Develop Reference

How to save this list to a text file in Python

import urllib.request
import re
import json
import csv
x = 4
search_keyword = input("Enter the keyword ")
html = urllib.request.urlopen("https://www.youtube.com/results?
search_query=" + search_keyword + "&sp=EgIIAQ%253D%253D")
video_ids = re.findall(r"watch\?v=(\S{11})", html.read().decode())
for i in video_ids:
print("https://www.youtube.com/watch?v=" + i, end = ' ')
I am new to Python and recently found a tutorial how to find YouTube links in Python and I want to save the list of the links to a text file, but it gives me a error:
TypeError: 'NoneType' object is not iterable

It looks like you want to write the video links to a text file, but I noticed you're importing the csv module (technically you won't need that here)
To avoid type errors in case your object is None you could explicitly build out a list of the video links:
video_urls = ['https://www.youtube.com/watch?v=' + i for i in video_ids or []]
note the video_ids or [] part is in case the video_ids object itself is None.
then save it to an output text file:
with open('out1.txt', 'w') as out_file:
out_file.write('\n'.join(video_urls))
with Python 3 using the pathlib module, it becomes even easier (this is just a shortcut which wraps the same logic above):
from pathlib import Path
Path('out2.txt').write_text('\n'.join(video_urls))

How to input a list of URLs saved in a .txt to a Python program?

I have a list of URLs saved in a .txt file and I would like to feed them, one at a time, to a variable named url to which I apply methods from the newspaper3k python library. The program extracts the URL content, authors of the article, a summary of the text, etc, then prints the info to a new .txt file. The script works fine when you give it one URL as user input, but what should I do in order to read from a .txt with thousands of URLs?
I am only beginning with Python, as a matter of fact this is my first script, so I have tried to simply say url = (myfile.txt), but I realized this wouldn't work because I have to read the file one line at a time. So I have tried to apply read() and readlines() to it, but it wouldn't work properly because 'str' object has no attribute 'read' or 'readlines'. What should I use to read those URLs saved in a .txt file, each beginning in a new line, as the input of my simple script? Should I convert string to something else?
Extract from the code, lines 1-18:
from newspaper import Article
from newspaper import fulltext
import requests
url = input("Article URL: ")
a = Article(url, language='pt')
html = requests.get(url).text
text = fulltext(html)
download = a.download()
parse = a.parse()
nlp = a.nlp()
title = a.title
publish_date = a.publish_date
authors = a.authors
keywords = a.keywords
summary = a.summary
Later I have built some functions to display the info in a desired format and save it to a new .txt. I know this is a very basic one, but I am honestly stuck... I have read other similar questions here but I couldn't properly understand or apply the suggestions. So, what is the best way to read URLs from a .txt file in order to feed them, one at a time, to the url variable, to which other methods are them applied to extract its content?
This is my first question here and I understand the forum is aimed at more experienced programmers, but I would really appreciate some help. If I need to edit or clarify something in this post, please let me know and I will correct immediately.

Here is one way you could do it:
from newspaper import Article
from newspaper import fulltext
import requests
with open('myfile.txt',r) as f:
for line in f:
#do not forget to strip the trailing new line
url = line.rstrip("\n")
a = Article(url, language='pt')
html = requests.get(url).text
text = fulltext(html)
download = a.download()
parse = a.parse()
nlp = a.nlp()
title = a.title
publish_date = a.publish_date
authors = a.authors
keywords = a.keywords
summary = a.summary

This could help you:
url_file = open('myfile.txt','r')
for url in url_file.readlines():
print url
url_file.close()
You can apply it on your code as the following
from newspaper import Article
from newspaper import fulltext
import requests
url_file = open('myfile.txt','r')
for url in url_file.readlines():
a = Article(url, language='pt')
html = requests.get(url).text
text = fulltext(html)
download = a.download()
parse = a.parse()
nlp = a.nlp()
title = a.title
publish_date = a.publish_date
authors = a.authors
keywords = a.keywords
summary = a.summary
url_file.close()

Is there a better way to scrape this data?

