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Web scraping using selenium and beautifulsoup.. trouble in parsing and selecting button

I am trying to web scrape the following website "url='https://angel.co/life-sciences'
". The website contains more than 8000 data. From this page I need the information like company name and link, joined date and followers. Before that I need to sort the followers column by clicking the button. then load more information by clicking more hidden button. The page is clickable (more hidden) content at the max 20 times, after that it doesn't load more information. But I can take only top follower information by sorting it. Here I have implemented the click() event but it's showing error.
Unable to locate element: {"method":"xpath","selector":"//div[#class="column followers sortable sortable"]"} #before edit this was my problem, using wrong class name
So do I need to give more sleep time here?(tried giving that but same error)
I need to parse all the above information then visit individual link of those website to scrape content div of that html page only.
Please suggest me a way to do it
Here is my current code, I have not added html parsing part using beautifulsoup.
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
from time import sleep
from selenium import webdriver
from bs4 import BeautifulSoup
#import urlib2
driver = webdriver.Chrome()
url='https://angel.co/life-sciences'
driver.get(url)
sleep(10)
driver.find_element_by_xpath('//div[#class="column followers sortable"]').click()#edited
sleep(5)
for i in range(2):
driver.find_element_by_xpath('//div[#class="more hidden"]').click()
sleep(8)
sleep(8)
element = driver.find_element_by_id("root").get_attribute('innerHTML')
#driver.execute_script("return document.getElementsByTagName('html')[0].innerHTML")
#WebDriverWait(driver, 10).until(EC.visibility_of_element_located((By.CLASS_NAME, 'more hidden')))
'''
results = html.find_elements_by_xpath('//div[#class="name"]')
# wait for the page to load
for result in results:
startup = result.find_elements_by_xpath('.//a')
link = startup.get_attribute('href')
print(link)
'''
page_source = driver.page_source
html = BeautifulSoup(element, 'html.parser')
#for link in html.findAll('a', {'class': 'startup-link'}):
# print link
divs = html.find_all("div", class_=" dts27 frw44 _a _jm")
The above code was working and was giving me html source before I have added the Followers click event.
My final goal is to import all these five information like Name of the company, Its link, Joined date, No of Followers and the company description (which to be obtained after visiting their individual links) into a CSV or xls file.
Help and comments are apprecieted.
This is my first python work and selenium, so little confused and need guidance.
Thanks :-)

The click method is intended to emulate a mouse click; it's for use on elements that can be clicked, such as buttons, drop-down lists, check boxes, etc. You have applied this method to a div element which is not clickable. Elements like div, span, frame and so on are used to organise HTML and provide for decoration of fonts, etc.
To make this code work you will need to identify the elements in the page that are actually clickable.

Oops my typing mistake or some silly mistake here, I was using the div class name wrong it is "column followers sortable" instead I was using "column followers sortable selected". :-(
Now the above works pretty good.. but can anyone guide me with beautifulsoup HTML parsing part?

Python scrape table

I'm new to programming so it's very likely my idea of doing what I'm trying to do is totally not the way to do that.
I'm trying to scrape standings table from this site - http://www.flashscore.com/hockey/finland/liiga/ - for now it would be fine if I could even scrape one column with team names, so I try to find td tags with the class "participant_name col_participant_name col_name" but the code returns empty brackets:
import requests
from bs4 import BeautifulSoup
import lxml
def table(url):
teams = []
source = requests.get(url).content
soup = BeautifulSoup(source, "lxml")
for td in soup.find_all("td"):
team = td.find_all("participant_name col_participant_name col_name")
teams.append(team)
print(teams)
table("http://www.flashscore.com/hockey/finland/liiga/")
I tried using tr tag to retrieve whole rows, but no success either.

I think the main problem here is that you are trying to scrape a dynamically generated content using requests, note that there's no participant_name col_participant_name col_name text at all in the HTML source of the page, which means this is being generated with JavaScript by the website. For that job you should use something like selenium together with ChromeDriver or the driver that you find better, below is an example using both of the mentioned tools:
from bs4 import BeautifulSoup
from selenium import webdriver
url = "http://www.flashscore.com/hockey/finland/liiga/"
driver = webdriver.Chrome()
driver.get(url)
source = driver.page_source
soup = BeautifulSoup(source, "lxml")
elements = soup.findAll('td', {'class':"participant_name col_participant_name col_name"})
I think another issue with your code is the way you were trying to access the tags, if you want to match a specific class or any other specific attribute you can do so using a Python's dictionary as an argument of .findAll function.
Now we can use elements to find all the teams' names, try print(elements[0]) and notice that the team's name is inside an a tag, we can access it using .a.text, so something like this:
teams = []
for item in elements:
team = item.a.text
print(team)
teams.append(team)
print(teams)
teams now should be the desired output:
>>> teams
['Assat', 'Hameenlinna', 'IFK Helsinki', 'Ilves', 'Jyvaskyla', 'KalPa', 'Lukko', 'Pelicans', 'SaiPa', 'Tappara', 'TPS Turku', 'Karpat', 'KooKoo', 'Vaasan Sport', 'Jukurit']
teams could also be created using list comprehension:
teams = [item.a.text for item in elements]

Mr Aguiar beat me to it! I will just point out that you can do it all with selenium alone. Of course he is correct in pointing out that this is one of the many sites that loads most of its content dynamically.
You might be interested in observing that I have used an xpath expression. These often make for compact ways of saying what you want. Not too hard to read once you get used to them.
>>> from selenium import webdriver
>>> driver = webdriver.Chrome()
>>> driver.get('http://www.flashscore.com/hockey/finland/liiga/')
>>> items = driver.find_elements_by_xpath('.//span[#class="team_name_span"]/a[text()]')
>>> for item in items:
... item.text
...
'Assat'
'Hameenlinna'
'IFK Helsinki'
'Ilves'
'Jyvaskyla'
'KalPa'
'Lukko'
'Pelicans'
'SaiPa'
'Tappara'
'TPS Turku'
'Karpat'
'KooKoo'
'Vaasan Sport'
'Jukurit'

You're very close.
Start out being a little less ambitious, and just focus on "participant_name". Take a look at https://www.crummy.com/software/BeautifulSoup/bs4/doc/#find-all . I think you want something like:
for td in soup.find_all("td", "participant_name"):
Also, you must be seeing different web content than I am. After a wget of your URL, grep doesn't find "participant_name" in the text at all. You'll want to verify that your code is looking for an ID or a class that is actually present in the HTML text.

Achieving the same using css selector which will let you make the code more readable and concise:
from selenium import webdriver; driver = webdriver.Chrome()
driver.get('http://www.flashscore.com/hockey/finland/liiga/')
for player_name in driver.find_elements_by_css_selector('.participant_name'):
print(player_name.text)
driver.quit()

Get ALL links from driver.find_elements by href not working

New to python and selenium webdriver. I am trying to check all the links on my own webpage and use it's http status code to see if it is a broken link or not. The code that I am running (reduced from original)...
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
import requests
links = driver.find_elements_by_xpath("//a[#href]")
while len(links):
url = links.pop()
url = url.get_attribute("href")
print(url)
The html looks like...
<ul>
<li>visit google</li>
<li>broken link ex</li>
</ul>
When I run my script, the only link that gets printed is the google link and not the broken link. I have done some test cases and it seems that only the links that include the phrase "http://www" in the link get printed. Although I can change the href links on my webpage to include this phrase, I have specific reasons as to why they cannot be included.
If I can just get all the links (with or without the "http://www" phrase) using driver.find_elements_by_xpath("//a[#href]"), then I can convert these later in the script to include the phrase and then get the http status codes.
I saw other posts but none that helped me get over this obstacle. Any clarification/workaround/hint would be appreciated.

the following list comprehension should get you a list of all links. It locates all anchor tags and generates a list containing the 'href' attribute of each element.
links = [elem.get_attribute("href") for elem in driver.find_elements_by_tag_name('a')]
here is same thing broken down into small steps and used as a function:
def get_all_links(driver):
links = []
elements = driver.find_elements_by_tag_name('a')
for elem in elements:
href = elem.get_attribute("href")
links.append(href)
return links

Trying to select element by xpath with Selenium but getting error "Unable to locate element"

