We Keep Coding

how to make full path for url in python?

I am little bit confused about creating full url.
I have such code :
def flats(self):
return [JsonFlatPage(property_data = flat, url = flat['propertyUrl'])
for flat in self.data['properties']]
in flat['propertyUrl'] I have '/properties/75599853', but I need to get like this one:
'https://www.rightmove.co.uk/properties/75599853#/'
with full path and # at the end.
I know that I should make constant URI in settings file, but then how I can convert it? Should I use f-strings?

I think since the base url https://www.rightmove.co.uk/ is fixed, you can do something like below to get what you need:
def flats(self):
baseUrl = 'https://www.rightmove.co.uk/'
return [JsonFlatPage(property_data = flat, url = baseUrl + flat['propertyUrl'] + "#/")
for flat in self.data['properties']]
You can also use f-strings as you mentioned as:
def flats(self):
baseUrl = 'https://www.rightmove.co.uk/'
return [JsonFlatPage(property_data = flat, url = f"{baseUrl}{flat['propertyUrl']}#/")
for flat in self.data['properties']]

simple web scraper very slow

I'm fairly new to python and web-scraping in general. The code below works but it seems to be awfully slow for the amount of information its actually going through. Is there any way to easily cut down on execution time. I'm not sure but it does seem like I have typed out more/made it more difficult then I actually needed to, any help would be appreciated.
Currently the code starts at the sitemap then iterates through a list of additional sitemaps. Within the new sitemaps it pulls data information to construct a url for the json data of a webpage. From the json data I pull an xml link that I use to search for a string. If the string is found it appends it to a text file.
#global variable
start = 'https://www.govinfo.gov/wssearch/getContentDetail?packageId='
dash = '-'
urlSitemap="https://www.govinfo.gov/sitemap/PLAW_sitemap_index.xml"
old_xml=requests.get(urlSitemap)
print (old_xml)
new_xml= io.BytesIO(old_xml.content).read()
final_xml=BeautifulSoup(new_xml)
linkToBeFound = final_xml.findAll('loc')
for loc in linkToBeFound:
urlPLmap=loc.text
old_xmlPLmap=requests.get(urlPLmap)
print(old_xmlPLmap)
new_xmlPLmap= io.BytesIO(old_xmlPLmap.content).read()
final_xmlPLmap=BeautifulSoup(new_xmlPLmap)
linkToBeFound2 = final_xmlPLmap.findAll('loc')
for pls in linkToBeFound2:
argh = pls.text.find('PLAW')
theWanted = pls.text[argh:]
thisShallWork =eval(requests.get(start + theWanted).text)
print(requests.get(start + theWanted))
dict1 = (thisShallWork['download'])
finaldict = (dict1['modslink'])[2:]
print(finaldict)
url2='https://' + finaldict
try:
old_xml4=requests.get(url2)
print(old_xml4)
new_xml4= io.BytesIO(old_xml4.content).read()
final_xml4=BeautifulSoup(new_xml4)
references = final_xml4.findAll('identifier',{'type': 'Statute citation'})
for sec in references:
if sec.text == "106 Stat. 4845":
Print(dash * 20)
print(sec.text)
Print(dash * 20)
sec313 = open('sec313info.txt','a')
sec313.write("\n")
sec313.write(pls.text + '\n')
sec313.close()
except:
print('error at: ' + url2)

No idea why i spent so long on this, but i did. Your code was really hard to look through. So i started with that, I broke it up into 2 parts, getting the links from the sitemaps, then the other stuff. I broke out a few bits into separate functions too.
This is checking about 2 urls per second on my machine which seems about right.
How this is better (you can argue with me about this part).
Don't have to reopen and close the output file after each write
Removed a fair bit of unneeded code
gave your variables better names (this does not improve speed in any way but please do this especially if you are asking for help with it)
Really the main thing... once you break it all up it becomes fairly clear that whats slowing you down is waiting on the requests which is pretty standard for web-scraping, you can look into multi threading to avoid the wait. Once you get into multi threading, the benefit of breaking up your code will likely also become much more evident.
# returns sitemap links
def get_links(s):
old_xml = requests.get(s)
new_xml = old_xml.text
final_xml = BeautifulSoup(new_xml, "lxml")
return final_xml.findAll('loc')
# gets the final url from your middle url and looks through it for the thing you are looking for
def scrapey(link):
link_id = link[link.find("PLAW"):]
r = requests.get('https://www.govinfo.gov/wssearch/getContentDetail?packageId={}'.format(link_id))
print(r.url)
try:
r = requests.get("https://{}".format(r.json()["download"]["modslink"][2:]))
print(r.url)
soup = BeautifulSoup(r.text, "lxml")
references = soup.findAll('identifier', {'type': 'Statute citation'})
for ref in references:
if ref.text == "106 Stat. 4845":
return r.url
else:
return False
except:
print("bah" + r.url)
return False
sitemap_links_el = get_links("https://www.govinfo.gov/sitemap/PLAW_sitemap_index.xml")
sitemap_links = map(lambda x: x.text, sitemap_links_el)
nlinks_el = map(get_links, sitemap_links)
links = [num.text for elem in nlinks_el for num in elem]
with open("output.txt", "a") as f:
for link in links:
url = scrapey(link)
if url is False:
print("no find")
else:
print("found on: {}".format(url))
f.write("{}\n".format(url))

