我有一个巨大的术语列表，我想从文本文件中提取并将它们分组到以下组之一：动物，艺术，建筑物，车辆，人，人，食物，玻璃，瓶子，标牌，口号，DJ，派对。我目前在tester2文件中有四个单词：
I like sorbet
I am a man wearing a shirt
Pizza is my favorite meal
formula 1 racing is awesome
steak
这是我的代码：
keyword_dictionary = {
'Animal' : ['animal', 'dog', 'cat'],
'Art' : ['art', 'sculpture', 'fearns'],
'Buildings' : ['building', 'architecture', 'gothic', 'skyscraper'],
'Vehicle' : ['car','formula','f-1','f1','f 1','f one','f-one','moped','mo ped','mo-ped','scooter'],
'Person' : ['person','dress','shirt','woman','man','attractive','adult','smiling','sleeveless','halter','spectacles','button','bodycon'],
'People' : ['people','women','men','attractive','adults','smiling','group','two','three','four','five','six','seven','eight','nine','ten','2','3','4','5','6','7','8','9','10'],
'Food' : ['food','plate','chicken','steak','pizza','pasta','meal','asian','beef','cake','candy','food pyramid','spaghetti','curry','lamb','sushi','meatballs','biscuit','apples','meat','mushroom','jelly', 'sorbet','nacho','burrito','taco','cheese'],
'Glass' : ['glass','drink','container','glasses','cup'],
'Bottle' : ['bottle','drink'],
'Signage' : ['sign','martini','ad','advert','card','bottles','logo','mat','chalkboard','blackboard'],
'Slogan' : ['Luck is overrated'],
'DJ' : ['dj','disc','jockey','mixer','instrument','turntable'],
'Party' : ['party']
}
def matcher(keywords, searcher):
for key, words in keywords.iteritems():
if searcher in words:
print key
with open("tester2.txt") as termsdesk:
for line in termsdesk:
term = matcher(keyword_dictionary, line.strip())
我希望我的结果看起来像这样：
Food
Person
Food
Vehicle
Food
但我只是得到了这个：
Food
我想象的是，而是让我的代码完成它需要制作的精确匹配＆＃34;类似于＆＃34;火柴。我不确定如何实现这一点。是否可以使用＆＃34; if＆＃34;功能也许？
答案 0 :(得分：1)
反转映射更有意义，效率更高：
keyword_dictionary = {'mo-ped': 'Vehicle', 'group': 'People', 'spaghetti': 'Food', 'f-1': 'Vehicle', '6': 'People',
'5': 'People', 'five': 'People', 'gothic': 'Buildings', 'seven': 'People', 'adults': 'People',
'burrito': 'Food', 'martini': 'Signage', 'f one': 'Vehicle', 'ten': 'People', 'instrument': 'DJ',
'dress': 'Person', 'drink': 'Bottle', 'mushroom': 'Food', 'cat': 'Animal', 'glass': 'Glass',
'animal': 'Animal', 'pizza': 'Food', 'formula': 'Vehicle', 'meal': 'Food', 'curry': 'Food',
'3': 'People', 'sign': 'Signage', 'f1': 'Vehicle', 'biscuit': 'Food', 'bottles': 'Signage',
'pasta': 'Food', 'card': 'Signage', 'sculpture': 'Art', '8': 'People', 'apples': 'Food', '9':
'People', 'nacho': 'Food', 'mat': 'Signage', 'bottle': 'Bottle', 'shirt': 'Person', 'halter':
'Person', 'jockey': 'DJ', 'six': 'People', 'beef': 'Food', 'party': 'Party', 'container': 'Glass',
'women': 'People', 'four': 'People', '10': 'People', 'attractive': 'Person', 'mo ped': 'Vehicle',
'blackboard': 'Signage', 'two': 'People', 'f-one': 'Vehicle', '4': 'People', 'car': 'Vehicle',
'cheese': 'Food', 'plate': 'Food', 'food': 'Food', 'smiling': 'Person', 'bodycon': 'Person',
'jelly': 'Food', 'button': 'Person', 'men': 'People', 'people': 'People', 'eight': 'People',
'sushi': 'Food', 'chalkboard': 'Signage', 'cake': 'Food', 'sorbet': 'Food', 'turntable': 'DJ',
'2': 'People', 'skyscraper': 'Buildings', 'nine': 'People', 'meatballs': 'Food', '7': 'People',
'art': 'Art', 'building': 'Buildings', 'sleeveless': 'Person', 'lamb': 'Food', 'disc': 'DJ',
'scooter': 'Vehicle', 'asian': 'Food', 'chicken': 'Food', 'food pyramid': 'Food', 'person':
'Person', 'ad': 'Signage', 'spectacles': 'Person', 'glasses': 'Glass', 'dog': 'Animal',
'logo': 'Signage', 'mixer': 'DJ', 'dj': 'DJ', 'architecture': 'Buildings', 'three': 'People',
'fearns': 'Art', 'taco': 'Food', 'f 1': 'Vehicle', 'steak': 'Food', 'cup': 'Glass', 'man':
'Person', 'woman': 'Person', 'advert': 'Signage', 'candy': 'Food', 'meat': 'Food',
'adult': 'Person', 'moped': 'Vehicle', 'Luck is overrated': 'Slogan'}
with open("test.txt") as termsdesk:
for line in termsdesk:
for word in line.split():
if word in keyword_dictionary:
print(keyword_dictionary[word])
