问：Lancaster 应该是“激进的”词干提取器，但它与 replied 正常工作。为什么？
这是因为 https://github.com/nltk/nltk/pull/1654 中改进了 Lancaster 词干分析器的实现
如果我们看一下https://github.com/nltk/nltk/blob/develop/nltk/stem/lancaster.py#L62，有一个后缀规则，改变-ied > -y
default_rule_tuple = (
"ai*2.", # -ia > - if intact
"a*1.", # -a > - if intact
"bb1.", # -bb > -b
"city3s.", # -ytic > -ys
"ci2>", # -ic > -
"cn1t>", # -nc > -nt
"dd1.", # -dd > -d
"dei3y>", # -ied > -y
...)
该功能允许用户输入新规则，如果没有添加其他规则，则它将使用parseRules 中的self.default_rule_tuple，其中将应用rule_tuple https://github.com/nltk/nltk/blob/develop/nltk/stem/lancaster.py#L196
def parseRules(self, rule_tuple=None):
"""Validate the set of rules used in this stemmer.
If this function is called as an individual method, without using stem
method, rule_tuple argument will be compiled into self.rule_dictionary.
If this function is called within stem, self._rule_tuple will be used.
"""
# If there is no argument for the function, use class' own rule tuple.
rule_tuple = rule_tuple if rule_tuple else self._rule_tuple
valid_rule = re.compile("^[a-z]+\*?\d[a-z]*[>\.]?$")
# Empty any old rules from the rule set before adding new ones
self.rule_dictionary = {}
for rule in rule_tuple:
if not valid_rule.match(rule):
raise ValueError("The rule {0} is invalid".format(rule))
first_letter = rule[0:1]
if first_letter in self.rule_dictionary:
self.rule_dictionary[first_letter].append(rule)
else:
self.rule_dictionary[first_letter] = [rule]
default_rule_tuple 实际上来自 paice-husk stemmer 的 whoosh 实现，也就是兰开斯特词干分析器 https://github.com/nltk/nltk/pull/1661 =)
问：在 Porter 中，In 还是大写的 In，为什么？
这太有趣了！而且很可能是一个错误。
>>> from nltk.stem import PorterStemmer
>>> porter = PorterStemmer()
>>> porter.stem('In')
'In'
如果我们查看代码，PorterStemmer.stem() 的第一件事就是将其变为小写，https://github.com/nltk/nltk/blob/develop/nltk/stem/porter.py#L651
def stem(self, word):
stem = word.lower()
if self.mode == self.NLTK_EXTENSIONS and word in self.pool:
return self.pool[word]
if self.mode != self.ORIGINAL_ALGORITHM and len(word) <= 2:
# With this line, strings of length 1 or 2 don't go through
# the stemming process, although no mention is made of this
# in the published algorithm.
return word
stem = self._step1a(stem)
stem = self._step1b(stem)
stem = self._step1c(stem)
stem = self._step2(stem)
stem = self._step3(stem)
stem = self._step4(stem)
stem = self._step5a(stem)
stem = self._step5b(stem)
return stem
但是如果我们查看代码，其他所有内容都返回stem，它是小写的，但是有两个 if 子句返回某种形式的原始word 没有被小写！！！
if self.mode == self.NLTK_EXTENSIONS and word in self.pool:
return self.pool[word]
if self.mode != self.ORIGINAL_ALGORITHM and len(word) <= 2:
# With this line, strings of length 1 or 2 don't go through
# the stemming process, although no mention is made of this
# in the published algorithm.
return word
第一个 if 子句检查单词是否在包含不规则单词及其词干的 self.pool 内。
第二个检查len(word)
问：请注意，Lancaster 正在删除“来”中以 e 结尾的单词，为什么？
毫不奇怪，也来自default_rule_tuple
https://github.com/nltk/nltk/blob/develop/nltk/stem/lancaster.py#L67，有一条规则会改变-e > - =)
问：如何禁用default_rule_tuple 中的-e > - 规则？
(Un-)幸运的是，LancasterStemmer._rule_tuple 对象是一个不可变的元组，所以我们不能简单地从中删除一项，但我们可以覆盖它 =)
>>> from nltk.stem import LancasterStemmer
>>> lancaster = LancasterStemmer()
>>> lancaster.stem('came')
'cam'
# Create a new stemmer object to refresh the cache.
>>> lancaster = LancasterStemmer()
>>> temp_rule_list = list(lancaster._rule_tuple)
# Find the 'e1>' rule.
>>> lancaster._rule_tuple.index('e1>')
12
# Create a temporary rule list from the tuple.
>>> temp_rule_list = list(lancaster._rule_tuple)
# Remove the rule.
>>> temp_rule_list.pop(12)
'e1>'
# Override the `._rule_tuple` variable.
>>> lancaster._rule_tuple = tuple(temp_rule_list)
# Et voila!
>>> lancaster.stem('came')
'came'