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How to download file by using python? [duplicate]

I have a small utility that I use to download an MP3 file from a website on a schedule and then builds/updates a podcast XML file which I've added to iTunes.
The text processing that creates/updates the XML file is written in Python. However, I use wget inside a Windows .bat file to download the actual MP3 file. I would prefer to have the entire utility written in Python.
I struggled to find a way to actually download the file in Python, thus why I resorted to using wget.
So, how do I download the file using Python?

One more, using urlretrieve:
import urllib.request
urllib.request.urlretrieve("http://www.example.com/songs/mp3.mp3", "mp3.mp3")
(for Python 2 use import urllib and urllib.urlretrieve)

Use urllib.request.urlopen():
import urllib.request
with urllib.request.urlopen('http://www.example.com/') as f:
html = f.read().decode('utf-8')
This is the most basic way to use the library, minus any error handling. You can also do more complex stuff such as changing headers.
On Python 2, the method is in urllib2:
import urllib2
response = urllib2.urlopen('http://www.example.com/')
html = response.read()

In 2012, use the python requests library
>>> import requests
>>>
>>> url = "http://download.thinkbroadband.com/10MB.zip"
>>> r = requests.get(url)
>>> print len(r.content)
10485760
You can run pip install requests to get it.
Requests has many advantages over the alternatives because the API is much simpler. This is especially true if you have to do authentication. urllib and urllib2 are pretty unintuitive and painful in this case.
2015-12-30
People have expressed admiration for the progress bar. It's cool, sure. There are several off-the-shelf solutions now, including tqdm:
from tqdm import tqdm
import requests
url = "http://download.thinkbroadband.com/10MB.zip"
response = requests.get(url, stream=True)
with open("10MB", "wb") as handle:
for data in tqdm(response.iter_content()):
handle.write(data)
This is essentially the implementation #kvance described 30 months ago.

import urllib2
mp3file = urllib2.urlopen("http://www.example.com/songs/mp3.mp3")
with open('test.mp3','wb') as output:
output.write(mp3file.read())
The wb in open('test.mp3','wb') opens a file (and erases any existing file) in binary mode so you can save data with it instead of just text.

Python 3
urllib.request.urlopen
import urllib.request
response = urllib.request.urlopen('http://www.example.com/')
html = response.read()
urllib.request.urlretrieve
import urllib.request
urllib.request.urlretrieve('http://www.example.com/songs/mp3.mp3', 'mp3.mp3')
Note: According to the documentation, urllib.request.urlretrieve is a "legacy interface" and "might become deprecated in the future" (thanks gerrit)
Python 2
urllib2.urlopen (thanks Corey)
import urllib2
response = urllib2.urlopen('http://www.example.com/')
html = response.read()
urllib.urlretrieve (thanks PabloG)
import urllib
urllib.urlretrieve('http://www.example.com/songs/mp3.mp3', 'mp3.mp3')

use wget module:
import wget
wget.download('url')

import os,requests
def download(url):
get_response = requests.get(url,stream=True)
file_name = url.split("/")[-1]
with open(file_name, 'wb') as f:
for chunk in get_response.iter_content(chunk_size=1024):
if chunk: # filter out keep-alive new chunks
f.write(chunk)
download("https://example.com/example.jpg")

An improved version of the PabloG code for Python 2/3:
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from __future__ import ( division, absolute_import, print_function, unicode_literals )
import sys, os, tempfile, logging
if sys.version_info >= (3,):
import urllib.request as urllib2
import urllib.parse as urlparse
else:
import urllib2
import urlparse
def download_file(url, dest=None):
"""
Download and save a file specified by url to dest directory,
"""
u = urllib2.urlopen(url)
scheme, netloc, path, query, fragment = urlparse.urlsplit(url)
filename = os.path.basename(path)
if not filename:
filename = 'downloaded.file'
if dest:
filename = os.path.join(dest, filename)
with open(filename, 'wb') as f:
meta = u.info()
meta_func = meta.getheaders if hasattr(meta, 'getheaders') else meta.get_all
meta_length = meta_func("Content-Length")
file_size = None
if meta_length:
file_size = int(meta_length[0])
print("Downloading: {0} Bytes: {1}".format(url, file_size))
file_size_dl = 0
block_sz = 8192
while True:
buffer = u.read(block_sz)
if not buffer:
break
file_size_dl += len(buffer)
f.write(buffer)
status = "{0:16}".format(file_size_dl)
if file_size:
status += " [{0:6.2f}%]".format(file_size_dl * 100 / file_size)
status += chr(13)
print(status, end="")
print()
return filename
if __name__ == "__main__": # Only run if this file is called directly
print("Testing with 10MB download")
url = "http://download.thinkbroadband.com/10MB.zip"
filename = download_file(url)
print(filename)