I am trying to scrape the list of followings for a given instagram user. This requires using Selenium to navigate to the user's Instagram page and then clicking "following". However, I cannot seem to click the "following" button with Selenium.
driver = webdriver.Chrome()
url = 'https://www.instagram.com/beforeeesunrise/'
driver.get(url)
driver.find_element_by_xpath('//*[#id="react-root"]/section/main/article/header/div[2]/ul/li[3]/a').click()
However, this results in a NoSuchElementException. I copied the xpath from the html, tried using the class name, partial link and full link and cannot seem to get this to work! I've also made sure that the above xpath include the element with a "click" event listener.
UPDATE: By logging in I was able to get the above information. However (!), now I cannot get the resulting list of "followings". When I click on the button with the driver, the html does not include the information in the pop up dialog that you see on Instagram. My goal is to get all of the users that the given username is following.

Make sure you are using the correct X Path.
Use the following link to get perfect X Paths to access web elements and then try.
Selenium Command
Hope this helps to solve the problem!

Try a different XPath. I've verified this is unique on the page.
driver.find_element_by_xpath("//a[contains(.,'following')]")

It's not the main goal of selenium to provide rich functionalities, from a web-scraping perspective, to find elements on the page, so the better option is to delegate this task to a specific tool, like BeautifulSoup. After we find what we're looking for, then, we can ask for selenium to interact with the element.
The bridge between selenium and BeautifulSoup will be this amazing function below that I found here. The function gets a single BeautifulSoup element and generates a unique XPATH that we can use on selenium.
import os
import re
from selenium import webdriver
from bs4 import BeautifulSoup as bs
import itertools
def xpath_soup(element):
"""
Generate xpath of soup element
:param element: bs4 text or node
:return: xpath as string
"""
components = []
child = element if element.name else element.parent
for parent in child.parents:
"""
#type parent: bs4.element.Tag
"""
previous = itertools.islice(parent.children, 0, parent.contents.index(child))
xpath_tag = child.name
xpath_index = sum(1 for i in previous if i.name == xpath_tag) + 1
components.append(xpath_tag if xpath_index == 1 else '%s[%d]' % (xpath_tag, xpath_index))
child = parent
components.reverse()
return '/%s' % '/'.join(components)
driver = webdriver.Chrome(executable_path=YOUR_CHROMEDRIVER_PATH)
driver.get(url = 'https://www.instagram.com/beforeeesunrise/')
source = driver.page_source
soup = bs(source, 'html.parser')
button = soup.find('button', text=re.compile(r'Follow'))
xpath_for_the_button = xpath_soup(button)
elm = driver.find_element_by_xpath(xpath_for_the_button)
elm.click()
...and works!
( but you need writing some code to log in with an account)

Parsing HTML Content using BeautifulSoup & Selenium

from selenium import webdriver
from selenium.webdriver.support.ui import Select
from bs4 import BeautifulSoup
import csv
import requests
import re
driver2 = webdriver.Chrome()
driver2.get("http://www.squawka.com/match-results?ctl=10_s2015")
soup=BeautifulSoup(driver2.page_source)
print soup
driver2.quit()
I'm trying to get the HREF of every "td", "Class":"Match Centre" and I need to use selenium to navigate through the pages but im struggling to incorporate the two so I can change the menu options and navigate through the different pages while feeding the links into my other code.
I've researched and tried ('inner-html') and the page.source currently in the code, but it doesn't get any of the web links I need.
Does anyone have a solution to get these links and navigate on the page. Could there be a way to get the XML of this page to get all the links?

Not sure why would you need BeautifulSoup (BS) here. Selenium alone is capable of locating elements and navigating through links on a page. For example, to get all the links to the match details page you can do as follow :
>>> matches = driver.find_elements_by_xpath("//td[#class='match-centre']/a")
>>> print [match.get_attribute("href") for match in matches]
As for navigating through the pages, you can use the following XPath :
//span[contains(#class,'page-numbers')]/following-sibling::a[1]
The above XPath finds link to the next page. To navigate through all the pages, you can try using a while loop; while the link to the next page is found :
perform a click action on the link,
grab all the href from current page,
locate the next page link.