cogniteev/docido-python-sdk

import os


def to_unicode(text, charset=None):
    """Convert input to an `unicode` object.

    For a `str` object, we'll first try to decode the bytes using the given
    `charset` encoding (or UTF-8 if none is specified), then we fall back to
    the latin1 encoding which might be correct or not, but at least preserves
    the original byte sequence by mapping each byte to the corresponding
    unicode code point in the range U+0000 to U+00FF.

    For anything else, a simple `unicode()` conversion is attempted,
    with special care taken with `Exception` objects.
    """
    if isinstance(text, str):
        try:
            return unicode(text, charset or 'utf-8')
        except UnicodeDecodeError:
            return unicode(text, 'latin1')
    elif isinstance(text, Exception):
        if os.name == 'nt' and \
                isinstance(text, (OSError, IOError)):  # pragma: no cover
            # the exception might have a localized error string encoded with
            # ANSI codepage if OSError and IOError on Windows
            try:
                return unicode(str(text), 'mbcs')
            except UnicodeError:
                pass
        # two possibilities for storing unicode strings in exception data:
        try:
            # custom __str__ method on the exception (e.g. PermissionError)
            return unicode(text)
        except UnicodeError:
            # unicode arguments given to the exception (e.g. parse_date)
            return ' '.join([to_unicode(arg) for arg in text.args])
    return unicode(text)


def exception_to_unicode(e, traceback=False):
    """Convert an `Exception` to an `unicode` object.

    In addition to `to_unicode`, this representation of the exception
    also contains the class name and optionally the traceback.
    """
    message = '%s: %s' % (e.__class__.__name__, to_unicode(e))
    if traceback:
        from docido_sdk.toolbox import get_last_traceback
        traceback_only = get_last_traceback().split('\n')[:-2]
        message = '\n%s\n%s' % (to_unicode('\n'.join(traceback_only)), message)
    return message


def levenshtein(s, t):
    """ Compute the Levenshtein distance between 2 strings, which
    is the minimum number of operations required to perform on a string to
    get another one.

    code taken from https://en.wikibooks.org

    :param basestring s:
    :param basestring t:
    :rtype: int
    """
    ''' From Wikipedia article; Iterative with two matrix rows. '''
    if s == t:
        return 0
    elif len(s) == 0:
        return len(t)
    elif len(t) == 0:
        return len(s)
    v0 = [None] * (len(t) + 1)
    v1 = [None] * (len(t) + 1)
    for i in range(len(v0)):
        v0[i] = i
    for i in range(len(s)):
        v1[0] = i + 1
        for j in range(len(t)):
            cost = 0 if s[i] == t[j] else 1
            v1[j + 1] = min(v1[j] + 1, v0[j + 1] + 1, v0[j] + cost)
        for j in range(len(v0)):
            v0[j] = v1[j]

    return v1[len(t)]