问Python - Scrapy到Json的输出拆分
EN

Stack Overflow用户

提问于 2018-07-07 18:43:24

回答 2查看 882关注 0票数 0

我正在学习NLP，为此，我使用Scrapy抓取了一篇亚马逊书评。我已经提取了所需的字段，并将它们输出为Json文件格式。当此文件作为df加载时，每个字段都被记录为一个列表，而不是一个单独的逐行格式。我如何拆分这个列表，以便df对每个项目都有一行，而不是将所有项目条目记录在单独的列表中？代码：

import scrapy


class ReviewspiderSpider(scrapy.Spider):
    name = 'reviewspider'
    allowed_domains = ['amazon.co.uk']
    start_urls = ['https://www.amazon.com/Gone-Girl-Gillian-Flynn/product-reviews/0307588378/ref=cm_cr_othr_d_paging_btm_1?ie=UTF8&reviewerType=all_reviews&pageNumber=1']

def parse(self, response):
    users = response.xpath('//a[contains(@data-hook, "review-author")]/text()').extract()
    titles = response.xpath('//a[contains(@data-hook, "review-title")]/text()').extract()
    dates = response.xpath('//span[contains(@data-hook, "review-date")]/text()').extract()
    found_helpful = response.xpath('//span[contains(@data-hook, "helpful-vote-statement")]/text()').extract()
    rating = response.xpath('//i[contains(@data-hook, "review-star-rating")]/span[contains(@class, "a-icon-alt")]/text()').extract()
    content = response.xpath('//span[contains(@data-hook, "review-body")]/text()').extract()

    yield {
        'users' : users.extract(),
        'titles' : titles.extract(),
        'dates' : dates.extract(),
        'found_helpful' : found_helpful.extract(),
        'rating' : rating.extract(),
        'content' : content.extract()
    }

示例输出：

users = ['Lauren', 'James'...'John']
dates = ['on September 28, 2017', 'on December 26, 2017'...'on November 17, 2016']
rating = ['5.0 out of 5 stars', '2.0 out of 5 stars'...'5.0 out of 5 stars']

所需输出：

index 1: [users='Lauren', dates='on September 28, 2017', rating='5.0 out of 5 stars']
index 2: [users='James', dates='On December 26, 2017', rating='5.0 out of 5 stars']
...

我知道可能应该编辑与爬行器相关的Pipeline来实现这一点，但是我的Python知识有限，无法理解Scrapy文档。我也尝试过here和here的解决方案，但是我知道的还不够多，无法将答案与我自己的代码结合起来。任何帮助都将不胜感激。

scrapy

output

python

json

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回答 2

Stack Overflow用户

回答已采纳

发布于 2018-07-07 22:54:24

编辑:我能够通过使用.css方法而不是.xpath来提出解决方案。我用来从一家时装零售商那里抓取衬衫清单的蜘蛛：

import scrapy
from ..items import ProductItem

class SportsdirectSpider(scrapy.Spider):
    name = 'sportsdirect'
    allowed_domains = ['www.sportsdirect.com']
    start_urls = ['https://www.sportsdirect.com/mens/mens-shirts']

def parse(self, response):
    products = response.css('.s-productthumbbox')
    for p in products:
        brand = p.css('.productdescriptionbrand::text').extract_first()
        name = p.css('.productdescriptionname::text').extract_first()
        price = p.css('.curprice::text').extract_first()
        item = ProductItem()
        item['brand'] = brand
        item['name'] = name
        item['price'] = price
        yield item

Stack Overflow用户

发布于 2018-07-07 22:18:43

在重读你的问题后，我很确定这就是你想要的：

def parse(self, response):
    users = response.xpath('//a[contains(@data-hook, "review-author")]/text()').extract()
    titles = response.xpath('//a[contains(@data-hook, "review-title")]/text()').extract()
    dates = response.xpath('//span[contains(@data-hook, "review-date")]/text()').extract()
    found_helpful = response.xpath('//span[contains(@data-hook, "helpful-vote-statement")]/text()').extract()
    rating = response.xpath('//i[contains(@data-hook, "review-star-rating")]/span[contains(@class, "a-icon-alt")]/text()').extract()
    content = response.xpath('//span[contains(@data-hook, "review-body")]/text()').extract()

    for user, title, date, found_helpful, rating, content in zip(users, titles, dates, found_helpful, rating, content):
        yield {
            'user': user,
            'title': title,
            'date': date,
            'found_helpful': found_helpful,
            'rating': rating,
            'content': content
        }