输出：
Food # sorbet
Person # man
Person # shirt
Food # meal
Vehicle # formula
Food # steak
如果你要走你的路线，你应该制作列表集，你需要迭代每个单词然后每个k，v配对：
keyword_dictionary = {
'Animal' : {'animal', 'dog', 'cat'},
'Art' : {'art', 'sculpture', 'fearns'},
'Buildings' : {'building', 'architecture', 'gothic', 'skyscraper'},
'Vehicle' : {'car','formula','f-1','f1','f 1','f one','f-one','moped','mo ped','mo-ped','scooter'},
'Person' : {'person','dress','shirt','woman','man','attractive','adult','smiling','sleeveless','halter','spectacles','button','bodycon'},
'People' : {'people','women','men','attractive','adults','smiling','group','two','three','four','five','six','seven','eight','nine','ten','2','3','4','5','6','7','8','9','10'},
'Food' : {'food','plate','chicken','steak','pizza','pasta','meal','asian','beef','cake','candy','food pyramid','spaghetti','curry','lamb','sushi','meatballs','biscuit','apples','meat','mushroom','jelly', 'sorbet','nacho','burrito','taco','cheese'},
'Glass' : {'glass','drink','container','glasses','cup'},
'Bottle' : {'bottle','drink'},
'Signage' : {'sign','martini','ad','advert','card','bottles','logo','mat','chalkboard','blackboard'},
'Slogan' : {'Luck is overrated'},
'DJ' : {'dj','disc','jockey','mixer','instrument','turntable'},
'Party' : {'party'}
}
def matcher(keywords, searcher):
for word in searcher:
for key, words in keywords.items():
if word in words:
print(key)
break
with open("test.txt") as termsdesk:
for line in termsdesk:
matcher(keyword_dictionary, line.split())
输出：
Food
Person
Person
Food
Vehicle
Food
您的功能不会返回任何内容，因此设置
term = matcher(....设置的字词等于None
比较你的逻辑，使用set作为值并反转映射：
您的代码将涉及遍历每一行和单词，然后遍历您的dict中的每个键和值以及另一个0(n)循环，使用您的列表查找值列表中的每个单词。
使用sets作为值，除了删除最后O(n)次搜索，用O(1)集查找替换它时，所有内容都与您自己的逻辑相同。
第一个代码只是循环遍历每一行和单词，并且通过常量工作检查单词是否为dict，如果单词在dict中，则获取每个值，因此效率更高。
如果您认为任何数量的匹配只是一个，您可以看到每个值的单词列表不是disjoint：
keyword_dictionary = {
'Animal' : {'animal', 'dog', 'cat'},
'Art' : {'art', 'sculpture', 'fearns'},
'Buildings' : {'building', 'architecture', 'gothic', 'skyscraper'},
'Vehicle' : {'car','formula','f-1','f1','f 1','f one','f-one','moped','mo ped','mo-ped','scooter'},
'Person' : {'person','dress','shirt','woman','man','attractive','adult','smiling','sleeveless','halter','spectacles','button','bodycon'},
'People' : {'people','women','men','attractive','adults','smiling','group','two','three','four','five','six','seven','eight','nine','ten','2','3','4','5','6','7','8','9','10'},
'Food' : {'food','plate','chicken','steak','pizza','pasta','meal','asian','beef','cake','candy','food pyramid','spaghetti','curry','lamb','sushi','meatballs','biscuit','apples','meat','mushroom','jelly', 'sorbet','nacho','burrito','taco','cheese'},
'Glass' : {'glass','drink','container','glasses','cup'},
'Bottle' : {'bottle','drink'},
'Signage' : {'sign','martini','ad','advert','card','bottles','logo','mat','chalkboard','blackboard'},
'Slogan' : {'Luck is overrated'},
'DJ' : {'dj','disc','jockey','mixer','instrument','turntable'},
'Party' : {'party'}
}
def matcher(keywords, searcher):
for key, words in keywords.items():
if not words.isdisjoint(searcher):
print(key)
with open("test.txt") as termsdesk:
for line in termsdesk:
matcher(keyword_dictionary, line.split())
输出：
Food
Person
Food
Vehicle
Food
如果每行只能获得一个匹配项，要将相同的逻辑应用于反向映射方法，您只需添加break：
with open("test.txt") as termsdesk:
for line in termsdesk:
for word in line.split():
if word in keyword_dictionary:
print(keyword_dictionary[word])
break
输出：
Food
Person
Food
Vehicle
Food
答案 1 :(得分：-1)
您是否尝试过以下操作？
config.assets.raise_runtime_errors = true
config.action_mailer.default_url_options = { host: 'localhost', port: '3000' }
config.action_mailer.delivery_method = :smtp
config.action_mailer.perform_deliveries = true
config.action_mailer.smtp_settings = {
:enable_starttls_auto => true,
:address => "webmail.xxx.com",
:port => 25,
# :domain => "xxx.com",
:authentication => :login,
:user_name => 'xxx',
:password => 'xxx',
:openssl_verify_mode => 'none'
}