Python 3.x 如何在python3中使用lzma(*.warc.xz)压缩warc记录?

Python 3.x 如何在python3中使用lzma(*.warc.xz)压缩warc记录?,python-3.x,lzma,xz,warc,Python 3.x,Lzma,Xz,Warc,我有一份warc记录清单。列表中的每一项都是这样创建的: header = warc.WARCHeader({ "WARC-Type": "response", "WARC-Target-URI": "www.somelink.com", }, defaults=True) data = "Some string" record = warc.WARCRecord(header, data.encode('utf-8','replace')) output_file = war


我有一份warc记录清单。列表中的每一项都是这样创建的:

header = warc.WARCHeader({
    "WARC-Type": "response",
    "WARC-Target-URI": "www.somelink.com",
}, defaults=True)
data = "Some string"
record = warc.WARCRecord(header, data.encode('utf-8','replace'))
output_file = warc.open("my_file.warc.gz", 'wb')
output_file.write_record(record) # type of record is WARCRecord
现在,我使用*.warc.gz存储我的记录,如下所示:

header = warc.WARCHeader({
    "WARC-Type": "response",
    "WARC-Target-URI": "www.somelink.com",
}, defaults=True)
data = "Some string"
record = warc.WARCRecord(header, data.encode('utf-8','replace'))
output_file = warc.open("my_file.warc.gz", 'wb')
output_file.write_record(record) # type of record is WARCRecord
然后写下这样的记录:

header = warc.WARCHeader({
    "WARC-Type": "response",
    "WARC-Target-URI": "www.somelink.com",
}, defaults=True)
data = "Some string"
record = warc.WARCRecord(header, data.encode('utf-8','replace'))
output_file = warc.open("my_file.warc.gz", 'wb')
output_file.write_record(record) # type of record is WARCRecord
但是如何使用lzma压缩为*.warc.xz?我在调用warc.open时尝试过用xz替换gz,但是python3中的warc不支持这种格式。我发现了这个,但我无法用这个来拯救战争记录:

output_file = lzma.open("my_file.warc.xz", 'ab', preset=9)
header = warc.WARCHeader({
    "WARC-Type": "response",
    "WARC-Target-URI": "www.somelink.com",
}, defaults=True)
data = "Some string"
record = warc.WARCRecord(header, data.encode('utf-8','replace'))
output_file.write(record)
错误消息是:

TypeError:需要一个类似字节的对象,而不是“WARCRecord”


感谢您的帮助。

WARCRecord类有一个方法,可以将记录写入文件对象

您可以使用它将记录写入使用
lzma.open()
创建的文件中