Question

我在ScraperWiki上使用以下代码在Twitter上搜索特定的主题标签它工作得很好，并且正在挑选推文中提供的任何邮政编码（如果没有，则返回 false ）。这是通过行data['location'] = scraperwiki.geo.extract_gb_postcode(result['text'])实现的但我仅对包含邮政编码信息的推文感兴趣（这是因为它们将在稍后阶段添加到Google地图中）。
最简单的方法是什么？我对PHP很不错，但Python对我来说是一个全新的领域。在此先感谢您的帮助祝福，祝马丁

import scraperwiki
import simplejson
import urllib2

QUERY = 'enter_hashtag_here'
RESULTS_PER_PAGE = '100'
NUM_PAGES = 10

for page in range(1, NUM_PAGES+1):
    base_url = 'http://search.twitter.com/search.json?q=%s&rpp=%s&page=%s' \
         % (urllib2.quote(QUERY), RESULTS_PER_PAGE, page)
    try:
        results_json = simplejson.loads(scraperwiki.scrape(base_url))
        for result in results_json['results']:
            #print result
            data = {}
            data['id'] = result['id']
            data['text'] = result['text']
            data['location'] = scraperwiki.geo.extract_gb_postcode(result['text'])
            data['from_user'] = result['from_user']
            data['created_at'] = result['created_at']
            print data['from_user'], data['text']
            scraperwiki.sqlite.save(["id"], data)
    except:
        print 'Oh dear, failed to scrape %s' % base_url
        break

Answer 1

你想要这个吗？我尝试了免费的ScraperWiki测试页面，似乎做了你想要的。如果你正在寻找更复杂的东西，请告诉我。

import scraperwiki
import simplejson
import urllib2

QUERY = 'meetup'
RESULTS_PER_PAGE = '100'
NUM_PAGES = 10

for page in range(1, NUM_PAGES+1):
    base_url = 'http://search.twitter.com/search.json?q=%s&rpp=%s&page=%s' \
         % (urllib2.quote(QUERY), RESULTS_PER_PAGE, page)
    try:
        results_json = simplejson.loads(scraperwiki.scrape(base_url))
        for result in results_json['results']:
            #print result
            data = {}
            data['id'] = result['id']
            data['text'] = result['text']
            data['location'] = scraperwiki.geo.extract_gb_postcode(result['text'])
            data['from_user'] = result['from_user']
            data['created_at'] = result['created_at']
            if data['location']:
                print data['location'], data['from_user']
                scraperwiki.sqlite.save(["id"], data)
    except:
        print 'Oh dear, failed to scrape %s' % base_url
        break

输出：

P93JX VSDC
FV36RL Bootstrappers
Ci76fP Eli_Regalado
UN56fn JasonPalmer1971
iQ3H6zR GNOTP
Qr04eB fcnewtech
sE79dW melindaveee
ud08GT MariaPanlilio
c9B8EE akibantech
ay26th Thepinkleash

我已经对它进行了一些改进，所以它比使用scraperwiki检查提取gb postcodes有点挑选，这可以让很多误报。基本上我从here接受了接受的答案，并添加了一些负面的lookbehind / lookahead来过滤掉更多。看起来像刮刀维基检查没有负面的lookbehind / lookahead的正则表达式。希望有所帮助。

import scraperwiki
import simplejson
import urllib2
import re

QUERY = 'sw4'
RESULTS_PER_PAGE = '100'
NUM_PAGES = 10

postcode_match = re.compile('(?<![0-9A-Z])([A-PR-UWYZ0-9][A-HK-Y0-9][AEHMNPRTVXY0-9]?[ABEHMNPRVWXY0-9]? {0,2}[0-9][ABD-HJLN-UW-Z]{2}|GIR 0AA)(?![0-9A-Z])', re.I)

for page in range(1, NUM_PAGES+1):
    base_url = 'http://search.twitter.com/search.json?q=%s&rpp=%s&page=%s' \
         % (urllib2.quote(QUERY), RESULTS_PER_PAGE, page)
    try:
        results_json = simplejson.loads(scraperwiki.scrape(base_url))
        for result in results_json['results']:
            #print result
            data = {}
            data['id'] = result['id']
            data['text'] = result['text']
            data['location'] = scraperwiki.geo.extract_gb_postcode(result['text'])
            data['from_user'] = result['from_user']
            data['created_at'] = result['created_at']
            if data['location'] and postcode_match.search(data['text']):
                print data['location'], data['text']
                scraperwiki.sqlite.save(["id"], data)
    except:
        print 'Oh dear, failed to scrape %s' % base_url
        break

ScraperWiki / Python：当属性为false时过滤掉记录

1 个答案: