I want to download multiple .hdr files from a website called hdrihaven.com.
My knowledge of python is not that great but here is what I have tried so far:
import requests
url = 'https://hdrihaven.com/files/hdris/'
resolution = '4k'
file = 'pump_station' #would need to be every file
url_2k = url + file + '_' + resolution + '.hdr'
print(url_2k)
r = requests.get(url_2k, allow_redirects=True)
open(file + resolution + '.hdr', 'wb').write(r.content)
Idealy file would just loop over every file in the directory.
Thanks for your answers in advance!
EDIT
I found a script on github that does what I need: https://github.com/Alzy/hdrihaven_dl. I edited it to fit my needs here: https://github.com/ktkk/hdrihaven-downloader. It uses the technique of looping through a list of all available files as proposed in the comments.
I have found that the requests module as well as urllib are extremly slow compared to native downloading from eg. Chrome. If anyone has an idea as to how I can speed these up pls let me know.
There are 2 ways you can do this:
You can use an URLto fetch all the files and iterate through a loop to download them individually. This of course only works if there exists such a URL.
You can pass in individual URL to a function that can download them in parallel/bulk.
For example:
import os
import requests
from time import time
from multiprocessing.pool import ThreadPool
def url_response(url):
path, url = url
r = requests.get(url, stream = True)
with open(path, 'wb') as f:
for ch in r:
f.write(ch)
urls = [("Event1", "https://www.python.org/events/python-events/805/"),("Event2", "https://www.python.org/events/python-events/801/"),
("Event3", "https://www.python.org/events/python-user-group/816/")]
start = time()
for x in urls:
url_response (x)
print(f"Time to download: {time() - start}")
This code snippet is taken from here Download multiple files (Parallel/bulk download). Read on there for more information on how you can do this.
Related
I've been going through the Q&A on this site, for an answer to my question. However, I'm a beginner and I find it difficult to understand some of the solutions. I need a very basic solution.
Could someone please explain a simple solution to 'Downloading a file through http' and 'Saving it to disk, in Windows', to me?
I'm not sure how to use shutil and os modules, either.
The file I want to download is under 500 MB and is an .gz archive file.If someone can explain how to extract the archive and utilise the files in it also, that would be great!
Here's a partial solution, that I wrote from various answers combined:
import requests
import os
import shutil
global dump
def download_file():
global dump
url = "http://randomsite.com/file.gz"
file = requests.get(url, stream=True)
dump = file.raw
def save_file():
global dump
location = os.path.abspath("D:\folder\file.gz")
with open("file.gz", 'wb') as location:
shutil.copyfileobj(dump, location)
del dump
Could someone point out errors (beginner level) and explain any easier methods to do this?
A clean way to download a file is:
import urllib
testfile = urllib.URLopener()
testfile.retrieve("http://randomsite.com/file.gz", "file.gz")
This downloads a file from a website and names it file.gz. This is one of my favorite solutions, from Downloading a picture via urllib and python.
This example uses the urllib library, and it will directly retrieve the file form a source.
For Python3+ URLopener is deprecated.
And when used you will get error as below:
url_opener = urllib.URLopener() AttributeError: module 'urllib' has no
attribute 'URLopener'
So, try:
import urllib.request
urllib.request.urlretrieve(url, filename)
As mentioned here:
import urllib
urllib.urlretrieve ("http://randomsite.com/file.gz", "file.gz")
EDIT: If you still want to use requests, take a look at this question or this one.
Four methods using wget, urllib and request.
