Python：如何分割WARC文件？

问题描述

我的目标是将来自CommonCrawl的WARC文件拆分和排序成单独的记录。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 ([e-mail])
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

如何在“ WARC / 1.0”行中将文件拆分为不同的记录？

解决方法

您可以使用“warcio”库来实现。

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')

它会将 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。