使用请求在python中下载大文件

问题:使用请求在python中下载大文件

请求是一个非常不错的库。我想用它来下载大文件(> 1GB)。问题是不可能将整个文件保留在内存中,我需要分块读取它。这是以下代码的问题

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 

由于某种原因,它无法按这种方式工作。仍将响应加载到内存中,然后再将其保存到文件中。

更新

如果您需要一个小型客户端(Python 2.x /3.x),可以从FTP下载大文件,则可以在此处找到它。它支持多线程和重新连接(它确实监视连接),还可以为下载任务调整套接字参数。

Requests is a really nice library. I’d like to use it for download 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 

By some reason it doesn’t work this way. It still loads response into memory before save it 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.


回答 0

使用以下流代码,无论下载文件的大小如何,Python内存的使用都受到限制:

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

请注意,使用返回的字节数iter_content不完全是chunk_size; 它应该是一个通常更大的随机数,并且在每次迭代中都应该有所不同。

https://requests.readthedocs.io/en/latest/user/advanced/#body-content-workflowhttps://requests.readthedocs.io/en/latest/api/#requests.Response.iter_content进一步参考。

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 https://requests.readthedocs.io/en/latest/user/advanced/#body-content-workflow and https://requests.readthedocs.io/en/latest/api/#requests.Response.iter_content for further reference.


回答 1

如果使用Response.raw和,则容易得多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

这样就无需占用过多内存就可以将文件流式传输到磁盘,并且代码很简单。

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.


回答 2

OP并不是在问什么,但是…这样做很简单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)

或这样,如果您要将其保存到临时文件中:

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)

我看了看这个过程:

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

而且我看到文件在增长,但内存使用量保持在17 MB。我想念什么吗?

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?


回答 3

您的块大小可能太大,您是否尝试过删除它-一次一次可能是1024个字节?(同样,您可以with用来整理语法)

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 

顺便说一句,您如何推断响应已加载到内存中?

这听起来仿佛Python没有刷新数据文件,从其他SO问题,你可以尝试f.flush(),并os.fsync()迫使文件的写入和释放内存;

    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())

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())