Simple yet Python 2 & Python 3 compatible way comes with six library:
from six.moves import urllib
urllib.request.urlretrieve("http://www.example.com/songs/mp3.mp3", "mp3.mp3")

Following are the most commonly used calls for downloading files in python:
urllib.urlretrieve ('url_to_file', file_name)
urllib2.urlopen('url_to_file')
requests.get(url)
wget.download('url', file_name)
Note: urlopen and urlretrieve are found to perform relatively bad with downloading large files (size > 500 MB). requests.get stores the file in-memory until download is complete.

Wrote wget library in pure Python just for this purpose. It is pumped up urlretrieve with these features as of version 2.0.

In python3 you can use urllib3 and shutil libraires.
Download them by using pip or pip3 (Depending whether python3 is default or not)
pip3 install urllib3 shutil
Then run this code
import urllib.request
import shutil
url = "http://www.somewebsite.com/something.pdf"
output_file = "save_this_name.pdf"
with urllib.request.urlopen(url) as response, open(output_file, 'wb') as out_file:
shutil.copyfileobj(response, out_file)
Note that you download urllib3 but use urllib in code

I agree with Corey, urllib2 is more complete than urllib and should likely be the module used if you want to do more complex things, but to make the answers more complete, urllib is a simpler module if you want just the basics:
import urllib
response = urllib.urlopen('http://www.example.com/sound.mp3')
mp3 = response.read()
Will work fine. Or, if you don't want to deal with the "response" object you can call read() directly:
import urllib
mp3 = urllib.urlopen('http://www.example.com/sound.mp3').read()

If you have wget installed, you can use parallel_sync.
pip install parallel_sync
from parallel_sync import wget
urls = ['http://something.png', 'http://somthing.tar.gz', 'http://somthing.zip']
wget.download('/tmp', urls)
# or a single file:
wget.download('/tmp', urls[0], filenames='x.zip', extract=True)
Doc:
https://pythonhosted.org/parallel_sync/pages/examples.html
This is pretty powerful. It can download files in parallel, retry upon failure , and it can even download files on a remote machine.

You can get the progress feedback with urlretrieve as well:
def report(blocknr, blocksize, size):
current = blocknr*blocksize
sys.stdout.write("\r{0:.2f}%".format(100.0*current/size))
def downloadFile(url):
print "\n",url
fname = url.split('/')[-1]
print fname
urllib.urlretrieve(url, fname, report)

If speed matters to you, I made a small performance test for the modules urllib and wget, and regarding wget I tried once with status bar and once without. I took three different 500MB files to test with (different files- to eliminate the chance that there is some caching going on under the hood). Tested on debian machine, with python2.
First, these are the results (they are similar in different runs):
$ python wget_test.py
urlretrive_test : starting
urlretrive_test : 6.56
==============
wget_no_bar_test : starting
wget_no_bar_test : 7.20
==============
wget_with_bar_test : starting
100% [......................................................................] 541335552 / 541335552
wget_with_bar_test : 50.49
==============
The way I performed the test is using "profile" decorator. This is the full code:
import wget
import urllib
import time
from functools import wraps
def profile(func):
#wraps(func)
def inner(*args):
print func.__name__, ": starting"
start = time.time()
ret = func(*args)
end = time.time()
print func.__name__, ": {:.2f}".format(end - start)
return ret
return inner
url1 = 'http://host.com/500a.iso'
url2 = 'http://host.com/500b.iso'
url3 = 'http://host.com/500c.iso'
def do_nothing(*args):
pass
#profile
def urlretrive_test(url):
return urllib.urlretrieve(url)
#profile
def wget_no_bar_test(url):
return wget.download(url, out='/tmp/', bar=do_nothing)
#profile
def wget_with_bar_test(url):
return wget.download(url, out='/tmp/')
urlretrive_test(url1)
print '=============='
time.sleep(1)
wget_no_bar_test(url2)
print '=============='
time.sleep(1)
wget_with_bar_test(url3)
print '=============='
time.sleep(1)
urllib seems to be the fastest

