r_api= requests.get(url = urll, headers = hederr, stream=True)
request_data = r_api.content
print(request_data)
I got gzip bytes as output/response and I want to convert it into .wav file.
I tried this code and it's converting the file to gzip wav not to .wav.
with open('myfile_1.wav', mode='bx') as f:
f.write(request_data)
gzip bytes
how to convert it to gzip bytes to .wav?
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You'll want to unzip the content, and you can use the gzip module to do this. Use gzip.decompress to read the content into a wav file:
import requests
import gzip
r = requests.get(url = urll, headers = hederr, stream=True)
with open('myfile.wav', 'wb') as fh:
fh.write(gzip.decompress(r.content))
Edit
Since this is apparently a wav.tar.gz file, you'll want to un-tar the decompressed content:
import requests
import gzip
import tarfile
from io import BytesIO
r = requests.get(url = urll, headers = hederr, stream=True)
with open('myfile.wav.tar', 'wb') as fh:
fh.write(gzip.decompress(r.content))
with tarfile.open('myfile.wav.tar', 'rb') as fh:
fh.extractall()
Where extractall will extract to the current working directory by default, but you can specify a separate directory if you want via fh.extractall(path='/some/path'). To avoid intermediate files, you could do:
with BytesIO(gzip.decompress(r.content)) as fh:
tf = tarfile.TarFile(fileobj=fh)
tf.extractall()
Related
I am trying to download a PDF file from a website and save it to disk. My attempts either fail with encoding errors or result in blank PDFs.
In [1]: import requests
In [2]: url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
In [3]: response = requests.get(url)
In [4]: with open('/tmp/metadata.pdf', 'wb') as f:
...: f.write(response.text)
---------------------------------------------------------------------------
UnicodeEncodeError Traceback (most recent call last)
<ipython-input-4-4be915a4f032> in <module>()
1 with open('/tmp/metadata.pdf', 'wb') as f:
----> 2 f.write(response.text)
3
UnicodeEncodeError: 'ascii' codec can't encode characters in position 11-14: ordinal not in range(128)
In [5]: import codecs
In [6]: with codecs.open('/tmp/metadata.pdf', 'wb', encoding='utf8') as f:
...: f.write(response.text)
...:
I know it is a codec problem of some kind but I can't seem to get it to work.
You should use response.content in this case:
with open('/tmp/metadata.pdf', 'wb') as f:
f.write(response.content)
From the document:
You can also access the response body as bytes, for non-text requests:
>>> r.content
b'[{"repository":{"open_issues":0,"url":"https://github.com/...
So that means: response.text return the output as a string object, use it when you're downloading a text file. Such as HTML file, etc.
And response.content return the output as bytes object, use it when you're downloading a binary file. Such as PDF file, audio file, image, etc.
You can also use response.raw instead. However, use it when the file which you're about to download is large. Below is a basic example which you can also find in the document:
import requests
url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
r = requests.get(url, stream=True)
with open('/tmp/metadata.pdf', 'wb') as fd:
for chunk in r.iter_content(chunk_size):
fd.write(chunk)
chunk_size is the chunk size which you want to use. If you set it as 2000, then requests will download that file the first 2000 bytes, write them into the file, and do this again, again and again, unless it finished.
So this can save your RAM. But I'd prefer use response.content instead in this case since your file is small. As you can see use response.raw is complex.
Relates:
How to download large file in python with requests.py?
How to download image using requests
In Python 3, I find pathlib is the easiest way to do this. Request's response.content marries up nicely with pathlib's write_bytes.
from pathlib import Path
import requests
filename = Path('metadata.pdf')
url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
response = requests.get(url)
filename.write_bytes(response.content)
You can use urllib:
import urllib.request
urllib.request.urlretrieve(url, "filename.pdf")
Please note I'm a beginner. If My solution is wrong, please feel free to correct and/or let me know. I may learn something new too.
My solution:
Change the downloadPath accordingly to where you want your file to be saved. Feel free to use the absolute path too for your usage.
Save the below as downloadFile.py.
Usage: python downloadFile.py url-of-the-file-to-download new-file-name.extension
Remember to add an extension!
Example usage: python downloadFile.py http://www.google.co.uk google.html
import requests
import sys
import os
def downloadFile(url, fileName):
with open(fileName, "wb") as file:
response = requests.get(url)
file.write(response.content)
scriptPath = sys.path[0]
downloadPath = os.path.join(scriptPath, '../Downloads/')
url = sys.argv[1]
fileName = sys.argv[2]
print('path of the script: ' + scriptPath)
print('downloading file to: ' + downloadPath)
downloadFile(url, downloadPath + fileName)
print('file downloaded...')
print('exiting program...')
Generally, this should work in Python3:
import urllib.request
..
urllib.request.get(url)
Remember that urllib and urllib2 don't work properly after Python2.
If in some mysterious cases requests don't work (happened with me), you can also try using
wget.download(url)
Related:
Here's a decent explanation/solution to find and download all pdf files on a webpage:
https://medium.com/#dementorwriter/notesdownloader-use-web-scraping-to-download-all-pdfs-with-python-511ea9f55e48
regarding Kevin answer to write in a folder tmp, it should be like this:
with open('./tmp/metadata.pdf', 'wb') as f:
f.write(response.content)
he forgot . before the address and of-course your folder tmp should have been created already
I have been trying to download a zipped csv using the requests library from a server host URL.
When I download a smaller file not requiring compression from the same server it has no problem reading in the CSV, but with this one I return encoding errors.
I have tried multiple types of encoding, reading in as pandas csv, reading in as zip file and opening (at which point I get the error that file is not a zip file).
I have additionally tried using the zipfile library as sugggested here: Reading csv zipped files in python
and have also tried setting both encoding and compression in read_csv.
The code which works for the non-zipped server file is below:
response = requests.get(url, auth=HTTPBasicAuth(un, pw), stream=True, verify = False)
dfs = pd.read_csv(response.raw)
but returns 'utf-8' codec can't decode byte 0xfd in position 0: invalid start byte when used for this file.
I have also tried:
request = get(url, auth=HTTPBasicAuth(un, pw), stream=True, verify=False)
zip_file = ZipFile(BytesIO(request.content))
files = zip_file.namelist()
with gzip.open(files[0], 'rb') as csvfile:
csvreader = csv.reader(csvfile)
for row in csvreader:
print(row)
which returns a seek attribute error.
Here is one way to do it:
import pandas as pd
import requests
from requests.auth import HTTPBasicAuth
from zipfile import ZipFile
import io
# Example dataset
url = 'https://www.stats.govt.nz/assets/Uploads/Retail-trade-survey/Retail-trade-survey-September-2020-quarter/Download-data/retail-trade-survey-september-2020-quarter-csv.zip'
response = requests.get(url, auth=HTTPBasicAuth(un, pw), stream=True, verify=False)
with ZipFile(io.BytesIO(response.content)) as myzip:
with myzip.open(myzip.namelist()[0]) as myfile:
df = pd.read_csv(myfile)
print(df)
If you want to read a specific csv in a multiple-csv zip file, replace myzip.namelist()[0] with the file you want to read. If you don't know its name, you can check the zip file content with print(ZipFile(io.BytesIO(response.content)))
I am trying to use tarfile to add a file in memory and then writing it back to disk, but the issue i am having is that in my final output, when i extract the newly created tar.gz file, i am getting an empty file. What am I doing wrong in my code?
import tarfile
import io
with open('logo.png', 'rb') as f:
data = f.read()
fh = io.BytesIO()
with tarfile.open(fileobj=fh, mode='w:gz') as tar:
info = tarfile.TarInfo('some.png')
tar.addfile(info, data)
with open('/tmp/test/test.tar.gz', 'wb') as f:
f.write(fh.getvalue())
I also tried doing tar.addfile(info, fh.write(data)), but that just creates a corrupted tar file.
TarFile.addfile() takes a file-like object.
When the documentation says:
tarinfo.size bytes are read from it and added to the archive.
It means that tarinfo.size is used to determine how many bytes to read. Therefore, you need set tarinfo.size appropriately.
The only thing you need to do is read the data from the source, count the length, then load that data into a BytesIO object:
E.g.
import tarfile
import io
with open('logo.png', 'rb') as f:
data = f.read()
source_f = io.BytesIO(initial_bytes=data)
fh = io.BytesIO()
with tarfile.open(fileobj=fh, mode='w:gz') as tar:
info = tarfile.TarInfo('logo.png')
info.size = len(data)
tar.addfile(info, source_f)
with open('test.tar.gz', 'wb') as f:
f.write(fh.getvalue())
or a more memory efficient way, seek the source file:
f = open('logo.png', 'rb')
f.seek(0,2) # go to the end
source_len = f.tell()
f.seek(0)
fh = io.BytesIO()
with tarfile.open(fileobj=fh, mode='w:gz') as tar:
info = tarfile.TarInfo('logo.png')
info.size = source_len
tar.addfile(info, f)
with open('test.tar.gz', 'wb') as f:
f.write(fh.getvalue())
f.close()
I want to stream a large file into a gzip file directly, instead of downloading it all into memory and then compressing. This is how far I have gotten (does not work). I know how to just download a file in python and save and I know how to compress one, it is the streaming part that does not work.
Note: this linked csv is not large, it is just an example url.
import requests
import zlib
url = f"http://samplecsvs.s3.amazonaws.com/Sacramentorealestatetransactions.csv"
with requests.get(url, stream=True) as r:
compressor = zlib.compressobj()
with open(save_file_path, 'wb') as f:
f.write(compressor.compress(r.raw))
Alright I figured it out:
with requests.get(url, stream=True, verify=False) as r:
if save_file_path.endswith('gz'):
compressor = zlib.compressobj(9, zlib.DEFLATED, zlib.MAX_WBITS | 16)
with open(save_file_path, 'wb') as f:
for chunk in r.iter_content(chunk_size=1024*1024):
f.write(compressor.compress(chunk))
f.write(compressor.flush())
else:
with open(save_file_path, 'wb') as f:
shutil.copyfileobj(r.raw, f)
If I have a URL that, when submitted in a web browser, pops up a dialog box to save a zip file, how would I go about catching and downloading this zip file in Python?
As far as I can tell, the proper way to do this is:
import requests, zipfile, StringIO
r = requests.get(zip_file_url, stream=True)
z = zipfile.ZipFile(StringIO.StringIO(r.content))
z.extractall()
of course you'd want to check that the GET was successful with r.ok.
For python 3+, sub the StringIO module with the io module and use BytesIO instead of StringIO: Here are release notes that mention this change.
import requests, zipfile, io
r = requests.get(zip_file_url)
z = zipfile.ZipFile(io.BytesIO(r.content))
z.extractall("/path/to/destination_directory")
Most people recommend using requests if it is available, and the requests documentation recommends this for downloading and saving raw data from a url:
import requests
def download_url(url, save_path, chunk_size=128):
r = requests.get(url, stream=True)
with open(save_path, 'wb') as fd:
for chunk in r.iter_content(chunk_size=chunk_size):
fd.write(chunk)
Since the answer asks about downloading and saving the zip file, I haven't gone into details regarding reading the zip file. See one of the many answers below for possibilities.
If for some reason you don't have access to requests, you can use urllib.request instead. It may not be quite as robust as the above.
import urllib.request
def download_url(url, save_path):
with urllib.request.urlopen(url) as dl_file:
with open(save_path, 'wb') as out_file:
out_file.write(dl_file.read())
Finally, if you are using Python 2 still, you can use urllib2.urlopen.
from contextlib import closing
def download_url(url, save_path):
with closing(urllib2.urlopen(url)) as dl_file:
with open(save_path, 'wb') as out_file:
out_file.write(dl_file.read())
With the help of this blog post, I've got it working with just requests.
The point of the weird stream thing is so we don't need to call content
on large requests, which would require it to all be processed at once,
clogging the memory. The stream avoids this by iterating through the data
one chunk at a time.
url = 'https://www2.census.gov/geo/tiger/GENZ2017/shp/cb_2017_02_tract_500k.zip'
response = requests.get(url, stream=True)
with open('alaska.zip', "wb") as f:
for chunk in response.iter_content(chunk_size=512):
if chunk: # filter out keep-alive new chunks
f.write(chunk)
Here's what I got to work in Python 3:
import zipfile, urllib.request, shutil
url = 'http://www....myzipfile.zip'
file_name = 'myzip.zip'
with urllib.request.urlopen(url) as response, open(file_name, 'wb') as out_file:
shutil.copyfileobj(response, out_file)
with zipfile.ZipFile(file_name) as zf:
zf.extractall()
Super lightweight solution to save a .zip file to a location on disk (using Python 3.9):
import requests
url = r'https://linktofile'
output = r'C:\pathtofolder\downloaded_file.zip'
r = requests.get(url)
with open(output, 'wb') as f:
f.write(r.content)
Either use urllib2.urlopen, or you could try using the excellent Requests module and avoid urllib2 headaches:
import requests
results = requests.get('url')
#pass results.content onto secondary processing...
I came here searching how to save a .bzip2 file. Let me paste the code for others who might come looking for this.
url = "http://api.mywebsite.com"
filename = "swateek.tar.gz"
response = requests.get(url, headers=headers, auth=('myusername', 'mypassword'), timeout=50)
if response.status_code == 200:
with open(filename, 'wb') as f:
f.write(response.content)
I just wanted to save the file as is.
Thanks to #yoavram for the above solution,
my url path linked to a zipped folder, and encounter an error of BADZipfile
(file is not a zip file), and it was strange if I tried several times it
retrieve the url and unzipped it all of sudden so I amend the solution a little
bit. using the is_zipfile method as per here
r = requests.get(url, stream =True)
check = zipfile.is_zipfile(io.BytesIO(r.content))
while not check:
r = requests.get(url, stream =True)
check = zipfile.is_zipfile(io.BytesIO(r.content))
else:
z = zipfile.ZipFile(io.BytesIO(r.content))
z.extractall()
Use requests, zipfile and io python packages.
Specially BytesIO function is used to keep the unzipped file in memory rather than saving it into the drive.
import requests
from zipfile import ZipFile
from io import BytesIO
r = requests.get(zip_file_url)
z = ZipFile(BytesIO(r.content))
file = z.extract(a_file_to_extract, path_to_save)
with open(file) as f:
print(f.read())