Python 对包含B或I标记的连续单词进行分组
我有以下数据:Python 对包含B或I标记的连续单词进行分组,python,python-3.x,nlp,Python,Python 3.x,Nlp,我有以下数据: [[('Natural', 'JJ', 'B'), ('language', 'NN', 'I'), ('processing', 'NN', 'I'), ('is', 'VBZ', 'O'), ('one', 'CD', 'O'), ('of', 'IN', 'O'), ('the', 'DT', 'O'), ('important', 'JJ', 'O'), ('branch', 'NN', 'O'), ('of', 'IN', 'O'), ('CS', 'NNP', 'B'
[[('Natural', 'JJ', 'B'), ('language', 'NN', 'I'), ('processing', 'NN', 'I'), ('is', 'VBZ', 'O'), ('one', 'CD', 'O'), ('of', 'IN', 'O'), ('the', 'DT', 'O'), ('important', 'JJ', 'O'), ('branch', 'NN', 'O'), ('of', 'IN', 'O'), ('CS', 'NNP', 'B'), ('.', '.', 'I')] ... ...]]
我想对带有标记B或I的连续单词进行分组,忽略带有“O”标记的单词
输出关键字应如下所示:
自然语言处理,
CS,
机器学习,
深度学习
我做了如下代码:
data=[[('Natural', 'JJ', 'B'), ('language', 'NN', 'I'), ('processing', 'NN', 'I'), ('is', 'VBZ', 'O'), ('one', 'CD', 'O'), ('of', 'IN', 'O'), ('the', 'DT', 'O'), ('important', 'JJ', 'O'), ('branch', 'NN', 'O'), ('of', 'IN', 'O'), ('CS', 'NNP', 'B'), ('.', '.', 'I')],
[('Machine', 'NN', 'B'), ('learning', 'NN', 'I'), (',', ',', 'I'), ('deep', 'JJ', 'I'), ('learning', 'NN', 'I'), ('are', 'VBP', 'O'), ('heavily', 'RB', 'O'), ('used', 'VBN', 'O'), ('in', 'IN', 'O'), ('natural', 'JJ', 'B'), ('language', 'NN', 'I'), ('processing', 'NN', 'I'), ('.', '.', 'I')],
[('It', 'PRP', 'O'), ('is', 'VBZ', 'O'), ('too', 'RB', 'O'), ('cool', 'JJ', 'O'), ('.', '.', 'O')]]
Key_words = []
index = 0
for sen in data:
for i in range(len(sen)):
while index < len(sen):
data=[('Natural','JJ','B'),('language','NN','I'),('processing','NN','I'),('is','VBZ','O'),('one','CD','O'),('of','IN','O'),('DT','O'),('important','JJ','O'),('branch','NN','O'),('of','IN','O'),('CS','NNP','B'),('IN','I'),',
[('Machine','NN','B'),('learning','NN','I'),('deep','JJ','I'),('learning','NN','I'),('are','VBP','O'),('throughly','RB','O'),('in','in','O'),('natural','JJ','B'),('language','NN I'),('processing','NN I','I'),('processing','NN I',',
[('It','PRP','O'),('is','VBZ','O'),('too','RB','O'),('cool','JJ','O'),(','O')]
关键词=[]
索引=0
对于数据中的sen:
对于范围内的i(len(sen)):
当指数
我不知道下一步该怎么办。谁能帮帮我吗
谢谢希望这有帮助
remove_o = list(filter(lambda x: x[2] in ['I', 'B'], data))
words = [item[0] for item in remove_o]
reuslt = ' '.join(words)
您应该使用
itertools.groupby
来获得相当紧凑的解决方案:
import itertools
import string
data = [[('Natural', 'JJ', 'B'), ('language', 'NN', 'I'), ('processing', 'NN', 'I'), ('is', 'VBZ', 'O'), ('one', 'CD', 'O'), ('of', 'IN', 'O'), ('the', 'DT', 'O'), ('important', 'JJ', 'O'), ('branch', 'NN', 'O'), ('of', 'IN', 'O'), ('CS', 'NNP', 'B'), ('.', '.', 'I')],
[('Machine', 'NN', 'B'), ('learning', 'NN', 'I'), (',', ',', 'I'), ('deep', 'JJ', 'I'), ('learning', 'NN', 'I'), ('are', 'VBP', 'O'), ('heavily', 'RB', 'O'), ('used', 'VBN', 'O'), ('in', 'IN', 'O'), ('natural', 'JJ', 'B'), ('language', 'NN', 'I'), ('processing', 'NN', 'I'), ('.', '.', 'I')],
[('It', 'PRP', 'O'), ('is', 'VBZ', 'O'), ('too', 'RB', 'O'), ('cool', 'JJ', 'O'), ('.', '.', 'O')]]
punctuation = set(string.punctuation)
keywords = [[' '.join(w[0] for w in g) for k, g in itertools.groupby(sen, key=lambda x: x[0] not in punctuation and x[2] != 'O') if k] for sen in data]
print(keywords)
# [['Natural language processing', 'CS'],
# ['Machine learning', 'deep learning', 'natural language processing'],
# []]
当“O”不作为第三个元素出现时,需要获取元组中的第一个值,对吗?你可以这样做
output = [j[0] for i in data for j in i if(j[2]!='O')]
以上代码与
for i in data:
for j in i:
if(j[2]!='O'): # if(j[2] in ['I','B']) also works
print(j[0]) # Or append to the output list