Just for the sake of completeness, it is also possible to call any program for retrieving files using the subprocess package. Programs dedicated to retrieving files are more powerful than Python functions like urlretrieve. For example, wget can download directories recursively (-R), can deal with FTP, redirects, HTTP proxies, can avoid re-downloading existing files (-nc), and aria2 can do multi-connection downloads which can potentially speed up your downloads.
import subprocess
subprocess.check_output(['wget', '-O', 'example_output_file.html', 'https://example.com'])
In Jupyter Notebook, one can also call programs directly with the ! syntax:
!wget -O example_output_file.html https://example.com

Late answer, but for python>=3.6 you can use:
import dload
dload.save(url)
Install dload with:
pip3 install dload

Source code can be:
import urllib
sock = urllib.urlopen("http://diveintopython.org/")
htmlSource = sock.read()
sock.close()
print htmlSource

I wrote the following, which works in vanilla Python 2 or Python 3.
import sys
try:
import urllib.request
python3 = True
except ImportError:
import urllib2
python3 = False
def progress_callback_simple(downloaded,total):
sys.stdout.write(
"\r" +
(len(str(total))-len(str(downloaded)))*" " + str(downloaded) + "/%d"%total +
" [%3.2f%%]"%(100.0*float(downloaded)/float(total))
)
sys.stdout.flush()
def download(srcurl, dstfilepath, progress_callback=None, block_size=8192):
def _download_helper(response, out_file, file_size):
if progress_callback!=None: progress_callback(0,file_size)
if block_size == None:
buffer = response.read()
out_file.write(buffer)
if progress_callback!=None: progress_callback(file_size,file_size)
else:
file_size_dl = 0
while True:
buffer = response.read(block_size)
if not buffer: break
file_size_dl += len(buffer)
out_file.write(buffer)
if progress_callback!=None: progress_callback(file_size_dl,file_size)
with open(dstfilepath,"wb") as out_file:
if python3:
with urllib.request.urlopen(srcurl) as response:
file_size = int(response.getheader("Content-Length"))
_download_helper(response,out_file,file_size)
else:
response = urllib2.urlopen(srcurl)
meta = response.info()
file_size = int(meta.getheaders("Content-Length")[0])
_download_helper(response,out_file,file_size)
import traceback
try:
download(
"https://geometrian.com/data/programming/projects/glLib/glLib%20Reloaded%200.5.9/0.5.9.zip",
"output.zip",
progress_callback_simple
)
except:
traceback.print_exc()
input()
Notes:
Supports a "progress bar" callback.
Download is a 4 MB test .zip from my website.

You can use PycURL on Python 2 and 3.
import pycurl
FILE_DEST = 'pycurl.html'
FILE_SRC = 'http://pycurl.io/'
with open(FILE_DEST, 'wb') as f:
c = pycurl.Curl()
c.setopt(c.URL, FILE_SRC)
c.setopt(c.WRITEDATA, f)
c.perform()
c.close()

Use Python Requests in 5 lines
import requests as req
remote_url = 'http://www.example.com/sound.mp3'
local_file_name = 'sound.mp3'
data = req.get(remote_url)
# Save file data to local copy
with open(local_file_name, 'wb')as file:
file.write(data.content)
Now do something with the local copy of the remote file

