我正在使用API,其中的信息将是这样的:
"id": "17",
"address": "Av. Nossa Senhora de Copacabana",
"addressComplement": "A",
"number": "945",
"cityId": "2",
"cityName": "Rio de Janeiro",
"state": "Rio de Janeiro",
"uf": "RJ",
"neighborhood": "Copacabana",
"properties": {},
"telephones": [],
"geolocation": {
"lat": -22.97625,
"lng": -43.19002
},
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但是,在某些记录中,它不包含该geolocation字段,因此我必须检查是否geolocation存在于我的代码中.
我试图用hasattr这个技巧,但我认为我做错了.
这是我的代码的一部分:
if hasattr(i, 'geolocation'):
address_lat = i['geolocation']['lat']
address_lng = i['geolocation']['lng']
else:
address_lat = 0.0
address_lng = 0.0
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我的想法是,它将检查索引的位置是否i存在内部的东西geolocation.如果有东西,则返回true并进入条件,否则var将接收0.0.
那么,我做错了什么?这是正确的使用方式hasattr吗?
@commands.command(aliases=['lookup'])
async def define(self, message, *, arg):
dictionary=PyDictionary()
Define = dictionary.meaning(arg)
length = len(arg.split())
if length == 1:
embed = discord.Embed(description="**Noun:** " + Define["Noun"][0] + "\n\n**Verb:** " + Define["Verb"][0], color=0x00ff00)
embed.set_author(name = ('Defenition of the word: ' + arg),
icon_url=message.author.avatar_url)
await message.send(embed=embed)
else:
CommandError = discord.Embed(description= "A Term must be only a single word" , color=0xfc0328)
await message.channel.send(embed=CommandError)
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我想检查 和 是否Noun在Verb字典中Define,因为当一个单词在其定义中只包含一个名词时,它会抛出一个错误,因为我试图通过机器人输出名词和动词,看看我得到了什么。我是字典新手,非常感谢任何帮助
我正在尝试获取输入,如果字典logins有一个与我的输入匹配的键,我想返回该键的值。
logins = {
'admin':'admin',
'turtle':'password123'
}
logging = input('username: ')
for logging in logins:
print(logins.values())
Run Code Online (Sandbox Code Playgroud) 这是我的代码
if "value" not in dictionary():
do something
else:
do something else
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我收到错误'TypeError:'dict'对象不可调用.
我已经尝试将第一行更改为
if dictionary["value"]:
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但得到一个不同的错误.我在哪里错了?
我有这样的映射关键字.
categories_mapping = {
'comics': 'Comic Books',
'cartoons': 'Comic Books',
'manga': 'Comic Books',
'video and computer games': 'Video Games',
'role playing games': 'Video Games',
'immigration': 'Immigration',
'police': 'Police',
'environmental': 'Environment',
'celebrity fan and gossip': 'Celebrity',
'space and technology': 'NASA / Space',
'movies and tv': 'TV and Movies',
'elections': 'Elections',
'referendums': 'Elections',
'sex': 'Sex',
'music': 'Music',
'technology and computing': 'Technology'}
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和这样的清单.
labels = ['technology and computing', 'arts and technology']
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如果列表中的任何单词位于字典的键中,我想返回字典的值.
这就是我想出来的,但我认为这不是非常pythonic.
cats = []
for k,v in categories_mapping.items():
for l in …Run Code Online (Sandbox Code Playgroud) 我试图在歌词典中搜索关键词.它们的键是歌曲标题,值是歌曲的长度.我想在字典中搜索这首歌,然后打印出那首歌和它的时间.我已经想出要搜索这首歌,但是不记得如何突出它的价值.这是我现在拥有的.
def getSongTime(songDictionary):
requestedSong=input("Enter song from playlist: ")
for song in list(songDictionary.keys()):
if requestedSong in songDictionary.keys():
print(requestedSong,value)
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