Question

假设我有模板来填充dict中的值：

我有这样的模板：

templates = [
   "I have four {fruit} in {place}", 
   "I have four {fruit} and {grain} in {place}",
   ...
]

使用这样的字典：

my_dict = {'fruit': ['apple', 'banana', 'mango'], 
           'place': ['kitchen', 'living room'],
           'grain' : ['wheat', 'rice']
          }

说我有这样一句话：

sentence = "I have four apple in kitchen"

鉴于这句话，模板和字典，我想知道这句话匹配其中一个模板并返回它匹配的值：

{'fruit': 'apple', 'place': 'kitchen'}

与上述相似如果：

Input: "I have four apple and wheat in kitchen"
Output: {'fruit': 'apple', 'grain': 'wheat', 'place': 'kitchen'}

如果能够处理这个问题会很棒：

Input: "I have four apple in bedroom" 
Output: {'fruit': 'apple'}

注意它只返回水果，而不是卧室，因为卧室不在地方的价值观。

Answer 1

将格式化的字符串转换为正则表达式：

import re

words = {k: '(?P<{}>{})'.format(k, '|'.join(map(re.escape, v))) for k, v in my_dict.items()}
patterns = [re.compile(template.format(**words)) for template in templates]

这会生成I have four (?P<fruit>apple|banana|mango) in (?P<place>kitchen|living room)"形式的模式。匹配这些将为您提供预期的输出：

for pattern in patterns:
    match = pattern.match(sentence)
    if match:
        matched_words = match.groupdict()

这是一种非常快速的O（N）方法来完全匹配句子：

>>> import re
>>> templates = [
...    "I have four {fruit} in {place}",
...    "I have four {fruit} and {grain} in {place}",
... ]
>>> my_dict = {'fruit': ['apple', 'banana', 'mango'],
...            'place': ['kitchen', 'living room'],
...            'grain' : ['wheat', 'rice']
...           }
>>> def find_matches(sentence):
...     for pattern in patterns:
...         match = pattern.match(sentence)
...         if match:
...             return match.groupdict()
...
>>> find_matches("I have four apple in kitchen")
{'fruit': 'apple', 'place': 'kitchen'}
>>> find_matches("I have four apple and wheat in kitchen")
{'fruit': 'apple', 'grain': 'wheat', 'place': 'kitchen'}

如果您需要模板匹配部分句子，请将可选部分包含在(?...)组中：

"I have four {fruit} in (?{place})"

或将\w+添加到单词列表中（除了有效单词），然后在匹配后对groupdict()验证my_dict结果。对于in bedroom案例，\w+将与bedroom部分匹配，但不会在my_dict的{{1}}列表中找到。{ / p>

Python填充字符串，其格式为字典

1 个答案: