在python和scrapy中检查另一个数组与另一个数组

Question

我想在Python中将一个变量设置为一个数组中的字符串元素，这是基于另一个数组中使用的字符串元素。我很难过如何去做。

这里有两个阵列：

genre = ["Dance", 
    "Festivals", 
    "Rock/pop" 
    ] 

我试图基于在另一个阵列即这三个要素来打印类型时start_urls = [0]，流派= [0]：

start_urls = [ 
    "http://www.allgigs.co.uk/whats_on/London/clubbing-1.html", 
    "http://www.allgigs.co.uk/whats_on/London/festivals-1.html", 
    "http://www.allgigs.co.uk/whats_on/London/tours-1.html" 
] 

全码：

genre = ["Dance", 
    "Festivals", 
    "Rock/pop" 
    ] 

class AllGigsSpider(CrawlSpider): 
    name = "allGigs" # Name of the Spider. In command promt, when in the correct folder, enter "scrapy crawl Allgigs". 
    allowed_domains = ["www.allgigs.co.uk"] # Allowed domains is a String NOT a URL. 
    start_urls = [ 
     "http://www.allgigs.co.uk/whats_on/London/clubbing-1.html", 
     "http://www.allgigs.co.uk/whats_on/London/festivals-1.html", 
     "http://www.allgigs.co.uk/whats_on/London/tours-1.html" 
    ] 

    rules = [ 
     Rule(SgmlLinkExtractor(restrict_xpaths='//div[@class="more"]'), # Search the start URL's for 
     callback="parse_item", 
     follow=True), 
    ] 

    def parse_start_url(self, response): 
     return self.parse_item(response) 

    def parse_item(self, response):#http://stackoverflow.com/questions/15836062/scrapy-crawlspider-doesnt-crawl-the-first-landing-page 
     for info in response.xpath('//div[@class="entry vevent"]'): 
      item = TutorialItem() # Extract items from the items folder. 
      item ['artist'] = info.xpath('.//span[@class="summary"]//text()').extract() # Extract artist information. 
      item ['date'] = info.xpath('.//span[@class="dates"]//text()').extract() # Extract date information. 
      preview = ''.join(str(s)for s in item['artist']) 
      #item ['genre'] = i.xpath('.//li[@class="style"]//text()').extract() 
      client = soundcloud.Client(client_id='401c04a7271e93baee8633483510e263', client_secret='b6a4c7ba613b157fe10e20735f5b58cc', callback='http://localhost:9000/#/callback.html') 
      tracks = client.get('/tracks', q = preview, limit=1) 
      for track in tracks: 
       print track.id 
       for i, val in enumerate(genre): 
         print '{} {}'.format(genre[i], start_urls[i]) 
       print genre 
       #for i, val in enumerate(genre): 
       #  print '{} {}'.format(genre[i], start_urls[i]) 
       item ['trackz'] = track.id 
       yield item 

任何帮助表示赞赏。

Answer 1

for i, val in enumerate(genre): 
    print '{} {}'.format(genre[i], start_urls[i]) 

应该工作