Question

我有一个这样的清单：

fmt_string="I am a smoker male of 25 years who wants a policy for 30 
yrs with a sum assured amount of 1000000 rupees"

从上面的列表中我删除了停用词然后得到了这个现在我有一个列表如下：

['smoker', 'male', '25', 'years', 'wants', 'policy', '30', 'yrs', 
'sum', 'assured', 'amount', '1000000', 'rupees']

从这个列表中我想提取25,30和1000000，但代码应该是25之前或之后的年份。 30可以是在政策之后，1000000可以在任何位置

最后输出应该是：

'1000000 30 25 male smoker'

我只想要一个健壮的代码，无论我在哪里找到这些值，我都会返回一个这样的列表。

Answer 1

这应该有用

import re
# Variation in places of the numbers in strings:
str1 = "I am a smoker male of 25 years who wants a policy for 30  yrs with a sum assured amount of 1000000 rupees"
str2 = "I am a smoker male of 25 years who wants a for 30 policy  yrs with a sum assured amount of 1000000 rupees"
str3 = "I am a smoker male of years 25 who wants a for 30 policy yrs with a sum assured amount of 1000000 rupees"
str4 = "I am a smoker male of 25 years who wants a for 30 policy yrs with a sum assured amount of 1000000 rupees"

regex = r".*?(((\d{2})\s?years)|(years\s?(\d{2}))).*(policy.*?(\d{2})|(\d{2}).*?policy).*(\d{7}).*$"
replacements = r"\9 \7 \8 \3 \5"

res_str1 = re.sub(regex, replacements, str1)
res_str2 = re.sub(regex, replacements, str2)
res_str3 = re.sub(regex, replacements, str3)
res_str4 = re.sub(regex, replacements, str4)


def clean_spaces(string):
    return re.sub(r"\s{1,2}", ' ', string)


print(clean_spaces(res_str1))
print(clean_spaces(res_str2))
print(clean_spaces(res_str3))
print(clean_spaces(res_str4))

输出：

1000000 30 25 
1000000 30 25 
1000000 30 25
1000000 30 25

<强>更新

上面的正则表达式有一些错误。当我试图改进它时，我注意到它是低效和丑陋的，因为它每次都会解析每个单个字符。如果我们坚持你原来解析单词的方法，我们可以做得更好。所以我的新解决方案是：

# Algorithm
# for each_word in the_list:
#     maintain a pre_list of terms that come before a number
#     if each_word is number:
#         if there is any element of desired_terms_list exists in pre_list:
#             pair the number & the desired_term and insert into the_dictionary
#             remove this desired_term from desired_terms_list
#             reset the pre_list
#         else:
#             put the number in number_at_hand
#     else:
#         if no number_at_hand:
#             add the current word into pre_list
#         else:
#             if the current_word an element of desired_terms_list:
#                 pair the number & the desired_term and insert into the_dictionary
#                 remove this desired_term from desired_terms_list
#                 reset number_at_hand

代码：

from pprint import pprint


class Extractor:
    def __init__(self, search_terms, string):
        self.pre_list = list()
        self.list = string.split()
        self.terms_to_look_for = search_terms
        self.dictionary = {}

    @staticmethod
    def is_number(string):
        try:
            int(string)
            return True
        except ValueError:
            return False

    def check_pre_list(self):
        for term in self.terms_to_look_for:
            if term in self.pre_list:
                return term
            else:
                return None

    def extract(self):
        number_at_hand = str()
        for word in self.list:
            if Extractor.is_number(word):
                check_result = self.check_pre_list()
                if check_result is not None:
                    self.dictionary[check_result] = word
                    self.terms_to_look_for.remove(check_result)
                    self.pre_list = list()
                else:
                    number_at_hand = word
            else:
                if number_at_hand == '':
                    self.pre_list.append(word)
                else:
                    if word in self.terms_to_look_for:
                        self.dictionary[word] = number_at_hand
                        self.terms_to_look_for.remove(word)
                        number_at_hand = str()
        return self.dictionary

用法：

ex1 = Extractor(['years', 'policy', 'amount'],
                'I am a smoker male of 25 years who wants a policy for 30 yrs with a sum assured amount of 1000000 rupees')
ex2 = Extractor(['years', 'policy', 'amount'],
                'I am a smoker male of 25 years who wants a for 30 yrs policy with a sum assured amount of 1000000 rupees')
ex3 = Extractor(['years', 'policy', 'amount'],
                'I am a smoker male of years 25 who wants a policy for 30 yrs with a sum assured amount of 1000000 rupees')
ex4 = Extractor(['years', 'policy', 'amount'],
                'I am a smoker male of years 25 who wants a for 30 yrs policy with a sum assured amount of 1000000 rupees')
pprint(ex1.extract())
pprint(ex2.extract())
pprint(ex3.extract())
pprint(ex4.extract())

输出：

{'amount': '1000000', 'policy': '30', 'years': '25'}
{'amount': '1000000', 'policy': '30', 'years': '25'}
{'amount': '1000000', 'policy': '30', 'years': '25'}
{'amount': '1000000', 'policy': '30', 'years': '25'}

我希望现在能有更好的表现。

Answer 2

使用re在列表中findall出现，join用逗号split来列出并将reverse()应用于列表，然后join ' '再次

您的数据：

li = ['smoker', 'male', '25', 'years', 'wants', 'policy', '30', 'yrs', 'sum', 'assured', 'amount', '1000000', 'rupees']

temp=",".join([l for l in li if re.findall('1000000|30|25|male|smoker',l)]).split(",")

temp.reverse()
temp = " ".join(temp)

输出：

'1000000 30 25 male smoker'

希望这个答案有用。

从列表中提取整数

2 个答案: