从S3使用博托解压my_file.zip拉 [英] Unzip my_file.zip pulled from s3 using boto

问题描述

我想使用博托打开的.zip 文件我在 S3 。我想直接用数据进行工作，我想避免创建临时文件。

I am trying to use boto to open a .zip file I have in s3. I am trying to work with the data directly, I want to avoid creating temporary files.

In [201]: import StringIO

In [202]: import boto

In [203]: conn = boto.connect_s3()

In [204]: my_bucket = conn.get_bucket('my_bucket')

In [205]: my_list = [ele for ele in my_bucket.list('my_file.zip')]

In [206]: f = StringIO.StringIO()

In [207]: my_list[0].get_file(f)

In [208]: f.seek(0)

如果该文件没有被压缩，我只想用：

If the file was not zipped I would just use:

my_content = my_list[0].get_contents_as_string()

但因为它是压缩，我收到的垃圾。

but since it is zipped, I am getting garbage.

这个问题的回答做什么，我想（我借了一点使用了拉链我试图从它），使用 GZIP ，但我无法找到任何东西。我试图用 zipfileZipFile ，而读，提取物和 extractall 方法似乎没有做我想做的。

An answer to this question does what I want (I borrowed a bit of my attempt from it) using gzip, but I can't find anything using for zip. I tried using zipfileZipFile, but read, extract and extractall methods don't seem to do what I want.

推荐答案

您应该看看Python模块GZIP：

You should look into the python module gzip :

https://docs.python.org/2/library/gzip.html

您应该能够StringIO的用gzip。

you should be able to stringIO with gzip. .

from boto.s3.connection import S3Connection
import gzip
from StringIO import StringIO

S3Conn = S3Connection() # assuming your .boto has been setup
Bucket = S3Conn.get_bucket('my_bucket')
my_list = [gzip.GzipFile(fileobj=(StringIO.StringIO(ele.get_contents_as_string()))) for ele in Bucket.list()]
#for readability I pulled this out
for item in my_list:
    item.read()

有关可读性名单COM prehension或许应该被打破了 - 但我跟你原来的帖子比较。

for readability the list comprehension should probably be broken up - but I followed your original posting to compare.

祝你好运！

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