mirror of
https://github.com/SickGear/SickGear.git
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221 lines
6 KiB
Python
221 lines
6 KiB
Python
#!/usr/bin/env python2
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# -*- coding: utf-8 -*-
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#
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# Smewt - A smart collection manager
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# Copyright (c) 2008-2012 Nicolas Wack <wackou@gmail.com>
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#
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# Smewt is free software; you can redistribute it and/or modify
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# it under the terms of the GNU General Public License as published by
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# the Free Software Foundation; either version 3 of the License, or
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# (at your option) any later version.
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#
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# Smewt is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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# GNU General Public License for more details.
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#
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# You should have received a copy of the GNU General Public License
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# along with this program. If not, see <http://www.gnu.org/licenses/>.
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#
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from __future__ import unicode_literals
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from guessit import s
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from guessit.patterns import sep
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import functools
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import unicodedata
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import re
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# string-related functions
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def normalize_unicode(s):
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return unicodedata.normalize('NFC', s)
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def strip_brackets(s):
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if not s:
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return s
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if ((s[0] == '[' and s[-1] == ']') or
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(s[0] == '(' and s[-1] == ')') or
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(s[0] == '{' and s[-1] == '}')):
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return s[1:-1]
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return s
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def clean_string(s):
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for c in sep[:-2]: # do not remove dashes ('-')
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s = s.replace(c, ' ')
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parts = s.split()
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result = ' '.join(p for p in parts if p != '')
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# now also remove dashes on the outer part of the string
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while result and result[0] in sep:
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result = result[1:]
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while result and result[-1] in sep:
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result = result[:-1]
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return result
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_words_rexp = re.compile(r'\w+', re.UNICODE)
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def find_words(s):
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return _words_rexp.findall(s.replace('_', ' '))
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def reorder_title(title):
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ltitle = title.lower()
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if ltitle[-4:] == ',the':
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return title[-3:] + ' ' + title[:-4]
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if ltitle[-5:] == ', the':
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return title[-3:] + ' ' + title[:-5]
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return title
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def str_replace(string, pos, c):
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return string[:pos] + c + string[pos+1:]
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def str_fill(string, region, c):
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start, end = region
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return string[:start] + c * (end - start) + string[end:]
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def levenshtein(a, b):
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if not a:
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return len(b)
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if not b:
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return len(a)
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m = len(a)
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n = len(b)
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d = []
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for i in range(m+1):
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d.append([0] * (n+1))
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for i in range(m+1):
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d[i][0] = i
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for j in range(n+1):
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d[0][j] = j
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for i in range(1, m+1):
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for j in range(1, n+1):
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if a[i-1] == b[j-1]:
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cost = 0
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else:
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cost = 1
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d[i][j] = min(d[i-1][j] + 1, # deletion
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d[i][j-1] + 1, # insertion
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d[i-1][j-1] + cost # substitution
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)
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return d[m][n]
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# group-related functions
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def find_first_level_groups_span(string, enclosing):
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"""Return a list of pairs (start, end) for the groups delimited by the given
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enclosing characters.
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This does not return nested groups, ie: '(ab(c)(d))' will return a single group
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containing the whole string.
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>>> find_first_level_groups_span('abcd', '()')
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[]
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>>> find_first_level_groups_span('abc(de)fgh', '()')
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[(3, 7)]
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>>> find_first_level_groups_span('(ab(c)(d))', '()')
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[(0, 10)]
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>>> find_first_level_groups_span('ab[c]de[f]gh(i)', '[]')
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[(2, 5), (7, 10)]
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"""
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opening, closing = enclosing
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depth = [] # depth is a stack of indices where we opened a group
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result = []
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for i, c, in enumerate(string):
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if c == opening:
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depth.append(i)
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elif c == closing:
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try:
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start = depth.pop()
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end = i
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if not depth:
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# we emptied our stack, so we have a 1st level group
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result.append((start, end+1))
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except IndexError:
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# we closed a group which was not opened before
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pass
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return result
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def split_on_groups(string, groups):
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"""Split the given string using the different known groups for boundaries.
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>>> s(split_on_groups('0123456789', [ (2, 4) ]))
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['01', '23', '456789']
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>>> s(split_on_groups('0123456789', [ (2, 4), (4, 6) ]))
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['01', '23', '45', '6789']
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>>> s(split_on_groups('0123456789', [ (5, 7), (2, 4) ]))
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['01', '23', '4', '56', '789']
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"""
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if not groups:
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return [ string ]
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boundaries = sorted(set(functools.reduce(lambda l, x: l + list(x), groups, [])))
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if boundaries[0] != 0:
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boundaries.insert(0, 0)
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if boundaries[-1] != len(string):
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boundaries.append(len(string))
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groups = [ string[start:end] for start, end in zip(boundaries[:-1],
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boundaries[1:]) ]
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return [ g for g in groups if g ] # return only non-empty groups
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def find_first_level_groups(string, enclosing, blank_sep=None):
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"""Return a list of groups that could be split because of explicit grouping.
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The groups are delimited by the given enclosing characters.
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You can also specify if you want to blank the separator chars in the returned
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list of groups by specifying a character for it. None means it won't be replaced.
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This does not return nested groups, ie: '(ab(c)(d))' will return a single group
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containing the whole string.
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>>> s(find_first_level_groups('', '()'))
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['']
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>>> s(find_first_level_groups('abcd', '()'))
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['abcd']
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>>> s(find_first_level_groups('abc(de)fgh', '()'))
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['abc', '(de)', 'fgh']
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>>> s(find_first_level_groups('(ab(c)(d))', '()', blank_sep = '_'))
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['_ab(c)(d)_']
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>>> s(find_first_level_groups('ab[c]de[f]gh(i)', '[]'))
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['ab', '[c]', 'de', '[f]', 'gh(i)']
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>>> s(find_first_level_groups('()[]()', '()', blank_sep = '-'))
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['--', '[]', '--']
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"""
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groups = find_first_level_groups_span(string, enclosing)
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if blank_sep:
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for start, end in groups:
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string = str_replace(string, start, blank_sep)
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string = str_replace(string, end-1, blank_sep)
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return split_on_groups(string, groups)
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