Question

我有一些数据看起来：

[[('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')] ... ...]]

我想对具有标签B或I的连续单词进行分组，而忽略具有'O'标签的连续单词。

输出关键字应类似于：

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):

我不知道下一步该怎么做。谁能帮我吗？

谢谢

Answer 1

您应该使用itertools.groupby作为一个紧凑的解决方案：

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'],
#  []]

Answer 2

希望这会有所帮助。

remove_o = list(filter(lambda x: x[2] in ['I', 'B'], data))
words = [item[0] for item in remove_o]
reuslt = ' '.join(words)

Answer 3

您需要在不存在“ O”作为第三个元素的地方获取元组中的第一个值，对吗？您可以通过这种方式进行。

output = [j[0] for i in data for j in i if(j[2]!='O')]

上面的代码与

相同

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

分组包含B或I标签的连续单词

3 个答案: