我正在尝试提取围绕给定单词的选定数量的单词。我将举一个例子以清楚：
string ="教育应针对人格的全面发展，并加强对人权和基本自由的尊重。"
1(选定的单词是开发，我需要获取周围的6个单词，然后获取：[to，to，to，fult，of，of，the，human]
2(但是，如果选定的单词在开始或第二个位置，我仍然需要获得6个单词，例如：
选定的单词是应，我应该得到：[教育，be，timed，to to to，the，full]
我应该使用" re"模块。到目前为止，我设法找到的是：
def search(text,n):
'''Searches for text, and retrieves n words either side of the text, which are retuned seperatly'''
word = r"W*([w]+)"
groups = re.search(r'{}W*{}{}'.format(word*n,'place',word*n), text).groups()
return groups[:n],groups[n:]
，但它只有第一种情况才能对我有所帮助。有人可以帮助我，我真的很感激。预先感谢您！
这将在您的文本中提取目标单词的所有出现，并在上下文中：
import re
text = ("Education shall be directed to the full development of the human personality "
"and to the strengthening of respect for human rights and fundamental freedoms.")
def search(target, text, context=6):
# It's easier to use re.findall to split the string,
# as we get rid of the punctuation
words = re.findall(r'w+', text)
matches = (i for (i,w) in enumerate(words) if w.lower() == target)
for index in matches:
if index < context //2:
yield words[0:context+1]
elif index > len(words) - context//2 - 1:
yield words[-(context+1):]
else:
yield words[index - context//2:index + context//2 + 1]
print(list(search('the', text)))
# [['be', 'directed', 'to', 'the', 'full', 'development', 'of'],
# ['full', 'development', 'of', 'the', 'human', 'personality', 'and'],
# ['personality', 'and', 'to', 'the', 'strengthening', 'of', 'respect']]
print(list(search('shall', text)))
# [['Education', 'shall', 'be', 'directed', 'to', 'the', 'full']]
print(list(search('freedoms', text)))
# [['respect', 'for', 'human', 'rights', 'and', 'fundamental', 'freedoms']]
棘手的棘手可能会出现偏离错误，但我认为这符合您的规格。我已经删除了标点符号，可能最好在发送字符串进行分析之前将其删除。我认为案件并不重要。
test_str = "Education shall be directed to the full development of the human personality and to the strengthening of respect for human rights and fundamental freedoms."
def get_surrounding_words(search_word, s, n_words):
words = s.lower().split(' ')
try:
i = words.index(search_word)
except ValueError:
return []
# Word is near start
if i < n_words/2:
words.pop(i)
return words[:n_words]
# Word is near end
elif i >= len(words) - n_words/2:
words.pop(i)
return words[-n_words:]
# Word is in middle
else:
words.pop(i)
return words[i-n_words/2:i+n_words/2]
def test(word):
print('{}: {}'.format(word, get_surrounding_words(word, test_str, 6)))
test('notfound')
test('development')
test('shall')
test('education')
test('fundamental')
test('for')
test('freedoms')
import sys, os
args = sys.argv[1:]
if len(args) != 2:
os.exit("Use with <string> <query>")
text = args[0]
query = args[1]
words = text.split()
op = []
left = 3
right = 3
try:
index = words.index(query)
if index <= left:
start = 0
else:
start = index - left
if start + left + right + 1 > len(words):
start = len(words) - left - right - 1
if start < 0:
start = 0
while len(op) < left + right and start < len(words):
if start != index:
op.append(words[start])
start += 1
except ValueError:
pass
print op
- 这是如何工作的？
- 在字符串中找到单词
- 看看我们是否可以从索引中左右 右单词
- 左 正确的单词数并将其保存在Op 中
- 打印op
解决问题的简单方法。首先将所有单词分开，然后从左右选择单词。
def custom_search(sentence, word, n):
given_string = sentence
given_word = word
total_required = n
word_list = given_string.strip().split(" ")
length_of_words = len(word_list)
output_list = []
given_word_position = word_list.index(given_word)
word_from_left = 0
word_from_right = 0
if given_word_position + 1 > total_required / 2:
word_from_left = total_required / 2
if given_word_position + 1 + (total_required / 2) <= length_of_words:
word_from_right = total_required / 2
else:
word_from_right = length_of_words - (given_word_position + 1)
remaining_words = (total_required / 2) - word_from_right
word_from_left += remaining_words
else:
word_from_right = total_required / 2
word_from_left = given_word_position
if word_from_left + word_from_right < total_required:
remaining_words = (total_required / 2) - word_from_left
word_from_right += remaining_words
required_words = []
for i in range(given_word_position - word_from_left, word_from_right +
given_word_position + 1):
if i != given_word_position:
required_words.append(word_list[i])
return required_words
sentence = "Education shall be directed to the full development of the human personality and to the strengthening of respect for human rights and fundamental freedoms."
custom_search(sentence, "shall", 6)
>>[Education, be, directed, to , the , full]
custom_search(sentence, "development", 6)
>>['to', 'the', 'full', 'of', 'the', 'human']
我认为这里不需要正则表达式。假设文本是构造良好的，只需将其分为一系列单词，然后写一对If-else语句以确保其检索必要的周围单词：
def search(text, word, n):
# text is the string you are searching
# word is the word you are looking for
# n is the TOTAL number of words you want surrounding the word
words = text.split(" ") # Create an array of words from the string
position = words.index(word) # Find the position of the desired word
distance_from_end = len(words) - position # How many words are after the word in the text
if position < n // 2 + n % 2: # If there aren't enough words before...
return words[:position], words[position + 1:n + 1]
elif distance_from_end < n // 2 + n % 2: # If there aren't enough words after...
return words[position - n + distance_from_end:position], words[position + 1:]
else: # Otherwise, extract an equal number of words from both sides (take from the right if odd)
return words[position - n // 2 - n % 2:position], words[position + 1:position + 1 + n//2]
string = "Education shall be directed to the full development of the human personality and to the
strengthening of respect for human rights and fundamental freedoms."
print search(string, "shall", 6)
# >> (['Education'], ['be', 'directed', 'to', 'the', 'full'])
print search(string, "human", 5)
# >> (['development', 'of', 'the'], ['personality', 'and'])
在您的示例中，您没有输出中包含目标单词，因此我也将其拒之门外。如果您想包含目标单词，请简单地组合两个数组，函数返回(在position上加入它们(。
希望这有帮助！