我有一个列表列表,其中包含需要被词形化的单词。我收到一个错误,指出字符串是必需的而不是列表,因为我正在使用Spacy。
如果我转换为字符串,即nlp(str(list_1))
,那么列表分隔符(如:“,”和“[”)被标记化并包含在我的输出中。
如何将列表列表中的项目变为lemmatize并将其恢复为相同的形式,即列表列表?
需要被词形化的词可以在列表列表中的任何位置。
我想要这样的东西:
输入:
[["flower", "grows", "garden"], [["boy", "running", "playground"]]
输出:
[["flower", "grow", "garden"], ["boy", "run", "playground"]]
import spacy
nlp = spacy.load("en_core_web_sm")
list_1 = [["flower", "grows", "garden"], ["boy", "running", "playground"]]
for item in nlp(str(list_1)):
print(item.lemma_)
我将此任务分为以下几部分:
你已经完成了它,但后代:
nlp = spacy.load("en_core_web_sm")
words = [["flower", "grows", "garden"], ["boy", "running", "playground"]]
我们需要有一个每个列表的长度,以便我们稍后可以迭代它(为了重塑输出)。使用numpy.cumsum我们可以创建一个数组,这将允许我们在O(n)
时间这样做。
# remember about importing numpy
lengths = np.cumsum([0] + list(map(len, words)))
print(lengths)
这将为我们提供以下数组(针对您的情况):
[0 3 6]
我们稍后将使用从该数组创建的范围,例如令牌[0:3]
构成第一阵列和令牌[3:6]
构成第二个。
flat_words = [item for sublist in words for item in sublist]
doc = spacy.tokens.Doc(nlp.vocab, words=flat_words)
最好将flat_words
作为列表传递,因此spacy
不必执行不必要的标记化操作。
最后迭代spacy.tokens.Span
对象,在他们的标记上,然后将这些(当然是lemmatized
)添加到列表中。
lemmatized = []
# Iterate starting with 1
for index in range(1, len(lengths)):
# Slice doc as described in the first point, so [0:3] and [3:6]
span = doc[lengths[index - 1] : lengths[index]]
# Add lemmatized tokens as list to the outer list
lemmatized.append([token.lemma_ for token in span])
print(lemmatized)
的输出将如你所愿:
[['flower', 'grow', 'garden'], ['boy', 'run', 'playground']]
为了让您更轻松,下面的整个代码:
import numpy as np
import spacy
nlp = spacy.load("en_core_web_sm")
words = [["flower", "grows", "garden"], ["boy", "running", "playground"]]
lengths = np.cumsum([0] + list(map(len, words)))
print(lengths)
flat_words = [item for sublist in words for item in sublist]
doc = spacy.tokens.Doc(nlp.vocab, words=flat_words)
lemmatized = []
# Iterate starting with 1
for index in range(1, len(lengths)):
# Slice doc as described in the first point, so [0:3] and [3:6]
span = doc[lengths[index - 1] : lengths[index]]
# Add lemmatized tokens as list to the list
lemmatized.append([token.lemma_ for token in span])
print(lemmatized)
在处理列表列表时,您可以加入列表中的项目,然后使用nlp()
。接下来,获取这些项目中的每个项目的引理。要再次返回列表列表,只需将项目发生在项目出现的索引处。
for item in list_1:
doc = nlp(' '.join(item))
for indexer,i in enumerate(doc):
item[indexer] = i.lemma_
print(list_1)
#Output:
[['flower', 'grow', 'garden'], ['boy', 'run', 'playground']]
我不认为这是最好的解决方案,但你可以做到这一点
import spacy
nlp = spacy.load("en_core_web_sm")
list_1 = [["flower", "grows", "garden"], ["boy", "running", "playground"]]
s=""
for item in nlp(str(list_1)):
s+=item.lemma_
ss=s[2:-2].replace('\'','').split('],[')
l=[]
for sss in ss :
l.append(sss.split(','))
print(l)
#output
[['flower', 'grow', 'garden'], ['boy', 'run', 'playground']]
这里:如果只改变这些特定的作品,这可以工作
main = [["flower", "grows", "garden"], [["boy", "running", "playground"]]
main[0][1] = "grow"
main[1][1] = "run"
# main = [["flower", "grow", "garden"], ["boy", "run", "playground"]]