unicode - Generate random UTF-8 string in Python

Question

I'd like to test the Unicode handling of my code. Is there anything I can put in random.choice() to select from the entire Unicode range, preferably not an external module? Neither Google nor StackOverflow seems to have an answer.

Edit: It looks like this is more complex than expected, so I'll rephrase the question - Is the following code sufficient to generate all valid non-control characters in Unicode?

unicode_glyphs = ''.join(
    unichr(char)
    for char in xrange(1114112) # 0x10ffff + 1
    if unicodedata.category(unichr(char))[0] in ('LMNPSZ')
    )

Answer 1

People may find their way here based mainly on the question title, so here's a way to generate a random string containing a variety of Unicode characters. To include more (or fewer) possible characters, just extend that part of the example with the code point ranges that you want.

import random

def get_random_unicode(length):

    try:
        get_char = unichr
    except NameError:
        get_char = chr

    # Update this to include code point ranges to be sampled
    include_ranges = [
        ( 0x0021, 0x0021 ),
        ( 0x0023, 0x0026 ),
        ( 0x0028, 0x007E ),
        ( 0x00A1, 0x00AC ),
        ( 0x00AE, 0x00FF ),
        ( 0x0100, 0x017F ),
        ( 0x0180, 0x024F ),
        ( 0x2C60, 0x2C7F ),
        ( 0x16A0, 0x16F0 ),
        ( 0x0370, 0x0377 ),
        ( 0x037A, 0x037E ),
        ( 0x0384, 0x038A ),
        ( 0x038C, 0x038C ),
    ]

    alphabet = [
        get_char(code_point) for current_range in include_ranges
            for code_point in range(current_range[0], current_range[1] + 1)
    ]
    return ''.join(random.choice(alphabet) for i in range(length))

if __name__ == '__main__':
    print('A random string: ' + get_random_unicode(10))