[英]extracting sentences from pos-tagged corpus with certain word, tag combos
我正在玩棕色語料庫,特別是“新聞”中帶標簽的句子。 我發現“ to”是帶有最多歧義詞標簽的詞(TO,IN,TO-HL,IN-HL,IN-TL,NPS)。 我正在嘗試編寫一個代碼,該代碼將從與“ to”關聯的每個標簽的語料庫中打印一個句子。 句子不需要“清除”標簽,而只包含“ to”和每個相關的pos標簽。
brown_sents = nltk.corpus.brown.tagged_sents(categories="news")
for sent in brown_sents:
for word, tag in sent:
if (word == 'to' and tag == "IN"):
print sent
我僅使用pos標簽之一嘗試了上面的代碼,以查看其是否有效,但它會打印所有示例。 我需要它僅打印找到的第一個與單詞,標記匹配的句子,然后停止。 我嘗試了這個:
for sent in brown_sents:
for word, tag in sent:
if (word == 'to' and tag == 'IN'):
print sent
if (word != 'to' and tag != 'IN'):
break
這是與pos-tag一起使用的,因為它是第一個與“ to”相關的標簽,但是如果我使用:
for sent in brown_sents:
for word, tag in sent:
if (word == 'to' and tag == 'TO-HL'):
print sent
if (word != 'to' and tag != 'TO-HL'):
break
它什么也不返回。 我想我太貼心了-願意幫忙嗎?
您可以繼續添加到當前代碼中,但是您的代碼沒有考慮以下因素:
如果您想堅持使用代碼,請嘗試以下操作:
from nltk.corpus import brown
brown_sents = brown.tagged_sents(categories="news")
def to_pos_sent(pos):
for sent in brown_sents:
for word, tag in sent:
if word == 'to' and tag == pos:
yield sent
for sent in to_pos_sent('TO'):
print sent
for sent in to_pos_sent('IN'):
print sent
我建議您將句子存儲在defaultdict(list)
,然后可以隨時檢索它們。
from nltk.corpus import brown
from collections import Counter, defaultdict
sents_with_to = defaultdict(list)
to_counts = Counter()
for i, sent in enumerate(brown.tagged_sents(categories='news')):
# Check if 'to' is in sentence.
uniq_words = dict(sent)
if 'to' in uniq_words or 'To' in uniq_words:
# Iterate through the sentence to find 'to'
for word, pos in sent:
if word.lower()=='to':
# Flatten the sentence into a string
sents_with_to[pos].append(sent)
to_counts[pos]+=1
for pos in sents_with_to:
for sent in sents_with_to[pos]:
print pos, sent
要訪問特定POS的句子,請執行以下操作:
for sent in sents_with_to['TO']:
print sent
您將意識到,如果對特定POS使用“ to”在句子中出現兩次。 它在sents_with_to[pos]
記錄了兩次。 如果要刪除它們,請嘗試:
sents_with_to_and_TO = set(" ".join(["#".join(word, pos) for word, pos in sent] for sent in sents_with_to['TO']))
關於為什么這不起作用:
for sent in brown_sents:
for word, tag in sent:
if (word == 'to' and tag == 'TO-HL'):
print sent
if (word != 'to' and tag != 'TO-HL'):
break
在進行解釋之前,您的代碼並沒有真正接近您想要的輸出。 這是因為您的if-else
語句並未真正滿足您的需求。
首先,您需要了解多個條件(即“ if”)在做什么。
# Loop through the sentence
for sent in brown_sents:
# Loop through each word with its POS
for word, tag in sent:
# For each sentence checks whether word and tag is in sentence:
if word == 'to' and tag == 'TO-HL':
print sent # If the condition is true, print sent
# After checking the first if, you continue to check the second if
# if word is not 'to' and tag is not 'TO-HL',
# you want to break out of the sentence. Note that you are still
# in the same iteration as the previous condition.
if word != 'to' and tag != 'TO-HL':
break
現在讓我們從一些基本的if-else
語句開始:
>>> from nltk.corpus import brown
>>> first_sent = brown.tagged_sents()[0]
>>> first_sent
[(u'The', u'AT'), (u'Fulton', u'NP-TL'), (u'County', u'NN-TL'), (u'Grand', u'JJ-TL'), (u'Jury', u'NN-TL'), (u'said', u'VBD'), (u'Friday', u'NR'), (u'an', u'AT'), (u'investigation', u'NN'), (u'of', u'IN'), (u"Atlanta's", u'NP$'), (u'recent', u'JJ'), (u'primary', u'NN'), (u'election', u'NN'), (u'produced', u'VBD'), (u'``', u'``'), (u'no', u'AT'), (u'evidence', u'NN'), (u"''", u"''"), (u'that', u'CS'), (u'any', u'DTI'), (u'irregularities', u'NNS'), (u'took', u'VBD'), (u'place', u'NN'), (u'.', u'.')]
>>> for word, pos in first_sent:
... if word != 'to' and pos != 'TO-HL':
... break
... else:
... print 'say hi'
...
>>>
從上面的示例中,我們遍歷了句子中的每個單詞+ POS和每對單詞pos, if
條件將檢查它是否不是單詞'to'而不是pos'TO-HL',如果是這樣的話就壞了,永遠不要對你say hi
。
因此,如果將代碼保持在if-else
條件下,則始終會中斷而不會繼續循環,因為to
不是句子中的第一個單詞,匹配的pos是不正確的。
實際上,您的if
條件試圖檢查每個單詞是否為“ to”以及其POS標簽是否為“ TO-HL”。
您要做的是檢查:
因此,條件(1)所需的if
條件為:
>>> from nltk.corpus import brown
>>> three_sents = brown.tagged_sents()[:3]
>>> for sent in three_sents:
... if 'to' in dict(sent):
... print sent
...
[(u'The', u'AT'), (u'September-October', u'NP'), (u'term', u'NN'), (u'jury', u'NN'), (u'had', u'HVD'), (u'been', u'BEN'), (u'charged', u'VBN'), (u'by', u'IN'), (u'Fulton', u'NP-TL'), (u'Superior', u'JJ-TL'), (u'Court', u'NN-TL'), (u'Judge', u'NN-TL'), (u'Durwood', u'NP'), (u'Pye', u'NP'), (u'to', u'TO'), (u'investigate', u'VB'), (u'reports', u'NNS'), (u'of', u'IN'), (u'possible', u'JJ'), (u'``', u'``'), (u'irregularities', u'NNS'), (u"''", u"''"), (u'in', u'IN'), (u'the', u'AT'), (u'hard-fought', u'JJ'), (u'primary', u'NN'), (u'which', u'WDT'), (u'was', u'BEDZ'), (u'won', u'VBN'), (u'by', u'IN'), (u'Mayor-nominate', u'NN-TL'), (u'Ivan', u'NP'), (u'Allen', u'NP'), (u'Jr.', u'NP'), (u'.', u'.')]
現在您知道if 'to' in dict(sent)
“ to”是否檢查句子中的“ to”。
然后檢查條件(2):
>>> for sent in three_sents:
... if 'to' in dict(sent):
... if dict(sent)['to'] == 'TO':
... print sent
...
[(u'The', u'AT'), (u'September-October', u'NP'), (u'term', u'NN'), (u'jury', u'NN'), (u'had', u'HVD'), (u'been', u'BEN'), (u'charged', u'VBN'), (u'by', u'IN'), (u'Fulton', u'NP-TL'), (u'Superior', u'JJ-TL'), (u'Court', u'NN-TL'), (u'Judge', u'NN-TL'), (u'Durwood', u'NP'), (u'Pye', u'NP'), (u'to', u'TO'), (u'investigate', u'VB'), (u'reports', u'NNS'), (u'of', u'IN'), (u'possible', u'JJ'), (u'``', u'``'), (u'irregularities', u'NNS'), (u"''", u"''"), (u'in', u'IN'), (u'the', u'AT'), (u'hard-fought', u'JJ'), (u'primary', u'NN'), (u'which', u'WDT'), (u'was', u'BEDZ'), (u'won', u'VBN'), (u'by', u'IN'), (u'Mayor-nominate', u'NN-TL'), (u'Ivan', u'NP'), (u'Allen', u'NP'), (u'Jr.', u'NP'), (u'.', u'.')]
>>> for sent in three_sents:
... if 'to' in dict(sent):
... if dict(sent)['to'] == 'TO-HL':
... print sent
...
>>>
現在您看到, if dict(sent)['to'] == 'TO-HL'
之后,您已經檢查if 'to' in dict(sent)
控制檢查pos限制的條件。
但是您意識到,如果在dict(sent)['to']
句子中有2個“ to”,則只能捕獲最后一個“ to”的POS。 這就是為什么您需要上一個答案中建議的defaultdict(list)
原因。
確實,沒有一種干凈的方法可以執行檢查,最有效的方法是前面的答案,嘆了口氣。
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