This may be a little late, But I saw pabloG's code and couldn't help adding a os.system('cls') to make it look AWESOME! Check it out :
import urllib2,os
url = "http://download.thinkbroadband.com/10MB.zip"
file_name = url.split('/')[-1]
u = urllib2.urlopen(url)
f = open(file_name, 'wb')
meta = u.info()
file_size = int(meta.getheaders("Content-Length")[0])
print "Downloading: %s Bytes: %s" % (file_name, file_size)
os.system('cls')
file_size_dl = 0
block_sz = 8192
while True:
buffer = u.read(block_sz)
if not buffer:
break
file_size_dl += len(buffer)
f.write(buffer)
status = r"%10d [%3.2f%%]" % (file_size_dl, file_size_dl * 100. / file_size)
status = status + chr(8)*(len(status)+1)
print status,
f.close()
If running in an environment other than Windows, you will have to use something other then 'cls'. In MAC OS X and Linux it should be 'clear'.

urlretrieve and requests.get are simple, however the reality not.
I have fetched data for couple sites, including text and images, the above two probably solve most of the tasks. but for a more universal solution I suggest the use of urlopen. As it is included in Python 3 standard library, your code could run on any machine that run Python 3 without pre-installing site-package
import urllib.request
url_request = urllib.request.Request(url, headers=headers)
url_connect = urllib.request.urlopen(url_request)
#remember to open file in bytes mode
with open(filename, 'wb') as f:
while True:
buffer = url_connect.read(buffer_size)
if not buffer: break
#an integer value of size of written data
data_wrote = f.write(buffer)
#you could probably use with-open-as manner
url_connect.close()
This answer provides a solution to HTTP 403 Forbidden when downloading file over http using Python. I have tried only requests and urllib modules, the other module may provide something better, but this is the one I used to solve most of the problems.

New Api urllib3 based implementation
>>> import urllib3
>>> http = urllib3.PoolManager()
>>> r = http.request('GET', 'your_url_goes_here')
>>> r.status
200
>>> r.data
*****Response Data****
More info: https://pypi.org/project/urllib3/

You can python requests
import os
import requests
outfile = os.path.join(SAVE_DIR, file_name)
response = requests.get(URL, stream=True)
with open(outfile,'wb') as output:
output.write(response.content)
You can use shutil
import os
import requests
import shutil
outfile = os.path.join(SAVE_DIR, file_name)
response = requests.get(url, stream = True)
with open(outfile, 'wb') as f:
shutil.copyfileobj(response.content, f)
If you are downloading from restricted url, don't forget to include access token in headers

I wanted do download all the files from a webpage. I tried wget but it was failing so I decided for the Python route and I found this thread.
After reading it, I have made a little command line application, soupget, expanding on the excellent answers of PabloG and Stan and adding some useful options.
It uses BeatifulSoup to collect all the URLs of the page and then download the ones with the desired extension(s). Finally it can download multiple files in parallel.
Here it is:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from __future__ import (division, absolute_import, print_function, unicode_literals)
import sys, os, argparse
from bs4 import BeautifulSoup
# --- insert Stan's script here ---
# if sys.version_info >= (3,):
#...
#...
# def download_file(url, dest=None):
#...
#...
# --- new stuff ---
def collect_all_url(page_url, extensions):
"""
Recovers all links in page_url checking for all the desired extensions
"""
conn = urllib2.urlopen(page_url)
html = conn.read()
soup = BeautifulSoup(html, 'lxml')
links = soup.find_all('a')
results = []
for tag in links:
link = tag.get('href', None)
if link is not None:
for e in extensions:
if e in link:
# Fallback for badly defined links
# checks for missing scheme or netloc
if bool(urlparse.urlparse(link).scheme) and bool(urlparse.urlparse(link).netloc):
results.append(link)
else:
new_url=urlparse.urljoin(page_url,link)
results.append(new_url)
return results
if __name__ == "__main__": # Only run if this file is called directly
# Command line arguments
parser = argparse.ArgumentParser(
description='Download all files from a webpage.')
parser.add_argument(
'-u', '--url',
help='Page url to request')
parser.add_argument(
'-e', '--ext',
nargs='+',
help='Extension(s) to find')
parser.add_argument(
'-d', '--dest',
default=None,
help='Destination where to save the files')
parser.add_argument(
'-p', '--par',
action='store_true', default=False,
help="Turns on parallel download")
args = parser.parse_args()
# Recover files to download
all_links = collect_all_url(args.url, args.ext)
# Download
if not args.par:
for l in all_links:
try:
filename = download_file(l, args.dest)
print(l)
except Exception as e:
print("Error while downloading: {}".format(e))
else:
from multiprocessing.pool import ThreadPool
results = ThreadPool(10).imap_unordered(
lambda x: download_file(x, args.dest), all_links)
for p in results:
print(p)
An example of its usage is:
python3 soupget.py -p -e <list of extensions> -d <destination_folder> -u <target_webpage>
And an actual example if you want to see it in action:
python3 soupget.py -p -e .xlsx .pdf .csv -u https://healthdata.gov/dataset/chemicals-cosmetics