For work, I was asked to create a spreadsheet of the names and addresses of all allopathic medical schools in the United States. Being new to python, I thought that this would be the perfect situation to try web scraping. While I eventually wrote a program that returned the data I needed, I know that there is a better way to do it as there were some extraneous characters (eg: ", ], [) that I had to go into excel and manually remove. I would just like to know if there was a better way I could have written this code so I can get what I needed, minus the extraneous characters.
Edit: I have also attached an image of the csv file that was created to show the extraneous characters that I'm speaking about.
from bs4 import BeautifulSoup
import requests
import csv
link = "https://members.aamc.org/eweb/DynamicPage.aspx?site=AAMC&webcode=AAMCOrgSearchResult&orgtype=Medical%20School" # noqa
# link to the site we want to scrape from
page_response = requests.get(link)
# fetching the content using the requests library
soup = BeautifulSoup(page_response.text, "html.parser")
# Calling BeautifulSoup in order to parse our document
data = []
# Empty list for the first scrape. We only get one column with many rows.
# We still have the line break tags here </br>
for tr in soup.find_all('tr', {'valign': 'top'}):
values = [td.get_text('</b>', strip=True) for td in tr.find_all('td')]
data.append(values)
data2 = []
# New list that we'll use to have name on index i, address on index i+1
for i in data:
test = list(str(i).split('</b>'))
# Using the line breaks to our advantage.
name = test[0].strip("['")
'''Here we are saying that the name of the school is the first element
before the first line break'''
addy = test[1:]
# The address is what comes after this first line break
data2.append(name)
data2.append(addy)
# Append the name of the school and address to our new list.
school_name = data2[::2]
# Making a new list that consists of the school name
school_address = data2[1::2]
# Another list that consists of the school's address.
with open("Medschooltest.csv", 'w', encoding='utf-8') as toWrite:
writer = csv.writer(toWrite)
writer.writerows(zip(school_name, school_address))
'''Zip the two together making a 2 column table with the schools name and
it's address'''
print("CSV Completed!")
Created CSV file

It seems applying conditional statements along with string manipulation can do the trick. I think the following script will lead you real close to what you want.
from bs4 import BeautifulSoup
import requests
import csv
link = "https://members.aamc.org/eweb/DynamicPage.aspx?site=AAMC&webcode=AAMCOrgSearchResult&orgtype=Medical%20School" # noqa
res = requests.get(link)
soup = BeautifulSoup(res.text, "html.parser")
with open("membersInfo.csv","w",newline="") as infile:
writer = csv.writer(infile)
writer.writerow(["Name","Address"])
for tr in soup.find_all('table', class_='bodyTXT'):
items = ', '.join([item.string for item in tr.select_one('td') if item.string!="\n" and item.string!=None])
name = items.split(",")[0].strip()
address = items.split(name)[1].strip(",")
writer.writerow([name,address])

If you have knowledge of SQL AND the data is in such a structured manner, it would be the best solution to extract it to a database.

Parsing the file name from list of url links

Ok so I am using a script that is downloading a files from urls listed in a urls.txt.
import urllib.request
with open("urls.txt", "r") as file:
linkList = file.readlines()
for link in linkList:
urllib.request.urlretrieve(link)
Unfortunately they are saved as temporary files due to lack of second argument in my urllib.request.urlretrieve function. As there are thousand of links in my text file naming them separately is not an option. The thing is that the name of the file is contained in those links, i.e. /DocumentXML2XLSDownload.vm?firsttime=true&repengback=true&d‌​ocumentId=XXXXXX&xsl‌​FileName=rher2xml.xs‌​l&outputFileName=XXX‌​X_2017_06_25_4.xls where the name of the file comes after outputFileName=
Is there an easy way to parse the file names and then use them in urllib.request.urlretrieve function as secondary argument? I was thinking of extracting those names in excel and placing them in another text file that would be read in similar fashion as urls.txt but I'm not sure how to implement it in Python. Or is there a way to make it exclusively in python without using excel?

You could parse the link on the go.
Example using a regular expression:
import re
with open("urls.txt", "r") as file:
linkList = file.readlines()
for link in linkList:
regexp = '((?<=\?outputFileName=)|(?<=\&outputFileName=))[^&]+'
match = re.search(regexp, link.rstrip())
if match is None:
# Make the user aware that something went wrong, e.g. raise exception
# and/or just print something
print("WARNING: Couldn't find file name in link [" + link + "]. Skipping...")
else:
file_name = match.group(0)
urllib.request.urlretrieve(link, file_name)

You can use urlparse and parse_qs to get the query string
from urlparse import urlparse,parse_qs
parse = urlparse('http://www.cwi.nl:80/%7Eguido/Python.html?name=Python&version=2')
print(parse_qs(parse.query)['name'][0]) # prints Python

