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I'm trying to figure out how to print a random unicode character in Python 3 using the format \uXXXX where each X is a character in [0-F]. This is what I have so far:

import random
chars = '0123456789ABCDEF'
L = len(chars)
fourRandInts = [random.randint(0,L-1) for i in range(4)]
fourRandChars = [chars[i] for i in fourRandInts]
s = r'\u{}{}{}{}'.format(*fourRandChars)
string = "print(u'{}')".format(s)
exec(string)

It seems to work, but I would prefer to avoid using exec. Is there a more Pythonic way to do this?

EDIT: It may seem that this question is a duplicate of #1477294 "Generate random UTF-8 string in Python" judging by the title, but that question was rephrased in an edit such that the answers there don't generally answer the original question, nor do they answer this question.

MountainDrew
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    You know that there is \UXXXXXXXX too, not only \uXXXX in order to get the non-BMP characters of the Uncode standard? Is your program also supposed to be able to output these non-BMP characters? – quant Nov 15 '18 at 16:46
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    Is there a reason that you can't just do `chr(random.randint(0,65536))`? – CJR Nov 15 '18 at 16:46
  • Possible duplicate of [Generate random UTF-8 string in Python](https://stackoverflow.com/questions/1477294/generate-random-utf-8-string-in-python) – 0x5453 Nov 15 '18 at 16:48
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    I'd use the way CJ59 showed. If for any reason you want to exclude some values from 'chars' (e.g. `chars = '13579BDF`) you could do `hexvalue = ''.join(random.choice(chars) for _ in range(4)); print(chr(int(hexvalue, 16)))` – Matthias Nov 15 '18 at 16:54
  • @CJ59, yes, that is certainly a much better way of doing it. If you would like to post it as a solution then I will accept it. – MountainDrew Nov 15 '18 at 17:00
  • @Matthias good to know! I was not aware of `random.choice()` (just like I wasn't aware that `chr(i)` could print a unicode character) – MountainDrew Nov 15 '18 at 17:03
  • @quant I wasn't familiar with that, thank you for pointing this out. It looks like this can be done by expanding the range of CJ59's line. – MountainDrew Nov 15 '18 at 17:11
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    @MountainDrew depends if you are using python2 or python3. On python3 chr(65536+25) is working, but in python2 it will throw an "ValueError: chr() arg not in range(256)" Exception. For python2 you need unichr(65536+25) - this function however does not exist in python3 anymore! – quant Nov 15 '18 at 17:18
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    Do note that not every codepoint is printable or even *valid*. There are a few standard functions that deal with Unicode metadata; perhaps there is something useful in there. – Jongware Nov 15 '18 at 18:39

2 Answers2

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One-liner solution thanks to @CJ59:

# print random unicode character from the Basic Multilingual Plane (BMP)
import random
print(chr(random.randint(0,65536)))

From the Python 3 chr() documentation:

chr(i)

Return the string representing a character whose Unicode code point is the integer i. For example, chr(97) returns the string 'a', while chr(8364) returns the string '€'. This is the inverse of ord().

The valid range for the argument is from 0 through 1,114,111 (0x10FFFF in base 16). ValueError will be raised if i is outside that range.

Solution that preserves the use of `chars` in my original question, thanks to @Matthias, allowing selection of hex digits for creating the unicode character:

# print unicode character using select hex chars
import random
chars = '0123456789ABCDEF'
# create random 4 character string from the characters in chars
hexvalue = ''.join(random.choice(chars) for _ in range(4))
# convert string representation of hex value to int,
# then convert to unicode character for printing
print(chr(int(hexvalue, 16)))

Function that returns a random unicode character only if it is printable:

This function uses the str.isprintable() method to only return a character if it is printable. This is useful if you want to generate a series of characters. Also includes an option for the character range.

import random
def randomPrintableUnicode(charRange = None):
    if charRange is None:
        charRange = (0,1114112)
    while True:
        i = random.randint(*charRange)
        c = chr(i)
        if c.isprintable():
            return c
        # should add another conditional break
        # to avoid infinite loop

# Print random unicode character
print(randomPrintableUnicode())

# Print random unicode character from the BMP
print(randomPrintableUnicode(charRange = (0,65536)))

# Print random string of 20 characters
# from the Cyrillic alphabet
cyrillicRange = (int('0410',16),int('0450',16))
print(
    ''.join(
        [
            randomPrintableUnicode(charRange = cyrillicRange)
            for _ in range(20)
        ]
    )
)
MountainDrew
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-1

You can make a forever loop that would generate a random unicode character along with it's ID and number. Also, it never crashes. (Unless you do something insane.) Delete 'while True:' to stop the forever loop and delete the 'sleep (1)' to stop the wait time.

    from random import randint
    from time import sleep
    while True:
    try:
      sleep(1)
      a=(randint(1,65663))
      print('Character:')
      print(chr(a))
      print('ID:' + str(hex(a)))
      print('Number:' + str(a) + '\n\n\n\n\n\n\n\n\n\n\n\n\n')
    except UnicodeEncodeError:
      print('Character is not possible to print. Moving on.')
ExplorPython
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