Download large file in python with requests

Question

Requests is a really nice library. I'd like to use it for downloading big files (>1GB). The problem is it's not possible to keep whole file in memory; I need to read it in chunks. And this is a problem with the following code:

import requests

def DownloadFile(url)
    local_filename = url.split('/')[-1]
    r = requests.get(url)
    f = open(local_filename, 'wb')
    for chunk in r.iter_content(chunk_size=512 * 1024): 
        if chunk: # filter out keep-alive new chunks
            f.write(chunk)
    f.close()
    return 

For some reason it doesn't work this way: it still loads the response into memory before it is saved to a file.

UPDATE

If you need a small client (Python 2.x /3.x) which can download big files from FTP, you can find it here. It supports multithreading & reconnects (it does monitor connections) also it tunes socket params for the download task.

Answer 1

With the following streaming code, the Python memory usage is restricted regardless of the size of the downloaded file:

def download_file(url):
    local_filename = url.split('/')[-1]
    # NOTE the stream=True parameter below
    with requests.get(url, stream=True) as r:
        r.raise_for_status()
        with open(local_filename, 'wb') as f:
            for chunk in r.iter_content(chunk_size=8192): 
                # If you have chunk encoded response uncomment if
                # and set chunk_size parameter to None.
                #if chunk: 
                f.write(chunk)
    return local_filename

Note that the number of bytes returned using iter_content is not exactly the chunk_size; it's expected to be a random number that is often far bigger, and is expected to be different in every iteration.

See body-content-workflow and Response.iter_content for further reference.

Answer 2

It's much easier if you use Response.raw and shutil.copyfileobj():

import requests
import shutil

def download_file(url):
    local_filename = url.split('/')[-1]
    with requests.get(url, stream=True) as r:
        with open(local_filename, 'wb') as f:
            shutil.copyfileobj(r.raw, f)

    return local_filename

This streams the file to disk without using excessive memory, and the code is simple.

Answer 3

Not exactly what OP was asking, but... it's ridiculously easy to do that with urllib:

from urllib.request import urlretrieve
url = 'http://mirror.pnl.gov/releases/16.04.2/ubuntu-16.04.2-desktop-amd64.iso'
dst = 'ubuntu-16.04.2-desktop-amd64.iso'
urlretrieve(url, dst)

Or this way, if you want to save it to a temporary file:

from urllib.request import urlopen
from shutil import copyfileobj
from tempfile import NamedTemporaryFile
url = 'http://mirror.pnl.gov/releases/16.04.2/ubuntu-16.04.2-desktop-amd64.iso'
with urlopen(url) as fsrc, NamedTemporaryFile(delete=False) as fdst:
    copyfileobj(fsrc, fdst)

I watched the process:

watch 'ps -p 18647 -o pid,ppid,pmem,rsz,vsz,comm,args; ls -al *.iso'

And I saw the file growing, but memory usage stayed at 17 MB. Am I missing something?

Answer 4

Your chunk size could be too large, have you tried dropping that - maybe 1024 bytes at a time? (also, you could use with to tidy up the syntax)

def DownloadFile(url):
    local_filename = url.split('/')[-1]
    r = requests.get(url)
    with open(local_filename, 'wb') as f:
        for chunk in r.iter_content(chunk_size=1024): 
            if chunk: # filter out keep-alive new chunks
                f.write(chunk)
    return 

Incidentally, how are you deducing that the response has been loaded into memory?

It sounds as if python isn't flushing the data to file, from other SO questions you could try f.flush() and os.fsync() to force the file write and free memory;

    with open(local_filename, 'wb') as f:
        for chunk in r.iter_content(chunk_size=1024): 
            if chunk: # filter out keep-alive new chunks
                f.write(chunk)
                f.flush()
                os.fsync(f.fileno())

Answer 5

Based on the Roman's most upvoted comment above, here is my implementation, Including "download as" and "retries" mechanism:

def download(url: str, file_path='', attempts=2):
    """Downloads a URL content into a file (with large file support by streaming)

    :param url: URL to download
    :param file_path: Local file name to contain the data downloaded
    :param attempts: Number of attempts
    :return: New file path. Empty string if the download failed
    """
    if not file_path:
        file_path = os.path.realpath(os.path.basename(url))
    logger.info(f'Downloading {url} content to {file_path}')
    url_sections = urlparse(url)
    if not url_sections.scheme:
        logger.debug('The given url is missing a scheme. Adding http scheme')
        url = f'http://{url}'
        logger.debug(f'New url: {url}')
    for attempt in range(1, attempts+1):
        try:
            if attempt > 1:
                time.sleep(10)  # 10 seconds wait time between downloads
            with requests.get(url, stream=True) as response:
                response.raise_for_status()
                with open(file_path, 'wb') as out_file:
                    for chunk in response.iter_content(chunk_size=1024*1024):  # 1MB chunks
                        out_file.write(chunk)
                logger.info('Download finished successfully')
                return file_path
        except Exception as ex:
            logger.error(f'Attempt #{attempt} failed with error: {ex}')
    return ''

Answer 6

use wget module of python instead. Here is a snippet

import wget
wget.download(url)