在IBM Watson Personality Insights API于今年年底关闭之前,我正在尝试学习它的基本工作原理。我有一个基本的文本文件,我想分析,但我有麻烦让代码正常运行I have been trying to follow along on the official sits instructions,但我被卡住了。我做错了什么?(我在下面的代码中涂抹了我的钥匙)
from ibm_watson import PersonalityInsightsV3
from ibm_cloud_sdk_core.authenticators import IAMAuthenticator
authenticator = IAMAuthenticator('BlottedOutKey')
personality_insights = PersonalityInsightsV3(
version='2017-10-13',
authenticator=authenticator
)
personality_insights.set_service_url('https://api.us-west.personality-insights.watson.cloud.ibm.com')
with open(join(C:\Users\AWaywardShepherd\Documents\Data Science Projects\TwitterScraper-master\TwitterScraper-master\snscrape\python-wrapper\Folder\File.txt), './profile.json')) as profile_json:
profile = personality_insights.profile(
profile_json.read(),
content_type='text/plain',
consumption_preferences=True,
raw_scores=True)
.get_result()
print(json.dumps(profile, indent=2))
我一直收到以下难以描述的语法错误:
File "<ipython-input-1-1c7761f3f3ea>", line 11
with open(join(C:\Users\AWaywardShepherd\Documents\Data Science Projects\TwitterScraper-master\TwitterScraper-master\snscrape\python-wrapper\Folder\File.txt), './profile.json')) as profile_json:
^ SyntaxError: invalid syntax
那条
open
线有很多错误在Python中,使用
join
连接字符串以进行聚集。通常这是一个路径和文件名。从当前工作目录获取路径并将其与路径联接在代码中只传入一个字符串,因此不需要使用join
使用
open
传递文件名和模式。模式类似于'r'
表示读取模式。因此,带有连接的代码变为相关问题 更多 >
编程相关推荐