Another possibility is with built-in http.client:
from http import HTTPStatus, client
from shutil import copyfileobj
# using https
connection = client.HTTPSConnection("www.example.com")
with connection.request("GET", "/noise.mp3") as response:
if response.status == HTTPStatus.OK:
copyfileobj(response, open("noise.mp3")
else:
raise Exception("request needs work")
The HTTPConnection object is considered “low-level” in that it performs the desired request once and assumes the developer will subclass it or script in a way to handle the nuances of HTTP. Libraries such as requests tend to handle more special cases such as automatically following redirects and so on.

You can use keras.utils.get_file to do it:
from tensorflow import keras
path_to_downloaded_file = keras.utils.get_file(
fname="file name",
origin="https://www.linktofile.com/link/to/file",
extract=True,
archive_format="zip", # downloaded file format
cache_dir="/", # cache and extract in current directory
)

Another way is to call an external process such as curl.exe. Curl by default displays a progress bar, average download speed, time left, and more all formatted neatly in a table.
Put curl.exe in the same directory as your script
from subprocess import call
url = ""
call(["curl", {url}, '--output', "song.mp3"])
Note: You cannot specify an output path with curl, so do an os.rename afterwards

Python How to download multiple files in parallel using multiprocessing.pool

I am trying to download and extract zip files using multiprocessing.Pool.But every time I execute the script only 3 zips will be downloaded and remaining files are not seen in the directory(CPU % is also touching 100%). Can someone help me how to solve this problem/suggest better approach and following the snippet that I have tried. I am completely new to multiprocessing. My goal is to download multiple files in parallel without reaching max CPU.
import StringIO
import os
import sys
import zipfile
from multiprocessing import Pool, cpu_count
import requests
filePath = os.path.dirname(os.path.abspath(__file__))
print("filePath is %s " % filePath)
sys.path.append(filePath)
url = ["http://mlg.ucd.ie/files/datasets/multiview_data_20130124.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/bbcsport.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/3sources.zip"]
def download_zips(url):
file_name = url.split("/")[-1]
response = requests.get(url)
sourceZip = zipfile.ZipFile(StringIO.StringIO(response.content))
print("\n Downloaded {} ".format(file_name))
sourceZip.extractall(filePath)
print("extracted {} \n".format(file_name))
sourceZip.close()
if __name__ == "__main__":
print("There are {} CPUs on this machine ".format(cpu_count()))
pool = Pool(cpu_count())
results = pool.map(download_zips, url)
pool.close()
pool.join()
output below
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
There are 4 CPUs on this machine
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
Downloaded bbcsport.zip
extracted bbcsport.zip
Downloaded 3sources.zip
extracted 3sources.zip
Downloaded multiview_data_20130124.zip
Downloaded movielists_20130821.zip
Downloaded movielists_20130821.zip
extracted multiview_data_20130124.zip
extracted movielists_20130821.zip
extracted movielists_20130821.zip

