(n):
  P1 = Count(n[0], n[1])
  results = []
  for x, y in zip(n[2::2], n[3::2]):
    P2 = Count(int(x), int(y))
    results.append(str(P1.distance(P2)))
  print(", ".join(results))

You'll have to remove the comma from the returned string on line 8. Also note I fixed your error on line 15.

Pass the results as separate arguments and specify the separator:

 print(*result, sep=',')