Python：检查字典项是否在字符串中，然后返回匹配项

我有一个如下所示的数据库表：

id    phrases                    message
1     "social media, marketing"  "This person is in marketing!"
2     "finance, fintech          "This person is in finance!"

# phrase_list
{'social media': 'This person is in marketing!', 
 'marketing': 'This person is in marketing!', 
 'finance': 'This person is in finance!', 
 'fintech': 'This person is in finance!'}

我循环浏览了关键短语并将其添加到字典中，如下所示：

messages = self.db_access.execute("SELECT * FROM messages")
messages = self.db_access.fetchall()
print(len(messages))
if self.db_access.rowcount > 0:
    has_message = True
    phrase_list = {}
    for the_rule in messages:
        print(the_rule[1])
        rule = the_rule[1].split(',')
        for phrase in rule:
            phrase = str(phrase)
            phrase_list[phrase] = str(the_rule[2])
    print(phrase_list)
    print("\n")
else:
    has_message = False

descriptions = ["I am in marketing, and it is super awesome", "I am in finance, and it is super awesome"]

这将产生以下结果：

id    phrases                    message
1     "social media, marketing"  "This person is in marketing!"
2     "finance, fintech          "This person is in finance!"

# phrase_list
{'social media': 'This person is in marketing!', 
 'marketing': 'This person is in marketing!', 
 'finance': 'This person is in finance!', 
 'fintech': 'This person is in finance!'}

因此，每一个短语都有它自己的附带信息，在其他地方使用

现在，我可以将这些dict键与字符串进行比较，如下所示：

messages = self.db_access.execute("SELECT * FROM messages")
messages = self.db_access.fetchall()
print(len(messages))
if self.db_access.rowcount > 0:
    has_message = True
    phrase_list = {}
    for the_rule in messages:
        print(the_rule[1])
        rule = the_rule[1].split(',')
        for phrase in rule:
            phrase = str(phrase)
            phrase_list[phrase] = str(the_rule[2])
    print(phrase_list)
    print("\n")
else:
    has_message = False

descriptions = ["I am in marketing, and it is super awesome", "I am in finance, and it is super awesome"]

我的下一步是将该字符串与键进行比较，如果它包含任何关键字，则打印匹配的键及其值/消息。这就是我到目前为止所做的：

for description in descriptions:
    print(description)
    if has_message == True:
        if any(x in description for x in phrase_list):
            # print matching keyword and message
        else:
            print("No matches, but a phrase list exists")

---

因此，我的问题是，我需要用什么替换该注释才能输出1）它匹配的关键字，以及2）与该关键字关联的消息？

您只需稍微重新构造代码。需要这样做是因为使用了

any

，它不返回

x

使表达式计算为

True

的信息。它只是告诉你有人做过或者没有人做过。如果您确实关心必须循环使用哪一个，或者可能使用

next

。无论如何，这里有一种方法：

for description in descriptions:
    print(description)
    if has_message == True:
        for x in phrase_list:
            if x in description:
                print(x, phrase_list[x])
                break
        else:
            print("No matches, but a phrase list exists")

---

注:

---

如果的

上的
else
令人困惑，只需将其删除即可。仅当x不在任何描述中时，代码才会到达它。
您只需稍微重新构造代码即可。需要这样做是因为使用了
any
，它不返回
x
使表达式计算为
True
的信息。它只是告诉你有人做过或者没有人做过。如果您确实关心必须循环使用哪一个，或者可能使用
next
。无论如何，这里有一种方法：

for description in descriptions:
    print(description)
    if has_message == True:
        for x in phrase_list:
            if x in description:
                print(x, phrase_list[x])
                break
        else:
            print("No matches, but a phrase list exists")


注:

如果

的

上的
else
令人困惑，只需将其删除即可。仅当x不在任何描述中时，代码才会到达它。
可能需要对其进行一些调整，但您可以使用正则表达式搜索匹配的键，然后在字典中查找该键，例如：

import re

phrase_list = {'social media': 'This person is in marketing!', 
 'marketing': 'This person is in marketing!', 
 'finance': 'This person is in finance!', 
 'fintech': 'This person is in finance!'}

descriptions = ["I am in marketing, and it is super awesome", "I am in finance, and it is super awesome", 'john smith']

def find_department(lookup, text):
    m = re.search('|'.join(sorted(lookup, key=len, reverse=True)), text)
    if m:
        return lookup[m.group(0)]
    else:
        return 'This person is a mystery!'

然后运行此命令将为您提供：

for desc in descriptions:
    print(desc, '->', find_department(phrase_list, desc))

#I am in marketing, and it is super awesome -> This person is in marketing!
#I am in finance, and it is super awesome -> This person is in finance!
#john smith -> This person is a mystery!

可能需要对其进行一些调整，但您可以使用正则表达式搜索匹配的键，然后在字典中查找该键，例如：

import re

phrase_list = {'social media': 'This person is in marketing!', 
 'marketing': 'This person is in marketing!', 
 'finance': 'This person is in finance!', 
 'fintech': 'This person is in finance!'}

descriptions = ["I am in marketing, and it is super awesome", "I am in finance, and it is super awesome", 'john smith']

def find_department(lookup, text):
    m = re.search('|'.join(sorted(lookup, key=len, reverse=True)), text)
    if m:
        return lookup[m.group(0)]
    else:
        return 'This person is a mystery!'

然后运行此命令将为您提供：

for desc in descriptions:
    print(desc, '->', find_department(phrase_list, desc))

#I am in marketing, and it is super awesome -> This person is in marketing!
#I am in finance, and it is super awesome -> This person is in finance!
#john smith -> This person is a mystery!

因此，您需要查看
descriptions
列表中的字符串是否包含任何作为dict？@Ev.Kounis上的值存储的字符串。OP希望测试
说明
列表中的每个字符串，以查看它是否包含在
短语列表
中存储为键的任何字符串，如果是，则打印键及其相关值。如果所有键都是单字，这将很容易，但像“社交媒体”这样的键会让它变得有点棘手。@PM2Ring-ExactlySo您想要的是查看
descriptions
列表中的字符串是否包含任何存储为dict？@Ev.Kounis上的值的字符串。OP希望测试
说明
列表中的每个字符串，以查看它是否包含在
短语列表
中存储为键的任何字符串，如果是，则打印键及其相关值。如果所有的键都是单字，这将很容易，但像“社交媒体”这样的键会让它变得有点棘手。@PM2Ring-这正是我需要的，谢谢。它完全符合我现有的代码，不需要更多的函数或导入的库！这正是我需要的，谢谢。它完全符合我现有的代码，不需要更多的函数或导入的库！