I've made a few minor tweeks in your function and it works fine. Please note that:
the file ".../movielists_20130821.zip" appears on your list twice, so you're donwloading the same thing twice (maybe a typo?)
The files ".../multiview_data_20130124.zip", ".../movielists_20130821.zip" and ".../3sources.zip", when extracted, yield a new directory. The file ".../bbcsport.zip", though, when extracted, places it's files in the root folder, your current working directory (see image below). Maybe you missed this check?
I added a try/except block in the donwload function. Why? Multiprocessing works by creating new (sub)processes to run stuff. If a subprocess throws an exception, the parent process does not catch it. So if any erros occour in this subprocess, it must be logged/handled there.
import sys, os
import zipfile
import requests
from multiprocessing import Pool, cpu_count
from functools import partial
from io import BytesIO
def download_zip(url, filePath):
try:
file_name = url.split("/")[-1]
response = requests.get(url)
sourceZip = zipfile.ZipFile(BytesIO(response.content))
print(" Downloaded {} ".format(file_name))
sourceZip.extractall(filePath)
print(" extracted {}".format(file_name))
sourceZip.close()
except Exception as e:
print(e)
if __name__ == "__main__":
filePath = os.path.dirname(os.path.abspath(__file__))
print("filePath is %s " % filePath)
# sys.path.append(filePath) # why do you need this?
urls = ["http://mlg.ucd.ie/files/datasets/multiview_data_20130124.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/bbcsport.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/3sources.zip"]
print("There are {} CPUs on this machine ".format(cpu_count()))
pool = Pool(cpu_count())
download_func = partial(download_zip, filePath = filePath)
results = pool.map(download_func, urls)
pool.close()
pool.join()

i suggest you do it using multithreading since it's an I/O bound like the following :
import requests, zipfile, io
import concurrent.futures
urls = ["http://mlg.ucd.ie/files/datasets/multiview_data_20130124.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/bbcsport.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/3sources.zip"]
def download_zips(url):
file_name = url.split("/")[-1]
response = requests.get(url)
sourceZip = zipfile.ZipFile(io.BytesIO(response.content))
print("\n Downloaded {} ".format(file_name))
sourceZip.extractall(filePath)
print("extracted {} \n".format(file_name))
sourceZip.close()
with concurrent.futures.ThreadPoolExecutor() as exector :
exector.map(download_zip, urls)

Best way to create a download link for a file in Flask?

In my project, when a user clicks a link, an AJAX request sends the information required to create a CSV. The CSV takes a long time to generate and so I want to be able to include a download link for the generated CSV in the AJAX response. Is this possible?
Most of the answers I've seen return the CSV in the following way:
return Response(
csv,
mimetype="text/csv",
headers={"Content-disposition":
"attachment; filename=myplot.csv"})
However, I don't think this is compatible with the AJAX response I'm sending with:
return render_json(200, {'data': params})
Ideally, I'd like to be able to send the download link in the params dict. But I'm also not sure if this is secure. How is this problem typically solved?

