如何将 zip 文件中的文件读取为文本而不是字节?
一个用于读取 ZIP 存档内的 CSV 文件的简单程序:
import csv, sys, zipfile
zip_file = zipfile.ZipFile(sys.argv[1])
items_file = zip_file.open('items.csv', 'rU')
for row in csv.DictReader(items_file):
pass
适用于 Python 2.7:
$ python2.7 test_zip_file_py3k.py ~/data.zip
$
但不适用于 Python 3.2:
$ python3.2 test_zip_file_py3k.py ~/data.zip
Traceback (most recent call last):
File "test_zip_file_py3k.py", line 8, in <module>
for row in csv.DictReader(items_file):
File "/somedir/python3.2/csv.py", line 109, in __next__
self.fieldnames
File "/somedir/python3.2/csv.py", line 96, in fieldnames
self._fieldnames = next(self.reader)
_csv.Error: iterator should return strings, not bytes (did you open the file
in text mode?)
Python 3 中的 csv
模块想要查看文本文件,但是 zipfile。 ZipFile.open
返回始终被视为二进制数据的 zipfile.ZipExtFile
。
如何在 Python 3 中实现这一功能?
A simple program for reading a CSV file inside a ZIP archive:
import csv, sys, zipfile
zip_file = zipfile.ZipFile(sys.argv[1])
items_file = zip_file.open('items.csv', 'rU')
for row in csv.DictReader(items_file):
pass
works in Python 2.7:
$ python2.7 test_zip_file_py3k.py ~/data.zip
$
but not in Python 3.2:
$ python3.2 test_zip_file_py3k.py ~/data.zip
Traceback (most recent call last):
File "test_zip_file_py3k.py", line 8, in <module>
for row in csv.DictReader(items_file):
File "/somedir/python3.2/csv.py", line 109, in __next__
self.fieldnames
File "/somedir/python3.2/csv.py", line 96, in fieldnames
self._fieldnames = next(self.reader)
_csv.Error: iterator should return strings, not bytes (did you open the file
in text mode?)
The csv
module in Python 3 wants to see a text file, but zipfile.ZipFile.open
returns a zipfile.ZipExtFile
that is always treated as binary data.
How does one make this work in Python 3?
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评论(6)
我刚刚注意到 Lennart 的答案不适用于 Python 3.1,但它可以适用于 Python 3.2。他们增强了
zipfile.ZipExtFile
在 Python 3.2 中(请参阅发行说明)。这些更改似乎使 zipfile.ZipExtFile 与io.TextWrapper
。顺便说一句,它可以在 Python 3.1 中工作,如果你取消注释下面的 hacky 行来猴子补丁
zipfile.ZipExtFile
,并不是说我会推荐这种 hackery。我包含它只是为了说明 Python 3.2 中为使事情顺利运行而所做的工作的本质。如果我必须支持 py3k < 3.2,那么我会采用 我的其他答案。
3.6+ 更新
从 3.6 开始,删除了对
mode='U'
的支持^1:从 3.8 开始,添加了 Path 对象,它给出了我们可以像内置的
open()
函数一样调用open()
方法(在以下情况下传递newline=''
我们的 CSV),我们得到 csv 阅读器接受的 io.TextIOWrapper 对象。请参阅尤里的回答,此处。I just noticed that Lennart's answer didn't work with Python 3.1, but it does work with Python 3.2. They've enhanced
zipfile.ZipExtFile
in Python 3.2 (see release notes). These changes appear to makezipfile.ZipExtFile
work nicely withio.TextWrapper
.Incidentally, it works in Python 3.1, if you uncomment the hacky lines below to monkey-patch
zipfile.ZipExtFile
, not that I would recommend this sort of hackery. I include it only to illustrate the essence of what was done in Python 3.2 to make things work nicely.If I had to support py3k < 3.2, then I would go with the solution in my other answer.
Update for 3.6+
Starting w/3.6, support for
mode='U'
was removed^1:Starting w/3.8, a Path object was added which gives us an
open()
method that we can call like the built-inopen()
function (passingnewline=''
in the case of our CSV) and we get back an io.TextIOWrapper object the csv readers accept. See Yuri's answer, here.您可以将其包装在 io.TextIOWrapper 中。
应该有效。
You can wrap it in a io.TextIOWrapper.
Should work.
如果您只想将文件读入字符串:
And if you just like to read a file into a string:
Lennart 的答案是在正确的轨道上(谢谢,Lennart,我投票赞成你的答案)并且它几乎有效:
问题出现io.TextWrapper 的第一个必需参数是 <强>缓冲;不是文件对象。
这似乎有效:
这似乎有点复杂,而且必须将整个(可能很大)zip 文件读入内存似乎很烦人。还有更好的办法吗?
这是在行动中:
Lennart's answer is on the right track (Thanks, Lennart, I voted up your answer) and it almost works:
The problem appears to be that io.TextWrapper's first required parameter is a buffer; not a file object.
This appears to work:
This seems a little complex and also it seems annoying to have to read in a whole (perhaps huge) zip file into memory. Any better way?
Here it is in action:
从 Python 3.8 开始,zipfile 模块具有 Path 对象 ,我们可以将其与其 open() 方法一起使用来获取 io.TextIOWrapper 对象,该对象可以传递给 csv 读取器:
Starting with Python 3.8, the zipfile module has the Path object, which we can use with its open() method to get an io.TextIOWrapper object, which can be passed to the csv readers:
这是打开 zip 文件并读取该 zip 内的文本文件的最小方法。我发现窍门是 TextIOWrapper read() 方法,上面的任何答案中都没有提到(上面提到了 BytesIO.read(),但 Python 文档推荐 TextIOWrapper)。
我希望他们在 ZipFile open() 方法中保留 mode='U' 参数来执行相同的操作,因为这非常简洁,但是,唉,这已经过时了。
Here's a minimal recipe to open a zip file and read a text file inside that zip. I found the trick to be the TextIOWrapper read() method, not mentioned in any answers above (BytesIO.read() was mentioned above, but Python docs recommend TextIOWrapper).
I wish they kept the mode='U' parameter in the ZipFile open() method to do this same thing since that was so succinct but, alas, that is obsolete.