How to check if a string in Python is in ASCII?

Question

I want to I check whether a string is in ASCII or not.

I am aware of ord(), however when I try ord('é'), I have TypeError: ord() expected a character, but string of length 2 found. I understood it is caused by the way I built Python (as explained in ord()'s documentation).

Is there another way to check?

Answer 1

To improve Alexander's solution from the Python 2.6 (and in Python 3.x) you can use helper module curses.ascii and use curses.ascii.isascii() function or various other: https://docs.python.org/2.6/library/curses.ascii.html

from curses import ascii

def isascii(s):
    return all(ascii.isascii(c) for c in s)

Answer 2

To prevent your code from crashes, you maybe want to use a try-except to catch TypeErrors

>>> ord("¶")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: ord() expected a character, but string of length 2 found

For example

def is_ascii(s):
    try:
        return all(ord(c) < 128 for c in s)
    except TypeError:
        return False

Answer 3

Python 3 way:

isascii = lambda s: len(s) == len(s.encode())

Answer 4

New in Python 3.7 (bpo32677)

No more tiresome/inefficient ascii checks on strings, new built-in str/bytes/bytearray method - .isascii() will check if the strings is ascii.

print("is this ascii?".isascii())
# True

Answer 5

I use the following to determine if the string is ascii or unicode:

>> print 'test string'.__class__.__name__
str
>>> print u'test string'.__class__.__name__
unicode
>>> 

Then just use a conditional block to define the function:

def is_ascii(input):
    if input.__class__.__name__ == "str":
        return True
    return False

Answer 6

How about doing this?

import string

def isAscii(s):
    for c in s:
        if c not in string.ascii_letters:
            return False
    return True