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Hi I would like to know how you can extract html / webbrowser data
So I can add it to a string
import requests
r = requests.get(“http://somesite.com”)
print(r.text)
you can save r.text into a variable... perform split function on it to put into a list... or use regular expressions to extract data between tags... there's a million possibilities once you have that string. Have fun!
Related
I am trying to clean the data I scraped from their links. I have over 100 links in a CSV I'm trying to clean.
This is what a link looks like in the CSV:
"https://www.google.com/url?rct=j&sa=t&url=https://www.somenewswebsite.com/news/society/new-covid-variant-decline-across-the-nation/014465428&ct=ga&cd=SldisGkopisopiasenjA6Y28Ug&usg=AFQjaskdfYJkasKugowe896fsdgfsweF"
I've observed that scraping this for HTML data doesn't go well and I have to get the URL present inside this.
I want to get the substring which starts with &url= and ends at &ct as that's where the real URL resides.
I've read posts like this but couldn't find one for ending str too. I've tried an approach from this using the substring package but it doesn't work for more than one character.
How do I do this? Preferably without using third party packages?
I don't understand problem
If you have string then you can use string- functions like .find() and slice [start:end]
text = "https://www.google.com/url?rct=j&sa=t&url=https://www.somenewswebsite.com/news/society/new-covid-variant-decline-across-the-nation/014465428&ct=ga&cd=SldisGkopisopiasenjA6Y28Ug&usg=AFQjaskdfYJkasKugowe896fsdgfsweF"
start = text.find('url=') + len('url=')
end = text.find('&ct=')
text[start:end]
But it may have url= and ct= in different order so better search first & after url=
text = "https://www.google.com/url?rct=j&sa=t&url=https://www.somenewswebsite.com/news/society/new-covid-variant-decline-across-the-nation/014465428&ct=ga&cd=SldisGkopisopiasenjA6Y28Ug&usg=AFQjaskdfYJkasKugowe896fsdgfsweF"
start = text.find('url=') + len('url=')
end = text.find('&', start)
text[start:end]
EDIT:
There is also standard module urllib.parse to work with url - to split or join it.
text = "https://www.google.com/url?rct=j&sa=t&url=https://www.somenewswebsite.com/news/society/new-covid-variant-decline-across-the-nation/014465428&ct=ga&cd=SldisGkopisopiasenjA6Y28Ug&usg=AFQjaskdfYJkasKugowe896fsdgfsweF"
import urllib.parse
url, query = urllib.parse.splitquery(text)
data = urllib.parse.parse_qs(query)
data['url'][0]
In data you have dictionary
{'cd': ['SldisGkopisopiasenjA6Y28Ug'],
'ct': ['ga'],
'rct': ['j'],
'sa': ['t'],
'url': ['https://www.somenewswebsite.com/news/society/new-covid-variant-decline-across-the-nation/014465428'],
'usg': ['AFQjaskdfYJkasKugowe896fsdgfsweF']}
EDIT:
Python shows warning that splitquery() is deprecated as of 3.8 and code should use urlparse()
text = "https://www.google.com/url?rct=j&sa=t&url=https://www.somenewswebsite.com/news/society/new-covid-variant-decline-across-the-nation/014465428&ct=ga&cd=SldisGkopisopiasenjA6Y28Ug&usg=AFQjaskdfYJkasKugowe896fsdgfsweF"
import urllib.parse
parts = urllib.parse.urlparse(text)
data = urllib.parse.parse_qs(parts.query)
data['url'][0]
I am trying to scrape a list of dates from: https://ca.finance.yahoo.com/quote/AAPL/options
The dates are located within a drop down menu right above the option chain. I've scraped text from this website before but this text is using a 'select' & 'option' syntax. How would I adjust my code to gather this type of text? I have used many variations of the code below to try and scrape the text but am having no luck.
Thank you very much.
import bs4
import requests
datesLink = ('https://ca.finance.yahoo.com/quote/AAPL/options')
datesPage = requests.get(datesLink)
datesSoup = BeautifulSoup(datesPage.text, 'lxml')
datesQuote = datesSoup.find('div', {'class': 'Cf Pt(18px)controls'}).find('option').text
The reason you can't seem to extract this dropdown list is because this list is generated dynamically, and the easiest way to know this is by saving your html content into a file and giving it a manual look, in a text editor.