How to store the HTML within an opening and closing tag with Python

I am reading in an HTML document and want to store the HTML nested within a div tag of a certain name, while maintaining its structure (the spacing). This is for the ability convert an HTML doc into components for React. I am struggling with how to store the structure of the nested HTML, and locate the correct closing tag for the div the denotes that everything nested within it will become a React component (div class='rc-componentname' is the opening tag). Any help would be very appreciated. Thanks!
Edit: I assume regex are the best way to go about this. I haven't used regex before so if that is correct someone could point me in the right direction for the expression used in this context that would be great.
import os
components = []
class react_template():
def __init__(self, component_name): # add nested html as second element
self.Import = "import React, { Component } from ‘react’;"
self.Class = "Class " + component_name + ' extends Component {'
self.Render = "render() {"
self.Return = "return "
self.Export = "Default export " + component_name + ";"
def react(component):
r = react_template(component)
if not os.path.exists('components'): # create components folder
os.mkdir('components')
os.chdir('components')
if not os.path.exists(component): # create folder for component
os.mkdir(component)
os.chdir(component)
with open(component + '.js', 'wb') as f: # create js component file
for j_key, j_code in r.__dict__.items():
f.write(j_code.encode('utf-8') + '\n'.encode('utf-8'))
f.close()
def process_html():
with open('file.html', 'r') as f:
for line in f:
if 'rc-' in line:
char_soup = list(line)
for index, char in enumerate(char_soup):
if char == 'r' and char_soup[index+1] == 'c' and char_soup[index+2] == '-':
sliced_soup = char_soup[int(index+3):]
c_slice_index = sliced_soup.index("\'")
component = "".join(sliced_soup[:c_slice_index])
components.append(component)
innerHTML(sliced_soup)
# react(component)
def innerHTML(sliced_soup): # work in progress
first_closing = sliced_soup.index(">")
sliced_soup = "".join(sliced_soup[first_closing:]).split(" ")
def generate_components(components):
for c in components:
react(c)
if __name__ == "__main__":
process_html()

I see you've used the word soup in your code... maybe you've already tried and disliked BeautifulSoup? If you haven't tried it, I'd recommend you look at BeautifulSoup instead of attempting to parse HTML with regex. Although regex would be sufficient for a single tag or even a handful of tags, markup languages are deceptively simple. BeautifulSoup is a fine library and can make things easier for dealing with markup.
https://www.crummy.com/software/BeautifulSoup/bs4/doc/
This will allow you to treat the entirety of your html as a single object and enable you to:
# create a list of specific elements as objects
soup.find_all('div')
# find a specific element by id
soup.find(id="custom-header")

ValueError: can only parse strings python

I am trying to gather a bunch of links using xpath which need to be scraped from the next page however, I keep getting the error that can only parse strings? I tried looking at the type of lk and it was a string after I casted it? What seems to be wrong?
def unicode_to_string(types):
try:
types = unicodedata.normalize("NFKD", types).encode('ascii', 'ignore')
return types
except:
return types
def getData():
req = "http://analytical360.com/access-points"
page = urllib2.urlopen(req)
tree = etree.HTML(page.read())
i = 0
for lk in tree.xpath('//a[#class="sabai-file sabai-file-image sabai-file-type-jpg "]//#href'):
print "Scraping Vendor #" + str(i)
trees = etree.HTML(urllib2.urlopen(unicode_to_string(lk)))
for ll in trees.xpath('//table[#id="archived"]//tr//td//a//#href'):
final = etree.HTML(urllib2.urlopen(unicode_to_string(ll)))

You should pass in strings not urllib2.orlopen.
Perhaps change the code like so:
trees = etree.HTML(urllib2.urlopen(unicode_to_string(lk)).read())
for i, ll in enumerate(trees.xpath('//table[#id="archived"]//tr//td//a//#href')):
final = etree.HTML(urllib2.urlopen(unicode_to_string(ll)).read())
Also, you don't seem to increment i.

How do I access a dictionary value for use with the urllib module in python?

Example - I have the following dictionary...
URLDict = {'OTX2':'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=OTX2&action=view_all',
'RAB3GAP':'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=RAB3GAP1&action=view_all',
'SOX2':'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=SOX2&action=view_all',
'STRA6':'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=STRA6&action=view_all',
'MLYCD':'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=MLYCD&action=view_all'}
I would like to use urllib to call each url in a for loop, how can this be done?
I have successfully done this with with the urls in a list format like this...
OTX2 = 'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=OTX2&action=view_all'
RAB3GAP = 'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=RAB3GAP1&action=view_all'
SOX2 = 'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=SOX2&action=view_all'
STRA6 = 'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=STRA6&action=view_all'
MLYCD = 'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=MLYCD&action=view_all'
URLList = [OTX2,RAB3GAP,SOX2,STRA6,PAX6,MLYCD]
for URL in URLList:
sourcepage = urllib.urlopen(URL)
sourcetext = sourcepage.read()
but I want to also be able to print the key later when returning data. Using a list format the key would be a variable and thus not able to access it for printing, I would lonly be able to print the value.
Thanks for any help.
Tom

Have you tried (as a simple example):
for key, value in URLDict.iteritems():
print key, value

Doesn't look like a dictionary is even necessary.
dbs = ['OTX2', 'RAB3GAP', 'SOX2', 'STRA6', 'PAX6', 'MLYCD']
urlbase = 'http://lsdb.hgu.mrc.ac.uk/variants.php?select_db=%s&action=view_all'
for db in dbs:
sourcepage = urllib.urlopen(urlbase % db)
sourcetext = sourcepage.read()

I would go about it like this:
for url_key in URLDict:
URL = URLDict[url_key]
sourcepage = urllib.urlopen(URL)
sourcetext = sourcepage.read()
The url is obviously URLDict[url_key] and you can retain the key value within the name url_key. For exemple:
print url_key
On the first iteration will printOTX2.