模式匹配一个句子中的多个值
Pattern match multiple values in a sentence
我有一个具有特定格式的句子。
<subject> <action> <object> @ <price> ... // The sentence can continue
我想从句子中提取这些值。
限制条件:
- 主题总是
Bob
或Alice
- 操作是
bought
或 sold
- 对象可以是任何1-7个字母的单词 //
4apples
should return NULL
- 价格是 float/integer
subject
前可以有句子但保证没有
包含 Bob/Alice
.
@
之后可能有也可能没有 space
示例:
Hi there, Bob sold apples @2.0 dollars each
期望的输出:
Subject: Bob
Action: sold
Object: apples
Price: 2.0
目前,我用天真的方法来做到这一点:
#!/usr/bin/env python3
sentence = "Hi there, alice sold apples @2.0 dollars each"
sentence = sentence.lower()
if 'alice' in sentence or 'bob' in sentence:
s_list = sentence.split(" ")
s_idx = -1
if 'bob' in sentence:
s_idx = s_list.index('bob')
elif 'alice' in sentence:
s_idx = s_list.index('alice')
if s_idx > -1:
Subject = s_list[s_idx]
Action = s_list[s_idx+1]
Object = s_list[s_idx+2] #more if/else to validate Object contraints
Price = s_list[s_idx+3] #more if/else to extract 2.0 if we get @2.0
print("Subject: {}, Action: {}, Object: {}, Price: {}".format(Subject, Action, Object, Price))
我怎样才能做得更好?可能使用 re
您可以为每个元素使用带有命名捕获组的正则表达式:
import re
sentence = "Hi there, alice sold apples @2.0 dollars each"
values = re.search('(?P<subject>bob|alice)\s+(?P<action>bought|sold)\s+(?P<object>[A-Za-z]{1,7})\s+@\s*(?P<price>\d+(?:\.\d+)?)', sentence)
if values:
Subject = values['subject']
Action = values['action']
Object = values['object']
Price = values['price']
print("Subject: {}, Action: {}, Object: {}, Price: {}".format(Subject, Action, Object, Price))
这将输出
Subject: alice, Action: sold, Object: apples, Price: 2.0
请注意,您可能需要提供 re.I
flag to re.search
以允许匹配 bob
或 Bob
(或 Sold
或 sold
等) ;在这种情况下,您可以将 object
捕获组中的 [A-Za-z]
替换为 [a-z]
.
我有一个具有特定格式的句子。
<subject> <action> <object> @ <price> ... // The sentence can continue
我想从句子中提取这些值。
限制条件:
- 主题总是
Bob
或Alice
- 操作是
bought
或sold
- 对象可以是任何1-7个字母的单词 //
4apples
should return NULL - 价格是 float/integer
subject
前可以有句子但保证没有 包含Bob/Alice
.@
之后可能有也可能没有 space
示例:
Hi there, Bob sold apples @2.0 dollars each
期望的输出:
Subject: Bob
Action: sold
Object: apples
Price: 2.0
目前,我用天真的方法来做到这一点:
#!/usr/bin/env python3
sentence = "Hi there, alice sold apples @2.0 dollars each"
sentence = sentence.lower()
if 'alice' in sentence or 'bob' in sentence:
s_list = sentence.split(" ")
s_idx = -1
if 'bob' in sentence:
s_idx = s_list.index('bob')
elif 'alice' in sentence:
s_idx = s_list.index('alice')
if s_idx > -1:
Subject = s_list[s_idx]
Action = s_list[s_idx+1]
Object = s_list[s_idx+2] #more if/else to validate Object contraints
Price = s_list[s_idx+3] #more if/else to extract 2.0 if we get @2.0
print("Subject: {}, Action: {}, Object: {}, Price: {}".format(Subject, Action, Object, Price))
我怎样才能做得更好?可能使用 re
您可以为每个元素使用带有命名捕获组的正则表达式:
import re
sentence = "Hi there, alice sold apples @2.0 dollars each"
values = re.search('(?P<subject>bob|alice)\s+(?P<action>bought|sold)\s+(?P<object>[A-Za-z]{1,7})\s+@\s*(?P<price>\d+(?:\.\d+)?)', sentence)
if values:
Subject = values['subject']
Action = values['action']
Object = values['object']
Price = values['price']
print("Subject: {}, Action: {}, Object: {}, Price: {}".format(Subject, Action, Object, Price))
这将输出
Subject: alice, Action: sold, Object: apples, Price: 2.0
请注意,您可能需要提供 re.I
flag to re.search
以允许匹配 bob
或 Bob
(或 Sold
或 sold
等) ;在这种情况下,您可以将 object
捕获组中的 [A-Za-z]
替换为 [a-z]
.