ASCII to String

Question

I have this password generator:

from random import sample

character = '!"#$%&\'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{}~'
print(''.join(sample(character, 10)))

Example output: bvwy2%-?/N

However, I don't like it that the variable character is so long.

I got the content of character from this ASCII table:

I used every character except |.

My question is how can I shorten the content of character?

Something like this:

for i in range(33, 126):
    # get ASCII values ?

Answer 1

Since you know that your character values go from 33 to 126, you can create a string by

characters = "".join(chr(x) for x in range(33, 127))

As you noted, this contains a pipe. To remove this, simply use str.replace():

characters = characters.replace("|", "")

Which gives:

characters = '!"#$%&\'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{}~'

---

However, since random.sample() can take any iterable, you could do the same thing with making characters a set. If you do this, you can have another set of disallowed characters and easily subtract the disallowed set from the all_chars set to get the allowed_chars set. For example, if you wanted to disallow pipes and quotes, you could do:

all_chars = set(chr(x) for x in range(33, 127))
disallowed_chars = {'|', "'", '"'}
allowed_chars = all_chars - disallowed_chars

random.sample(allowed_chars, 10) # This also works

Answer 2

Like @9769953 told it your code could be:

from random import sample
import string

character = string.ascii_letters + string.digits + string.punctuation
character.replace('|', '') # remove the |
print(''.join(sample(character, 10)))