python - Ipython 在命令提示符下工作，但在浏览器中不工作

Question

我使用的是 Windows 8 64 位笔记本电脑。我的电脑上已经安装了 python 2.7。所以我使用 easy_install 安装了 Ipython。我现在可以在 C:\Python27\Scripts 中看到一个 ipython 应用程序文件。当我运行该文件时，它会打开我的 cmd，我可以编写代码。但是，当我在同一文件夹下的 cmd 中说 Ipython notebook 时，它不会在浏览器中打开笔记本并出现以下错误 -

C:\Users\UserName>cd C:\Python27\Scripts\

C:\Python27\Scripts>Ipython notebook
Traceback (most recent call last):
  File "C:\Python27\Scripts\ipython-script.py", line 9, in <module>
    load_entry_point('ipython==1.2.0', 'console_scripts', 'ipython')()
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\__init__.p
y", line 118, in start_ipython
    return launch_new_instance(argv=argv, **kwargs)
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\config\app
lication.py", line 544, in launch_instance
    app.initialize(argv)
  File "<string>", line 2, in initialize
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\config\app
lication.py", line 89, in catch_config_error
    return method(app, *args, **kwargs)
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\terminal\i
papp.py", line 312, in initialize
    super(TerminalIPythonApp, self).initialize(argv)
  File "<string>", line 2, in initialize
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\config\app
lication.py", line 89, in catch_config_error
    return method(app, *args, **kwargs)
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\core\appli
cation.py", line 373, in initialize
    self.parse_command_line(argv)
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\terminal\i
papp.py", line 307, in parse_command_line
    return super(TerminalIPythonApp, self).parse_command_line(argv)
  File "<string>", line 2, in parse_command_line
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\config\app
lication.py", line 89, in catch_config_error
    return method(app, *args, **kwargs)
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\config\app
lication.py", line 474, in parse_command_line
    return self.initialize_subcommand(subc, subargv)
  File "<string>", line 2, in initialize_subcommand
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\config\app
lication.py", line 89, in catch_config_error
    return method(app, *args, **kwargs)
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\config\app
lication.py", line 405, in initialize_subcommand
    subapp = import_item(subapp)
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\utils\impo
rtstring.py", line 42, in import_item
    module = __import__(package, fromlist=[obj])
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\html\noteb
ookapp.py", line 36, in <module>
    check_for_zmq('2.1.11', 'IPython.html')
  File "C:\Python27\lib\site-packages\ipython-1.2.0-py2.7.egg\IPython\utils\zmqr
elated.py", line 37, in check_for_zmq
    raise ImportError("%s requires pyzmq >= %s"%(required_by, minimum_version))
ImportError: IPython.html requires pyzmq >= 2.1.11

C:\Python27\Scripts>

Answer 1

简答:

确保 pyzmq 安装在与 ipython 相同的(或没有)虚拟环境中。对我来说，ipython notebook 是按常规安装的，但它的依赖项已放入我的虚拟环境中。在正确的位置快速安装为我解决了这个错误。

长答案:

我将此解决方案放在这里是因为错误消息是相同的，尽管场景与原始发布者的不同，我认为。

我在虚拟盒子上的虚拟环境中运行 iPython notebook。我认为虚拟盒子位是无关紧要的，但虚拟环境不是。我在虚拟环境中通过 pip 安装了 ipython[notebook] 并得到了同样的错误 ImportError: IPython.html requires pyzmq >= 2.1.11

但是，我可以清楚地看到 pyzmq 已安装。 pip freeze 和 pip install --upgrade pyzmq 都显示我有最新的 pyzmq 版本可用，它肯定是 >= 2.1.11 (pip freeze|grep pyzmq 给出 pyzmq==14.3.1).

我退出虚拟环境，再次查看pyzmq，发现在virtualenv中只安装了pyzmq。这应该是它应该工作的方式，但由于某种原因，ipython 正在寻找主要的 python 位置，而不是 virtualenv 的 python 位置。我怀疑 ipython 是安装在虚拟环境之外的。我不知道为什么 pyzmq 只安装在里面。无论如何，pip install pyzmq 在 我的虚拟环境之外修复了一切，现在一切正常。抱歉回答冗长。