#!/usr/bin/python
import requests
from StringIO import StringIO
from PIL import Image
import profile as profile
import urllib
import wget
url = 'https://tinypng.com/images/social/website.jpg'
def testRequest():
image_name = 'test1.jpg'
r = requests.get(url, stream=True)
with open(image_name, 'wb') as f:
for chunk in r.iter_content():
f.write(chunk)
def testRequest2():
image_name = 'test2.jpg'
r = requests.get(url)
i = Image.open(StringIO(r.content))
i.save(image_name)
def testUrllib():
image_name = 'test3.jpg'
testfile = urllib.URLopener()
testfile.retrieve(url, image_name)
def testwget():
image_name = 'test4.jpg'
wget.download(url, image_name)
if __name__ == '__main__':
profile.run('testRequest()')
profile.run('testRequest2()')
profile.run('testUrllib()')
profile.run('testwget()')
testRequest - 4469882 function calls (4469842 primitive calls) in 20.236 seconds
testRequest2 - 8580 function calls (8574 primitive calls) in 0.072 seconds
testUrllib - 3810 function calls (3775 primitive calls) in 0.036 seconds
testwget - 3489 function calls in 0.020 seconds
I use wget.
Simple and good library if you want to example?
import wget
file_url = 'http://johndoe.com/download.zip'
file_name = wget.download(file_url)
wget module support python 2 and python 3 versions
Exotic Windows Solution
import subprocess
subprocess.run("powershell Invoke-WebRequest {} -OutFile {}".format(your_url, filename), shell=True)
import urllib.request
urllib.request.urlretrieve("https://raw.githubusercontent.com/dnishimoto/python-deep-learning/master/list%20iterators%20and%20generators.ipynb", "test.ipynb")
downloads a single raw juypter notebook to file.
For text files, you can use:
import requests
url = 'https://WEBSITE.com'
req = requests.get(url)
path = "C:\\YOUR\\FILE.html"
with open(path, 'wb') as f:
f.write(req.content)
I started down this path because ESXi's wget is not compiled with SSL and I wanted to download an OVA from a vendor's website directly onto the ESXi host which is on the other side of the world.
I had to disable the firewall(lazy)/enable https out by editing the rules(proper)
created the python script:
import ssl
import shutil
import tempfile
import urllib.request
context = ssl._create_unverified_context()
dlurl='https://somesite/path/whatever'
with urllib.request.urlopen(durl, context=context) as response:
with open("file.ova", 'wb') as tmp_file:
shutil.copyfileobj(response, tmp_file)
ESXi libraries are kind of paired down but the open source weasel installer seemed to use urllib for https... so it inspired me to go down this path
Another clean way to save the file is this:
import csv
import urllib
urllib.retrieve("your url goes here" , "output.csv")
I'm trying to download images from a list of URL's. Each URL contains a txt file with jpeg information. The URL's are uniform except for an incremental change in the folder number. Below are example URL's
Min: https://marco.ccr.buffalo.edu/data/train/train-00001-of-00407
Max: https://marco.ccr.buffalo.edu/data/train/train-00407-of-00407
I want to read each of these URL's and store the their output to another folder. I was looking into the requests python library to do this but Im wondering how to iterate over the URL's and essentially write my loop to increment over that number in the URL. Apologize in advance if I misuse the terminology. Thanks!
# This may be terrible starting code
# imported the requests library
import requests
url = "https://marco.ccr.buffalo.edu/data/train/train-00001-of-00407"
# URL of the image to be downloaded is defined as image_url
r = requests.get(url) # create HTTP response object
# send a HTTP request to the server and save
# the HTTP response in a response object called r
with open("data.txt",'wb') as f:
# Saving received content as a png file in
# binary format
# write the contents of the response (r.content)
# to a new file in binary mode.
f.write(r.content)
You can generate urls like this and perform get for each
for i in range(1,408):
url = "https://marco.ccr.buffalo.edu/data/train/train-" + str(i).zfill(5) + "-of-00407"
print (url)
Also use a variable in the filename to keep a different copy of each. For eg, use this
with open("data" + str(i) + ".txt",'wb') as f:
Overall code may look something like this (not exactly this)
import requests
for i in range(1,408):
url = "https://marco.ccr.buffalo.edu/data/train/train-" + str(i).zfill(5) + "-of-00407"
r = requests.get(url)
# you might have to change the extension
with open("data" + str(i).zfill(5) + ".txt",'wb') as f:
f.write(r.content)
I've been going through the Q&A on this site, for an answer to my question. However, I'm a beginner and I find it difficult to understand some of the solutions. I need a very basic solution.