I think one solution may the futures library (pip install futures). The first endpoint can queue up the task and then send the file name back, and then another endpoint can be used to retrieve the file. I also included gzip because it might be a good idea if you are sending larger files. I think more robust solutions use Celery or Rabbit MQ or something along those lines. However, this is a simple solution that should accomplish what you are asking for.
from flask import Flask, jsonify, Response
from uuid import uuid4
from concurrent.futures import ThreadPoolExecutor
import time
import os
import gzip
app = Flask(__name__)
# Global variables used by the thread executor, and the thread executor itself
NUM_THREADS = 5
EXECUTOR = ThreadPoolExecutor(NUM_THREADS)
OUTPUT_DIR = os.path.dirname(os.path.abspath(__file__))
# this is your long running processing function
# takes in your arguments from the /queue-task endpoint
def a_long_running_task(*args):
time_to_wait, output_file_name = int(args[0][0]), args[0][1]
output_string = 'sleeping for {0} seconds. File: {1}'.format(time_to_wait, output_file_name)
print(output_string)
time.sleep(time_to_wait)
filename = os.path.join(OUTPUT_DIR, output_file_name)
# here we are writing to a gzipped file to save space and decrease size of file to be sent on network
with gzip.open(filename, 'wb') as f:
f.write(output_string)
print('finished writing {0} after {1} seconds'.format(output_file_name, time_to_wait))
# This is a route that starts the task and then gives them the file name for reference
#app.route('/queue-task/<wait>')
def queue_task(wait):
output_file_name = str(uuid4()) + '.csv'
EXECUTOR.submit(a_long_running_task, [wait, output_file_name])
return jsonify({'filename': output_file_name})
# this takes the file name and returns if exists, otherwise notifies it is not yet done
#app.route('/getfile/<name>')
def get_output_file(name):
file_name = os.path.join(OUTPUT_DIR, name)
if not os.path.isfile(file_name):
return jsonify({"message": "still processing"})
# read without gzip.open to keep it compressed
with open(file_name, 'rb') as f:
resp = Response(f.read())
# set headers to tell encoding and to send as an attachment
resp.headers["Content-Encoding"] = 'gzip'
resp.headers["Content-Disposition"] = "attachment; filename={0}".format(name)
resp.headers["Content-type"] = "text/csv"
return resp
if __name__ == '__main__':
app.run()

Multi-threading for downloading NCBI files in Python

So recently I have taken on the task of downloading large collection of files from the ncbi database. However I have run into times where I have to create multiple databases. This code here which works to downloads all the viruses from the ncbi website. My question is there any way to speed up the process of downloading these files.
Currently the runtime of this program is more than 5hours. I have looked into multi-threading and could never get it to work because some of these files take more than 10seconds to download and I do not know how to handle stalling. (new to programing) Also is there a way of handling urllib2.HTTPError: HTTP Error 502: Bad Gateway. I get this sometimes with with certain combinations of retstart and retmax. This crashes the program and I have to restart the download from a different location by changingthe 0 in the for statement.
import urllib2
from BeautifulSoup import BeautifulSoup
#This is the SearchQuery into NCBI. Spaces are replaced with +'s.
SearchQuery = 'viruses[orgn]+NOT+Retroviridae[orgn]'
#This is the Database that you are searching.
database = 'protein'
#This is the output file for the data
output = 'sample.fasta'
#This is the base url for NCBI eutils.
base = 'http://eutils.ncbi.nlm.nih.gov/entrez/eutils/'
#Create the search string from the information above
esearch = 'esearch.fcgi?db='+database+'&term='+SearchQuery+'&usehistory=y'
#Create your esearch url
url = base + esearch
#Fetch your esearch using urllib2
print url
content = urllib2.urlopen(url)
#Open url in BeautifulSoup
doc = BeautifulSoup(content)
#Grab the amount of hits in the search
Count = int(doc.find('count').string)
#Grab the WebEnv or the history of this search from usehistory.
WebEnv = doc.find('webenv').string
#Grab the QueryKey
QueryKey = doc.find('querykey').string
#Set the max amount of files to fetch at a time. Default is 500 files.
retmax = 10000
#Create the fetch string
efetch = 'efetch.fcgi?db='+database+'&WebEnv='+WebEnv+'&query_key='+QueryKey
#Select the output format and file format of the files.
#For table visit: http://www.ncbi.nlm.nih.gov/books/NBK25499/table/chapter4.chapter4_table1
format = 'fasta'
type = 'text'
#Create the options string for efetch
options = '&rettype='+format+'&retmode='+type
#For statement 0 to Count counting by retmax. Use xrange over range
for i in xrange(0,Count,retmax):
#Create the position string
poision = '&retstart='+str(i)+'&retmax='+str(retmax)
#Create the efetch URL
url = base + efetch + poision + options
print url
#Grab the results
response = urllib2.urlopen(url)
#Write output to file
with open(output, 'a') as file:
for line in response.readlines():
file.write(line)
#Gives a sense of where you are
print Count - i - retmax