You CAN, however, parse those dates out of the script source code, which is in the same html file, using some ugly regex way. For example, this seems to work:
import requests, re
from datetime import *
content = requests.get('https://ca.finance.yahoo.com/quote/AAPL/options').content.decode()
match = re.search(r'"OptionContractsStore".*?"expirationDates".*?\[(.*?)\]', content)
dates = [datetime.fromtimestamp(int(x), tz=timezone.utc) for x in match.group(1).split(',')]
for d in dates:
print(d.strftime('%Y-%m-%d'))
It should be obvious that parsing stuff in such a nasty way isn't fool-proof, and likely going to break sooner rather than later. But the same can be said about any kind of web scraping entirely.
You can simply read HTML directly to Pandas:
import pandas as pd
URI = 'https://ca.finance.yahoo.com/quote/AAPL/options'
df = pd.read_html(URI)[0] #[1] depending on the table you wish for
I've got the following HTML code :
Bad URL
This is a string \uff42\uff41\uff44\uff0e\uff43\uff4f\uff4d which both
Chrome and Firefox think http://ｂａｄ．ｃｏｍ is the same as http://bad.com.
I need to compare the collected URLs with a list of whitelisted URLs.
How do I convert http://ｂａｄ．ｃｏｍ to http://bad.com using Python? Do the browsers replace "confusible" characters?
Alternatively is it possible to compare two URLs?
You can use unicodedata:
import unicodedata
link = 'http://ｂａｄ．ｃｏｍ'
normalized = unicodedata.normalize('NFKC', link)
What is 'NFKC' you can find in official docs.
I want to scrape the data on this link
http://www.realclearpolitics.com/epolls/json/5491_historical.js?1453388629140&callback=return_json
I am not sure what type of this link is, is it html or json or something else. Sorry for my bad web knowledge. But I try to use the following code to scrape:
import requests
url='http://www.realclearpolitics.com/epolls/json/5491_historical.js?1453388629140&callback=return_json'
source=requests.get(url).text
The type of the source is unicode. I also try to use the urllib2 to scrape like:
source2=urllib2.urlopen(url).read()
The type of source2 is string. I am not sure which method is better. Because the link is not like the normal webpage contains different tags. If I want to clean the scraped data and form the dataframe data (like the pandas dataframe), what method or process I should follow/
Thanks.
The returned response is text containing valid JSON data within it. You can validate it on your own using a service like http://jsonlint.com/ if you want. For doing so just copy the code within the brackets
return_json("JSON code to copy")
In order to make use of that data you just need to parse it in your program. Here an example: https://docs.python.org/2/library/json.html
The response is text. It does contain JSON, just need to extract it
import json
strip_len = len("return_json(")
source=requests.get(url).text[strip_len:-2]
source = json.loads(source)
I am trying to make a simple python script to extract certain links from a webpage. I am able to extract link successfully but now I want to extract some more information like bitrate,size,duration given on that webpage.
I am using the below xpath to extract the above mentioned info
>>> doc = lxml.html.parse('http://mp3skull.com/mp3/linkin_park_faint.html')
>>> info = doc.xpath(".//*[#id='song_html']/div[1]/text()")
>>> info[0:7]
['\n\t\t\t', '\n\t\t\t\t3.71 mb\t\t\t', '\n\t\t\t', '\n\t\t\t\t3.49 mb\t\t\t', '\n\t\t\t', '\n\t\t\t\t192 kbps', '2:41']
Now what I need is that for a particular link the info I require is generated in a form of tuple like (bitrate,size,duration).
The xpath I mentioned above generates the required info but it is ill-formatted that is it is not possible to achieve my required format with any logic at least I am not able to that.
So, is there any way to achieve the output in my format.?
I think BeautifulSoup will do the job, it parses even badly formatted HTML:
http://www.crummy.com/software/BeautifulSoup/
parsing is quite easy with BeautifulSoup - for example:
import bs4
import urllib
soup = bs4.BeautifulSoup(urllib.urlopen('http://mp3skull.com/mp3/linkin_park_faint.html').read())
print soup.find_all('a')
and have quite good docs:
http://www.crummy.com/software/BeautifulSoup/bs4/doc/
You can actually strip everything out with XPath:
translate(.//*[#id='song_html']/div[1]/text(), "\n\t,'", '')
So for your additional question, either:
info[0, len(info)]
for altogether, or:
info.rfind(" ")
Since the translate leaves a space character, but you could replace that with whatever you wanted.
Addl info found here
How are you with regular expressions and python's re module?
http://docs.python.org/library/re.html may be essential.
As far as getting the data out of the array, re.match(regex,info[n]) should suffice, as far as the triple tuple goes, the python tuple syntax takes care of it. Simply match from members of your info array with re.match.
import re
matching_re = '.*' # this re matches whole strings, rather than what you need
incoming_value_1 = re.match(matching_re,info[1])
# etc.
var truple = (incoming_value_1, incoming_value_2, incoming_value_2