Could someone please explain a simple solution to 'Downloading a file through http' and 'Saving it to disk, in Windows', to me?
I'm not sure how to use shutil and os modules, either.
The file I want to download is under 500 MB and is an .gz archive file.If someone can explain how to extract the archive and utilise the files in it also, that would be great!
Here's a partial solution, that I wrote from various answers combined:
import requests
import os
import shutil
global dump
def download_file():
global dump
url = "http://randomsite.com/file.gz"
file = requests.get(url, stream=True)
dump = file.raw
def save_file():
global dump
location = os.path.abspath("D:\folder\file.gz")
with open("file.gz", 'wb') as location:
shutil.copyfileobj(dump, location)
del dump
Could someone point out errors (beginner level) and explain any easier methods to do this?
A clean way to download a file is:
import urllib
testfile = urllib.URLopener()
testfile.retrieve("http://randomsite.com/file.gz", "file.gz")
This downloads a file from a website and names it file.gz. This is one of my favorite solutions, from Downloading a picture via urllib and python.
This example uses the urllib library, and it will directly retrieve the file form a source.
For Python3+ URLopener is deprecated.
And when used you will get error as below:
url_opener = urllib.URLopener() AttributeError: module 'urllib' has no
attribute 'URLopener'
So, try:
import urllib.request
urllib.request.urlretrieve(url, filename)
As mentioned here:
import urllib
urllib.urlretrieve ("http://randomsite.com/file.gz", "file.gz")
EDIT: If you still want to use requests, take a look at this question or this one.
Four methods using wget, urllib and request.
#!/usr/bin/python
import requests
from StringIO import StringIO
from PIL import Image
import profile as profile
import urllib
import wget
url = 'https://tinypng.com/images/social/website.jpg'
def testRequest():
image_name = 'test1.jpg'
r = requests.get(url, stream=True)
with open(image_name, 'wb') as f:
for chunk in r.iter_content():
f.write(chunk)
def testRequest2():
image_name = 'test2.jpg'
r = requests.get(url)
i = Image.open(StringIO(r.content))
i.save(image_name)
def testUrllib():
image_name = 'test3.jpg'
testfile = urllib.URLopener()
testfile.retrieve(url, image_name)
def testwget():
image_name = 'test4.jpg'
wget.download(url, image_name)
if __name__ == '__main__':
profile.run('testRequest()')
profile.run('testRequest2()')
profile.run('testUrllib()')
profile.run('testwget()')
testRequest - 4469882 function calls (4469842 primitive calls) in 20.236 seconds
testRequest2 - 8580 function calls (8574 primitive calls) in 0.072 seconds
testUrllib - 3810 function calls (3775 primitive calls) in 0.036 seconds
testwget - 3489 function calls in 0.020 seconds
I use wget.
Simple and good library if you want to example?
import wget
file_url = 'http://johndoe.com/download.zip'
file_name = wget.download(file_url)
wget module support python 2 and python 3 versions
Exotic Windows Solution
import subprocess
subprocess.run("powershell Invoke-WebRequest {} -OutFile {}".format(your_url, filename), shell=True)
import urllib.request
urllib.request.urlretrieve("https://raw.githubusercontent.com/dnishimoto/python-deep-learning/master/list%20iterators%20and%20generators.ipynb", "test.ipynb")
downloads a single raw juypter notebook to file.
For text files, you can use:
import requests
url = 'https://WEBSITE.com'
req = requests.get(url)
path = "C:\\YOUR\\FILE.html"
with open(path, 'wb') as f:
f.write(req.content)
I started down this path because ESXi's wget is not compiled with SSL and I wanted to download an OVA from a vendor's website directly onto the ESXi host which is on the other side of the world.
I had to disable the firewall(lazy)/enable https out by editing the rules(proper)
created the python script:
import ssl
import shutil
import tempfile
import urllib.request
context = ssl._create_unverified_context()
dlurl='https://somesite/path/whatever'
with urllib.request.urlopen(durl, context=context) as response:
with open("file.ova", 'wb') as tmp_file:
shutil.copyfileobj(response, tmp_file)
ESXi libraries are kind of paired down but the open source weasel installer seemed to use urllib for https... so it inspired me to go down this path
Another clean way to save the file is this:
import csv
import urllib
urllib.retrieve("your url goes here" , "output.csv")
I'm working on a script that will automatically update an installed version of Calibre. Currently I have it downloading the latest portable version. I seem to be having trouble saving the zipfile. Currently my code is:
import urllib2
import re
import zipfile
#tell the user what is happening
print("Calibre is Updating")
#download the page
url = urllib2.urlopen ( "http://sourceforge.net/projects/calibre/files" ).read()
#determin current version
result = re.search('title="/[0-9.]*/([a-zA-Z\-]*-[0-9\.]*)', url).groups()[0][:-1]
#download file
download = "http://status.calibre-ebook.com/dist/portable/" + result
urllib2.urlopen( download )
#save
output = open('install.zip', 'w')
output.write(zipfile.ZipFile("install.zip", ""))
output.close()
You don't need to use zipfile.ZipFile for this (and the way you're using it, as well as urllib2.urlopen, has problems as well). Instead, you need to save the urlopen result in a variable, then read it and write that output to a .zip file. Try this code:
#download file
download = "http://status.calibre-ebook.com/dist/portable/" + result
request = urllib2.urlopen( download )
#save
output = open("install.zip", "w")
output.write(request.read())
output.close()
There also can be a one-liner:
open('install.zip', 'wb').write(urllib.urlopen('http://status.calibre-ebook.com/dist/portable/' + result).read())
which doesn't have a good memory-efficiency, but still works.
If you just want to download a file from the net, you can use urllib.urlretrieve:
Copy a network object denoted by a URL to a local file ...
Example using requests instead of urllib2:
import requests, re, urllib
print("Calibre is updating...")
content = requests.get("http://sourceforge.net/projects/calibre/files").content
# determine current version
v = re.search('title="/[0-9.]*/([a-zA-Z\-]*-[0-9\.]*)', content).groups()[0][:-1]
download_url = "http://status.calibre-ebook.com/dist/portable/{0}".format(v)
print("Downloading {0}".format(download_url))
urllib.urlretrieve(download_url, 'install.zip')
# file should be downloaded at this point
have you tryed
output = open('install.zip', 'wb') // note the "b" flag which means "binary file"
Is there a way I can download all/some the image files (e.g. JPG/PNG) from a Google Images search result?
I can use the following code to download one image that I already know its url:
import urllib.request
file = "Facts.jpg" # file to be written to
url = "http://www.compassion.com/Images/Hunger-Facts.jpg"
response = urllib.request.urlopen (url)
fh = open(file, "wb") #open the file for writing
fh.write(response.read()) # read from request while writing to file
To download multiple images, it has been suggested that I define a function and use that function to repeat the task for each image url that I would like to write to disk:
def image_request(url, file):
response = urllib.request.urlopen(url)
fh = open(file, "wb") #open the file for writing
fh.write(response.read())
And then loop over a list of urls with:
for i, url in enumerate(urllist):
image_request(url, str(i) + ".jpg")
However, what I really want to do is download all/some image files (e.g. JPG/PNG) from my own search result from Google Images without necessarily having a list of the image urls beforehand.
P.S.
Please I am a complete beginner and would favour an answer that breaks down the broad steps to achieve this over one that is bogs down on specific codes. Thanks.
You can use the Google API like this, where BLUE and DOG are your search parameters:
https://ajax.googleapis.com/ajax/services/search/images?v=1.0&q=BLUE%20DOG
There is a developer guide about this here:
https://developers.google.com/image-search/v1/jsondevguide
You need to parse this JSON format before you can use the links directly.
Here's a start to your JSON parsing:
import json
j = json.loads('{"one" : "1", "two" : "2", "three" : "3"}')
print(j['two'])