小编ger*_*ste的帖子

我是个新手,而且我在jupyter笔记本中生成它们时遇到问题.

每当我生成一个情节时,evrything工作正常,但是当我尝试保存笔记本时,我收到一条错误消息,告诉我笔记本验证失败,因为它在任何给定的模式下无效(例如直接从plot.ly webiste获取).

这是一个例子:

import plotly.plotly as py
iplot([{"x": [1, 2, 3], "y": [3, 1, 6]}])

当我尝试保存笔记本时,我收到以下消息:笔记本验证失败:{u'data':[{u'y':[3,1,6],u'x':[1,2,3] }],u'layout':{}}在任何给定的模式下无效:

{
 "data": [
  {
   "y": [
    3, 
    1, 
    6
   ], 
   "x": [
    1, 
    2, 
    3
   ]
  }
 ], 
 "layout": {}
}

谢谢!

我一直在尝试使用Scrapy从Google Analytics获取一些数据,尽管事实上我是一个完整的Python新手我已经取得了一些进展.我现在可以通过Scrapy登录Google Analytics,但我需要发出一个AJAX请求来获取我想要的数据.我试图用下面的代码复制我的浏览器的HTTP请求标题,但它似乎不起作用,我的错误日志说

解包的价值太多了

有人可以帮忙吗？我已经工作了两天,我觉得我非常接近,但我也很困惑.

这是代码:

from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector
from scrapy.http import FormRequest, Request
from scrapy.selector import Selector
import  logging
from super.items import SuperItem
from scrapy.shell import inspect_response
import json

class LoginSpider(BaseSpider):
    name = 'super'
    start_urls = ['https://accounts.google.com/ServiceLogin?service=analytics&passive=true&nui=1&hl=fr&continue=https%3A%2F%2Fwww.google.com%2Fanalytics%2Fweb%2F%3Fhl%3Dfr&followup=https%3A%2F%2Fwww.google.com%2Fanalytics%2Fweb%2F%3Fhl%3Dfr#identifier']

    def parse(self, response):
        return [FormRequest.from_response(response,
                    formdata={'Email': 'Email'},

                    callback=self.log_password)]


    def log_password(self, response):
        return [FormRequest.from_response(response,
                    formdata={'Passwd': 'Password'},

                    callback=self.after_login)]

    def after_login(self, response):
      if "authentication failed" in response.body:
        self.log("Login failed", level=logging.ERROR)
        return
    # We've successfully authenticated, let's have some fun!
      else: …