如何将类字段作为参数传递给类方法上的装饰器?我想做的是:
class Client(object):
def __init__(self, url):
self.url = url
@check_authorization("some_attr", self.url)
def get(self):
do_work()
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它抱怨自我不存在将self.url传递给装饰者.有没有解决的办法?
我试图弄清楚如何运行Python的line_profiler来逐行执行这个问题的答案中给出的格式.
我安装了模块并调用它的LineProfiler
对象如下所示,但我得到的输出只是一次,而不是逐行汇总.
有任何想法吗?此外,我如何获得numbers = [random.randint(1,100) for i in range(1000)]
任何功能之外的线路的时间?
from line_profiler import LineProfiler
import random
def do_stuff(numbers):
s = sum(numbers)
l = [numbers[i]/43 for i in range(len(numbers))]
m = ['hello'+str(numbers[i]) for i in range(len(numbers))]
numbers = [random.randint(1,100) for i in range(1000)]
profile = LineProfiler(do_stuff(numbers))
profile.print_stats()
[] Timer unit: 3.20721e-07 s
Run Code Online (Sandbox Code Playgroud) 我想使用 line profiler 来分析我的 django 项目以显示代码性能的分析。
我正在关注这些链接:
但这对我不起作用。我收到有关开发服务器的错误,如下所示:
Traceback (most recent call last):
File "manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "/.virtualenvs/test/lib/python3.4/site-packages/django/core/management/__init__.py", line 350, in execute_from_command_line
utility.execute()
File "/.virtualenvs/test/lib/python3.4/site-packages/django/core/management/__init__.py", line 342, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/.virtualenvs/test/lib/python3.4/site-packages/django/core/management/__init__.py", line 195, in fetch_command
klass = load_command_class(app_name, subcommand)
File "/.virtualenvs/test/lib/python3.4/site-packages/django/core/management/__init__.py", line 39, in load_command_class
module = import_module('%s.management.commands.%s' % (app_name, name))
File "/.virtualenvs/test/lib64/python3.4/importlib/__init__.py", line 109, in import_module
return _bootstrap._gcd_import(name[level:], package, level)
File "<frozen importlib._bootstrap>", line 2254, in _gcd_import
File "<frozen importlib._bootstrap>", line 2237, …
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