本文整理汇总了Python中languages.Alphabet.lowercase方法的典型用法代码示例。如果您正苦于以下问题：Python Alphabet.lowercase方法的具体用法？Python Alphabet.lowercase怎么用？Python Alphabet.lowercase使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类languages.Alphabet的用法示例。
在下文中一共展示了Alphabet.lowercase方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: process
# 需要导入模块: from languages import Alphabet [as 别名]
# 或者: from languages.Alphabet import lowercase [as 别名]
def process(self, rack):
""" Generate the data that will be shown to the user on the result page.
This includes a list of permutations of the rack, as well as combinations
of the rack with a single additional letter. High scoring words are also
tabulated. """
# Start with basic hygiene
if not rack:
return False
rack = rack.strip()
if not rack:
return False
# Make sure we reset all state in case we're called multiple times
self._counter = 0
self._allwords = [] # List of tuples: (word, score)
self._highscore = 0
self._highwords = []
self._combinations = { }
self._rack = u''
self._pattern = False
# Do a sanity check on the input by calculating its raw score, thereby
# checking whether all the letters are valid
score = 0
rack_lower = u'' # Rack converted to lowercase
wildcards = 0 # Number of wildcard characters
# If the rack starts with an equals sign ('=') we do a pattern match
# instead of a permutation search
if rack[0] == u'=':
self._pattern = True
rack = rack[1:]
# Sanitize the rack, converting upper case to lower case and
# catching invalid characters
try:
for c in rack:
ch = c
if ch in Alphabet.upper:
# Uppercase: find corresponding lowercase letter
ch = Alphabet.lowercase(ch)
if ch in u'?_*':
# This is one of the allowed wildcard characters
wildcards += 1
ch = u'?'
else:
score += Alphabet.scores[ch]
rack_lower += ch
except KeyError:
# A letter in the rack is not valid, even after conversion to lower case
return False
if not self._pattern and (wildcards > 2):
# Too many wildcards in a permutation search - need to constrain result set size
return False
# The rack contains only valid letters
self._rack = rack_lower
# Generate combinations
if not self._pattern and not wildcards:
# If no wildcards given, check combinations with one additional letter
query = self._rack + u'?'
# Permute the rack with one additional letter
p = self._word_db.find_permutations(query)
# Check the permutations to find valid words and their scores
if p is not None:
for word in p:
# Only interested in full-length permutations, i.e. with the additional letter
if len(word) == len(query):
# Find out which letter was added
addedletter = Alphabet.string_subtract(word, self._rack)
self._add_combination(addedletter, word)
# Check permutations
# The shortest possible rack to check for permutations is 2 letters
if len(self._rack) < 2:
return True
if self._pattern:
# Use pattern matching
p = self._word_db.find_matches(self._rack, True) # We'd like a sorted result
else:
# Find permutations
p = self._word_db.find_permutations(self._rack)
if p is None:
return True
for word in p:
if len(word) < 2:
# Don't show single letter words
continue
# Calculate the basic score of the word
score = self.score(word)
if wildcards and not self._pattern:
# Complication: Make sure we don't count the score of the wildcard tile(s)
wildchars = Alphabet.string_subtract(word, self._rack)
# What we have left are the wildcard substitutes: subtract'em
score -= self.score(wildchars)
self._add_permutation(word, score)
# Successful
return True