我的目标是将来自CommonCrawl的WARC文件拆分并排序为其单独的记录。示例文件：
WARC/1.0
WARC-Type: warcinfo
WARC-Date: 2020-08-04T01:43:40Z
WARC-Record-ID: <urn:uuid:959ea654-33fd-466b-b1bf-f08aa8abe774>
Content-Length: 500
Content-Type: application/warc-fields
WARC-Filename: CC-MAIN-20200804014340-20200804044340-00045.warc.gz
isPartOf: CC-MAIN-2020-34
publisher: Common Crawl
description: Wide crawl of the web for August 2020
operator: Common Crawl Admin (info@commoncrawl.org)
hostname: ip-10-67-67-22.ec2.internal
software: Apache Nutch 1.17 (modified, https://github.com/commoncrawl/nutch/)
robots: checked via crawler-commons 1.2-SNAPSHOT (https://github.com/crawler-commons/crawler-commons)
format: WARC File Format 1.1
conformsTo: http://iipc.github.io/warc-specifications/specifications/warc-format/warc-1.1/
WARC/1.0
WARC-Type: request
WARC-Date: 2020-08-04T03:25:25Z
WARC-Record-ID: <urn:uuid:6c0b749a-4d02-4a77-ab93-9bc4ba094cdc>
Content-Length: 303
Content-Type: application/http; msgtype=request
WARC-Warcinfo-ID: <urn:uuid:959ea654-33fd-466b-b1bf-f08aa8abe774>
WARC-IP-Address: 104.254.66.40
WARC-Target-URI: http://00.auto.sohu.com/d/details?cityCode=450100&planId=1450&trimId=145372
我如何在以下行将文件拆分为不同的记录："；WARC/1.0"；？
您可以使用"warcio"；lib。
示例代码：
import requests
from warcio.archiveiterator import ArchiveIterator
from warcio.warcwriter import WARCWriter
def split_records(url):
resp = requests.get(url, stream=True)
for record in ArchiveIterator(resp.raw, arc2warc=True):
if record.rec_type == 'warcinfo':
print(record.raw_stream.read())
elif record.rec_type == 'response':
id = record.rec_headers.get_header('WARC-Record-ID').rsplit(':', 1)[-1].rstrip('>')
print(id)
output = open('%s.warc.gz' % (id), 'wb')
writer = WARCWriter(output, gzip=True)
writer.write_record(record)
output.close()
split_records('https://cdn.ruarxive.org/public/webcollect2020/kgi/komitetgi.ru/komitetgi.ru.warc.gz')
它将把WARC文件拆分为单个WARC记录，存储在与脚本文件相同的路径中。