我正在尝试删除字符串中的所有单个字符
输入:“这是一辆大车,它有宽敞的座位”
我的输出应该是:
输出:“这是一辆大车,它有宽敞的座位”
这里我使用了表达式
import re
re.compile('\b(?<=)[a-z](?=)\b')
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这与字符串中的第一个单个字符匹配...
任何帮助将不胜感激......提前致谢
我正在尝试在Django中安装django-oauth2-provider.在安装和配置settings.py之后,在迁移期间,我收到的错误如下:
django.core.exceptions.AppRegistryNotReady:尚未加载应用.
settings.py
INSTALLED_APPS = [
'django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
'rest_framework',
'hello_api',
'rest_framework.authtoken',
'provider',
'provider.oauth2',
]
MIDDLEWARE = [
'django.middleware.security.SecurityMiddleware',
'django.contrib.sessions.middleware.SessionMiddleware',
'django.middleware.common.CommonMiddleware',
'django.middleware.csrf.CsrfViewMiddleware',
'django.contrib.auth.middleware.AuthenticationMiddleware',
'django.contrib.messages.middleware.MessageMiddleware',
'django.middleware.clickjacking.XFrameOptionsMiddleware',
#'corsheaders.middleware.CorsMiddleware',
]
ROOT_URLCONF = 'hello_api2.urls'
TEMPLATES = [
{
'BACKEND': 'django.template.backends.django.DjangoTemplates',
'DIRS': [],
'APP_DIRS': True,
'OPTIONS': {
'context_processors': [
'django.template.context_processors.debug',
'django.template.context_processors.request',
'django.contrib.auth.context_processors.auth',
'django.contrib.messages.context_processors.messages',
],
},
},
]
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错误回溯:
error: File "manage.py", line 22, in <module>
execute_from_command_line(sys.argv)
File "/home/ravi/PycharmProjects/hello-api2/venv/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 367, in execute_from_command_line
utility.execute()
File "/home/ravi/PycharmProjects/hello-api2/venv/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 341, in execute
django.setup()
File "/home/ravi/PycharmProjects/hello-api2/venv/local/lib/python2.7/site-packages/django/__init__.py", …Run Code Online (Sandbox Code Playgroud) 下面的代码片段以递归方式将嵌套列表和元素展平为新列表,但无法将所有元素附加到列表中.
预期产量:[1,2,4,5,6,7,5,8]
我的输出:[1,2,8]
def foo(l):
result = []
for i in l:
if type(i)==list:
foo(i)
else:
result.append(i)
return result
input_list = [1,2,[4,5,[6,7],5],8]
print (foo(input_list))
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