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#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
# Smewt - A smart collection manager
# Copyright (c) 2008 Nicolas Wack <wackou@gmail.com>
#
# Smewt is free software; you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation; either version 3 of the License, or
# (at your option) any later version.
#
# Smewt is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program. If not, see <http://www.gnu.org/licenses/>.
#
from guessit.patterns import sep
import unicodedata
import copy
# string-related functions
def strip_brackets(s):
if not s:
return s
if ((s[0] == '[' and s[-1] == ']') or
(s[0] == '(' and s[-1] == ')') or
(s[0] == '{' and s[-1] == '}')):
return s[1:-1]
return s
def clean_string(s):
for c in sep[:-2]: # do not remove dashes ('-')
s = s.replace(c, ' ')
parts = s.split()
result = ' '.join(p for p in parts if p != '')
# now also remove dashes on the outer part of the string
while result and result[0] in sep:
result = result[1:]
while result and result[-1] in sep:
result = result[:-1]
return result
def str_replace(string, pos, c):
return string[:pos] + c + string[pos+1:]
def str_fill(string, region, c):
start, end = region
return string[:start] + c * (end - start) + string[end:]
def to_utf8(o):
"""Convert all unicode strings found in the given object to utf-8
strings."""
if isinstance(o, unicode):
return o.encode('utf-8')
elif isinstance(o, list):
return [ to_utf8(i) for i in o ]
elif isinstance(o, dict):
# need to do it like that to handle Guess instances correctly
# FIXME: why is that necessary?
result = copy.deepcopy(o)
for key, value in o.items():
result[to_utf8(key)] = to_utf8(value)
return result
else:
return o
def to_unicode(o):
"""Convert all strings found in the given object to normalized
unicode strings, using the UTF-8 codec if needed."""
if isinstance(o, unicode):
return unicodedata.normalize('NFC', o)
if isinstance(o, str):
return unicodedata.normalize('NFC', o.decode('utf-8'))
elif isinstance(o, list):
return [ to_unicode(i) for i in o ]
elif isinstance(o, dict):
# need to do it like that to handle Guess instances correctly
#result = copy.deepcopy(o)
for key, value in o.items():
result[to_unicode(key)] = to_unicode(value)
return result
else:
return o
def levenshtein(a, b):
if not a:
return len(b)
if not b:
return len(a)
m = len(a)
n = len(b)
d = []
for i in range(m+1):
d.append([0] * (n+1))
for i in range(m+1):
d[i][0] = i
for j in range(n+1):
d[0][j] = j
for i in range(1, m+1):
for j in range(1, n+1):
if a[i-1] == b[j-1]:
cost = 0
else:
cost = 1
d[i][j] = min(d[i-1][j] + 1, # deletion
d[i][j-1] + 1, # insertion
d[i-1][j-1] + cost # substitution
)
return d[m][n]
# group-related functions
def find_first_level_groups_span(string, enclosing):
"""Return a list of pairs (start, end) for the groups delimited by the given
enclosing characters.
This does not return nested groups, ie: '(ab(c)(d))' will return a single group
containing the whole string.
>>> find_first_level_groups_span('abcd', '()')
[]
>>> find_first_level_groups_span('abc(de)fgh', '()')
[(3, 7)]
>>> find_first_level_groups_span('(ab(c)(d))', '()')
[(0, 10)]
>>> find_first_level_groups_span('ab[c]de[f]gh(i)', '[]')
[(2, 5), (7, 10)]
"""
opening, closing = enclosing
depth = [] # depth is a stack of indices where we opened a group
result = []
for i, c, in enumerate(string):
if c == opening:
depth.append(i)
elif c == closing:
try:
start = depth.pop()
end = i
if not depth:
# we emptied our stack, so we have a 1st level group
result.append((start, end+1))
except IndexError:
# we closed a group which was not opened before
pass
return result
def split_on_groups(string, groups):
"""Split the given string using the different known groups for boundaries.
>>> split_on_groups('0123456789', [ (2, 4) ])
['01', '23', '456789']
>>> split_on_groups('0123456789', [ (2, 4), (4, 6) ])
['01', '23', '45', '6789']
>>> split_on_groups('0123456789', [ (5, 7), (2, 4) ])
['01', '23', '4', '56', '789']
"""
if not groups:
return [ string ]
boundaries = sorted(set(reduce(lambda l, x: l + list(x), groups, [])))
if boundaries[0] != 0:
boundaries.insert(0, 0)
if boundaries[-1] != len(string):
boundaries.append(len(string))
groups = [ string[start:end] for start, end in zip(boundaries[:-1],
boundaries[1:]) ]
return [ g for g in groups if g ] # return only non-empty groups
def find_first_level_groups(string, enclosing, blank_sep=None):
"""Return a list of groups that could be split because of explicit grouping.
The groups are delimited by the given enclosing characters.
You can also specify if you want to blank the separator chars in the returned
list of groups by specifying a character for it. None means it won't be replaced.
This does not return nested groups, ie: '(ab(c)(d))' will return a single group
containing the whole string.
>>> find_first_level_groups('', '()')
['']
>>> find_first_level_groups('abcd', '()')
['abcd']
>>> find_first_level_groups('abc(de)fgh', '()')
['abc', '(de)', 'fgh']
>>> find_first_level_groups('(ab(c)(d))', '()', blank_sep = '_')
['_ab(c)(d)_']
>>> find_first_level_groups('ab[c]de[f]gh(i)', '[]')
['ab', '[c]', 'de', '[f]', 'gh(i)']
>>> find_first_level_groups('()[]()', '()', blank_sep = '-')
['--', '[]', '--']
"""
groups = find_first_level_groups_span(string, enclosing)
if blank_sep:
for start, end in groups:
string = str_replace(string, start, blank_sep)
string = str_replace(string, end-1, blank_sep)
return split_on_groups(string, groups)