小编tit*_*dam的帖子

我正在尝试使用PyInstaller创建一个独立的OSX应用程序,它运行我已经制作的GUI.当我在终端中输入以下内容时:

pyinstaller gui.py

一切似乎都有效,直到我收到以下错误:

File "/Users/username/anaconda/bin/PyInstaller", line 11, in <module>
    sys.exit(run())
File "/Users/username/anaconda/lib/python2.7/site-
packages/PyInstaller/__main__.py", line 90, in run
run_build(pyi_config, spec_file, **vars(args))
File "/Users/username/anaconda/lib/python2.7/site-
packages/PyInstaller/__main__.py", line 46, in run_build
PyInstaller.building.build_main.main(pyi_config, spec_file, **kwargs)
File "/Users/username/anaconda/lib/python2.7/site-
packages/PyInstaller/building/build_main.py", line 788, in main
build(specfile, kw.get('distpath'), kw.get('workpath'), 
kw.get('clean_build'))
File "/Users/username/anaconda/lib/python2.7/site-
packages/PyInstaller/building/build_main.py", line 734, in build
exec(text, spec_namespace)
File "<string>", line 16, in <module>
File "/Users/username/anaconda/lib/python2.7/site-
packages/PyInstaller/building/build_main.py", line 212, in __init__
self.__postinit__()
File "/Users/username/anaconda/lib/python2.7/site-
packages/PyInstaller/building/datastruct.py", line 178, in __postinit__
self.assemble()
File "/Users/username/anaconda/lib/python2.7/site-
packages/PyInstaller/building/build_main.py", line 470, in assemble …

我创建了以下数据帧 df

       col1  col2  col3
    0     4     5     2
    1     5     2     4
    2     3    10     3
    3     6     2     2
    4     3     2     4 

我现在想要的是翻转行,以便df看起来像这样:

         column_name  value
    0        col1      4
    1        col2      5
    2        col3      2
    3        col1      5
    4        col2      2
    5        col3      4
   ...       ...      ...

我想我需要使用stack(),但我不确定如何.我尝试了以下内容

df = df.stack().rename_axis(['column_name']).reset_index(name = 'value')

但是会返回以下错误

raise ValueError('Length of names must match number of levels in '
ValueError: Length of names must match number of levels in MultiIndex. …

我正在尝试使用 Excel 中的搜索公式来检查列是否包含字符串。但是，我不需要精确的匹配。所以 fi，如果列中的单元格中有 abcd，并且当我搜索 abc 时，我仍然想要返回一个值。不幸的是，目前我只有在精确搜索 abcd 时才会得到返回值。

我现在使用的公式如下所示：

  =SEARCH("abc";A2:A8)

即使 A3 中有“abcd”，也没有返回任何值。

提前致谢！

我正在尝试浏览名为 iens 的网站上的不同页面。我正在使用 selenium + python 单击“volgende”（在荷兰语中表示“下一个”），但我希望我的程序继续单击下一个，直到使用 while 循环没有更多页面为止。所以在这种情况下，我希望我的程序在第 23 页结束。现在我可以通过关闭 cookie 弹出消息进入第 2 页，等到它关闭然后单击“Volgende”按钮。

我的代码如下所示：

 from selenium.webdriver.support.ui import WebDriverWait
 from selenium import webdriver
 from selenium.webdriver.common.by import By
 from selenium.webdriver.support import expected_conditions as EC

 chrome_path = '/Users/user/Downloads/chromedriver'
 driver = webdriver.Chrome(chrome_path)
 driver.get('https://www.iens.nl/restaurant+utrecht')

 #wait for cookie message
 close_icon = WebDriverWait(driver, 5, 
 0.25).until(EC.visibility_of_element_located([By.CSS_SELECTOR, 
 '.cookiePolicy-close']))

 close_icon.click()

 #wait for cookie message to disappear
 WebDriverWait(driver, 5, 
 0.25).until(EC.invisibility_of_element_located([By.CSS_SELECTOR, 
 '.cookiePolicy-close']))

 click_icon = WebDriverWait(driver, 5, 
 0.25).until(EC.visibility_of_element_located([By.LINK_TEXT, 
 'Volgende']))
 click_icon.click()

该网站名为https://www.iens.nl/restaurant+utrecht

提前致谢！

我正在尝试使用QT Designer创建GUI.我使用以下教程将我的.ui设计器文件转换为.py文件:http://pyqt.sourceforge.net/Docs/PyQt5/designer.html.