Python Blog RSS Feed Scraping BeautifulSoup Output to .txt Files

Apologies in advance for the long block of code following. I'm new to BeautifulSoup, but found there were some useful tutorials using it to scrape RSS feeds for blogs. Full disclosure: this is code adapted from this video tutorial which has been immensely helpful in getting this off the ground: http://www.youtube.com/watch?v=Ap_DlSrT-iE.
Here's my problem: the video does a great job of showing how to print the relevant content to the console. I need to write out each article's text to a separate .txt file and save it to some directory (right now I'm just trying to save to my Desktop). I know the problem lies i the scope of the two for-loops near the end of the code (I've tried to comment this for people to see quickly--it's the last comment beginning # Here's where I'm lost...), but I can't seem to figure it out on my own.
Currently what the program does is takes the text from the last article read in by the program and writes that out to the number of .txt files that are indicated in the variable listIterator. So, in this case I believe there are 20 .txt files that get written out, but they all contain the text of the last article that's looped over. What I want the program to do is loop over each article and print the text of each article out to a separate .txt file. Sorry for the verbosity, but any insight would be really appreciated.
from urllib import urlopen
from bs4 import BeautifulSoup
import re
# Read in webpage.
webpage = urlopen('http://talkingpointsmemo.com/feed/livewire').read()
# On RSS Feed site, find tags for title of articles and
# tags for article links to be downloaded.
patFinderTitle = re.compile('<title>(.*)</title>')
patFinderLink = re.compile('<link rel.*href="(.*)"/>')
# Find the tags listed in variables above in the articles.
findPatTitle = re.findall(patFinderTitle, webpage)
findPatLink = re.findall(patFinderLink, webpage)
# Create a list that is the length of the number of links
# from the RSS feed page. Use this to iterate over each article,
# read it in, and find relevant text or <p> tags.
listIterator = []
listIterator[:] = range(len(findPatTitle))
for i in listIterator:
# Print each title to console to ensure program is working.
print findPatTitle[i]
# Read in the linked-to article.
articlePage = urlopen(findPatLink[i]).read()
# Find the beginning and end of articles using tags listed below.
divBegin = articlePage.find("<div class='story-teaser'>")
divEnd = articlePage.find("<footer class='article-footer'>")
# Define article variable that will contain all the content between the
# beginning of the article to the end as indicated by variables above.
article = articlePage[divBegin:divEnd]
# Parse the page using BeautifulSoup
soup = BeautifulSoup(article)
# Compile list of all <p> tags for each article and store in paragList
paragList = soup.findAll('p')
# Create empty string to eventually convert items in paragList to string to
# be written to .txt files.
para_string = ''
# Here's where I'm lost and have some sort of scope issue with my for-loops.
for i in paragList:
para_string = para_string + str(i)
newlist = range(len(findPatTitle))
for i in newlist:
ofile = open(str(listIterator[i])+'.txt', 'w')
ofile.write(para_string)
ofile.close()

The reason why it seems that only the last article is written down, is because all the articles are writer to 20 separate files over and over again. Lets have a look at the following:
for i in paragList:
para_string = para_string + str(i)
newlist = range(len(findPatTitle))
for i in newlist:
ofile = open(str(listIterator[i])+'.txt', 'w')
ofile.write(para_string)
ofile.close()
You are writing parag_string over and over again to the same 20 files for each iteration. What you need to be doing is this, append all your parag_strings to a separate list, say paraStringList, and then write all its contents to separate files, like so:
for i, var in enumerate(paraStringList): # Enumerate creates a tuple
with open("{0}.txt".format(i), 'w') as writer:
writer.write(var)
Now that this needs to be outside of your main loop i.e. for i in listIterator:(...). This is a working version of the program:
from urllib import urlopen
from bs4 import BeautifulSoup
import re
webpage = urlopen('http://talkingpointsmemo.com/feed/livewire').read()
patFinderTitle = re.compile('<title>(.*)</title>')
patFinderLink = re.compile('<link rel.*href="(.*)"/>')
findPatTitle = re.findall(patFinderTitle, webpage)[0:4]
findPatLink = re.findall(patFinderLink, webpage)[0:4]
listIterator = []
listIterator[:] = range(len(findPatTitle))
paraStringList = []
for i in listIterator:
print findPatTitle[i]
articlePage = urlopen(findPatLink[i]).read()
divBegin = articlePage.find("<div class='story-teaser'>")
divEnd = articlePage.find("<footer class='article-footer'>")
article = articlePage[divBegin:divEnd]
soup = BeautifulSoup(article)
paragList = soup.findAll('p')
para_string = ''
for i in paragList:
para_string += str(i)
paraStringList.append(para_string)
for i, var in enumerate(paraStringList):
with open("{0}.txt".format(i), 'w') as writer:
writer.write(var)