To download files using multiple threads:
#!/usr/bin/env python
import shutil
from contextlib import closing
from multiprocessing.dummy import Pool # use threads
from urllib2 import urlopen
def generate_urls(some, params): #XXX pass whatever parameters you need
for restart in range(*params):
# ... generate url, filename
yield url, filename
def download((url, filename)):
try:
with closing(urlopen(url)) as response, open(filename, 'wb') as file:
shutil.copyfileobj(response, file)
except Exception as e:
return (url, filename), repr(e)
else: # success
return (url, filename), None
def main():
pool = Pool(20) # at most 20 concurrent downloads
urls = generate_urls(some, params)
for (url, filename), error in pool.imap_unordered(download, urls):
if error is not None:
print("Can't download {url} to {filename}, "
"reason: {error}".format(**locals())
if __name__ == "__main__":
main()

You should use multithreading, it's the right way for downloading tasks.
"these files take more than 10seconds to download and I do not know how to handle stalling",
I don't think this would be a problem because Python's multithreading will handle this, or I'd rather say multithreading is just for this kind of I/O-bound work. When a thread is waiting for download to complete, CPU will let other threads do their work.
Anyway, you'd better at least try and see what happen.

Two ways to effect your task. 1. Using process instead of thread, multiprocess is the module you should use. 2. Using Event-based, gevent is the right module.
502 error is not your script's fault. Simply, following pattern could be used to do retry
try_count = 3
while try_count > 0:
try:
download_task()
except urllib2.HTTPError:
clean_environment_for_retry()
try_count -= 1
In the line of except, you can refine the detail to do particular things according to concrete HTTP status code.

Downloading videos in FLV format from YouTube

I can’t really understand how YouTube serves videos, but I have been reading through what I can.
It seems like the old method get_video is now obsolete and can't be used any more. Is there another Pythonic and simple method for collecting YouTube videos?

You might have some luck with youtube-dl
http://rg3.github.com/youtube-dl/documentation.html
I'm not sure if there's a good API, but it's written in Python, so theoretically you could do something a little better than Popen :)

Here is a quick Python script which downloads a YouTube video. No bells and whistles, just scrapes out the necessary URLs, hits the generate_204 URL and then streams the data to a file:
import lxml.html
import re
import sys
import urllib
import urllib2
_RE_G204 = re.compile('"(http:.+.youtube.com.*\/generate_204[^"]+")', re.M)
_RE_URLS = re.compile('"fmt_url_map": "(\d*[^"]+)",.*', re.M)
def _fetch_url(url, ref=None, path=None):
opener = urllib2.build_opener()
headers = {}
if ref:
headers['Referer'] = ref
request = urllib2.Request(url, headers=headers)
handle = urllib2.urlopen(request)
if not path:
return handle.read()
sys.stdout.write('saving: ')
# Write result to file
with open(path, 'wb') as out:
while True:
part = handle.read(65536)
if not part:
break
out.write(part)
sys.stdout.write('.')
sys.stdout.flush()
sys.stdout.write('\nFinished.\n')
def _extract(html):
tree = lxml.html.fromstring(html)
res = {'204': _RE_G204.findall(html)[0].replace('\\', '')}
for script in tree.findall('.//script'):
text = script.text_content()
if 'fmt_url_map' not in text:
continue
# Found it. Extract the URLs we need
for tmp in _RE_URLS.findall(text)[0].split(','):
url_id, url = tmp.split('|')
res[url_id] = url.replace('\\', '')
break
return res
def main():
target = sys.argv[1]
dest = sys.argv[2]
html = _fetch_url(target)
res = dict(_extract(html))
# Hit the 'generate_204' URL first and remove it
_fetch_url(res['204'], ref=target)
del res['204']
# Download the video. Now I grab the first 'download' URL and use it.
first = res.values()[0]
_fetch_url(first, ref=target, path=dest)
if __name__ == '__main__':
main()
Running it:
python youdown.py 'http://www.youtube.com/watch?v=Je_iqbgGXFw' stevegadd.flv
saving: ........................... finished.

I would recommend writing your own parser using urllib2 or Beautiful Soup. You can look at the source code for DownThemAll to see how that plugin finds the video URL.