我有以下代码:

#
# Created by: PyQt5 UI code generator 5.7
#
# WARNING! All changes made in this file will be lost!
import sys
from PyQt5 import QtCore, QtGui, QtWidgets
from PyQt5.QtWidgets import QApplication, QDialog

class Ui_ImageDialog(object):
    def setupUi(self, MainWindow):
        MainWindow.setObjectName("MainWindow")
        MainWindow.resize(800, 600)
        self.centralwidget = QtWidgets.QWidget(MainWindow)
        self.centralwidget.setObjectName("centralwidget")
        self.pushButton = QtWidgets.QPushButton(self.centralwidget)
        self.pushButton.setGeometry(QtCore.QRect(160, 200, 191, 151))
        self.pushButton.setObjectName("pushButton")
        self.pushButton_2 = QtWidgets.QPushButton(self.centralwidget)
        self.pushButton_2.setGeometry(QtCore.QRect(450, 200, 191, 151))
        self.pushButton_2.setObjectName("pushButton_2")
        self.label = QtWidgets.QLabel(self.centralwidget)
        self.label.setGeometry(QtCore.QRect(150, 70, 501, 101))
        palette = QtGui.QPalette()
        brush …

我正在学习如何使用scrapy.作为练习,我正在尝试访问https://www.ubereats.com/stores/,单击地址文本框,输入位置,然后按Enter按钮移动到包含可用于此的餐馆的下一页地点.我有以下lua代码:

function main(splash)
  local url = splash.args.url
  assert(splash:go(url))
  assert(splash:wait(5))

  local element = splash:select('.base_29SQWm')
  local bounds = element:bounds()
  assert(element:mouseclick{x = bounds.width/2, y = bounds.height/2})
    assert(element:send_text("Wall Street"))
  assert(splash:send_keys("<Return>"))
  assert(splash:wait(5))

  return {
  html = splash:html(),
  }
end

当我点击"渲染!" 在splash API中,我收到以下错误消息:

  {
      "info": {
          "message": "Lua error: [string \"function main(splash)\r...\"]:7: attempt to index local 'element' (a nil value)",
          "type": "LUA_ERROR",
          "error": "attempt to index local 'element' (a nil value)",
          "source": "[string \"function main(splash)\r...\"]",
          "line_number": 7
      },
      "error": 400,
      "type": "ScriptError",
      "description": "Error …

我正在尝试将scrapy与selenium一起使用，以便能够与javascript交互，并且仍然拥有scrapy提供的强大的抓取框架。我编写了一个访问http://www.iens.nl的脚本，在搜索栏中输入“阿姆斯特丹”，然后成功点击搜索按钮。单击搜索按钮后，我希望scrapy 从新呈现的页面中检索一个元素。不幸的是，scrapy 不返回任何值。

这是我的代码的样子：

from selenium import webdriver
from scrapy.loader import ItemLoader
from scrapy import Request
from scrapy.crawler import CrawlerProcess
from properties import PropertiesItem
import scrapy


class BasicSpider(scrapy.Spider):
    name = "basic"
    allowed_domains = ["web"]
    # Start on a property page
    start_urls = ['http://www.iens.nl']

    def __init__(self):
        chrome_path = '/Users/username/Documents/chromedriver'
        self.driver = webdriver.Chrome(chrome_path)

    def parse(self, response):
        self.driver.get(response.url)
        text_box = self.driver.find_element_by_xpath('//*[@id="searchText"]')
        submit_button = self.driver.find_element_by_xpath('//*[@id="button_search"]')
        text_box.send_keys("Amsterdam")
        submit_button.click()

        l = ItemLoader(item=PropertiesItem(), response=response)
        l.add_xpath('description', '//*[@id="results"]/ul/li[1]/div[2]/h3/a/')

        return l.load_item()


process = CrawlerProcess()
process.crawl(BasicSpider)
process.start()

“properties”是另一个脚本，看起来像这样：

from …

我想弄清楚scrapy和泼水.作为练习,我尝试在下面的javascript网站上点击按钮:http://thestlbrowns.com/然后返回新渲染页面的html.

我的代码看起来像这样:

import scrapy
import json
from scrapy import Request
class MySpider(scrapy.Spider):
    name = 'spiderman'
    domain = ['web']
    start_urls = ['http://thestlbrowns.com/']

    def start_requests(self):
        script = """
         function main(splash)
             local url = splash.args.url
             assert(splash:go(url))
             assert(splash:wait(1))

             assert(splash:runjs("$('#title.play-ball > a:first-child').click()"))
             assert(splash:wait(1))

             -- return result as a JSON object
             return {
                 html = splash:html(),
                 -- we don't need screenshot or network activity
                 --png = splash:png(),
                 --har = splash:har(),
             }
         end
        """

        for url in self.start_urls:
            yield Request(url, self.parse, meta={'splash': {'args':{'lua_source': …

我正在编写一个lua脚本，该脚本要与scrapy + splash一起用于网站。我想编写一个输入文本然后单击一个按钮的脚本。我有以下代码：

function main(splash)
   local url = splash.args.url
   assert(splash:go(url))
   assert(splash:wait(5))

   local element = splash:select('.input_29SQWm')
   assert(element:send_text("Wall Street, New York"))
   assert(splash:send_keys("<Return>"))
   assert(splash:wait(5))

   return {
     html = splash:html(),
   }
end

现在，我正在使用启动API来测试我的代码是否正常运行。当我单击“渲染！”时 我收到以下消息：

{
  "info": {
      "message": "Lua error: [string \"function main(splash)\r...\"]:7: attempt to index local 'element' (a nil value)",
      "type": "LUA_ERROR",
      "error": "attempt to index local 'element' (a nil value)",
      "source": "[string \"function main(splash)\r...\"]",
      "line_number": 7
  },
  "error": 400,
  "type": "ScriptError",
  "description": "Error happened while executing Lua script"
}

因此，由于某些原因，当我尝试发送“纽约华尔街”时，元素仍然为零。我不明白为什么；如果我在Chrome控制台中输入以下内容：

$('.input_29SQWm')

我找到了想要的元素！ …

我正在学习如何使用scrapy + splash.我已经在虚拟环境中创建了一个项目,现在我正在做这个教程:https://github.com/scrapy-plugins/scrapy-splash.

我跑过了:

$ docker run -p 8050:8050 scrapinghub/splash

结果导致:

2017-01-12 09:18:50+0000 [-] Log opened.
2017-01-12 09:18:50.225754 [-] Splash version: 2.3
2017-01-12 09:18:50.227033 [-] Qt 5.5.1, PyQt 5.5.1, WebKit 538.1, sip 4.17, Twisted 16.1.1, Lua 5.2
2017-01-12 09:18:50.227201 [-] Python 3.4.3 (default, Nov 17 2016, 01:08:31) [GCC 4.8.4]
2017-01-12 09:18:50.227645 [-] Open files limit: 1048576
2017-01-12 09:18:50.227882 [-] Can't bump open files limit
2017-01-12 09:18:50.333978 [-] Xvfb is started: ['Xvfb', ':1', '-screen', '0', '1024x768x24']
2017-01-12 …