Count word on the page - python-3.x

Question

How to scrap any site and search for the given word and displays how many times it occurred
class LinkedinScraper(scrapy.Spider):
name = "linked"
def start_requests(self):
urls = ['https://www.linkedin.com/']
for url in urls:
yield scrapy.Request(url=url, callback=self.parse)
def parse(self, response):
page = response.url.split("/")[-2]
filename = 'linkedin.html' % page
with open(filename, 'wb') as f:
f.write(response.body)
self.log('Saved file %s' % filename)

Answer 1

You can use regex with response.body to find all occurrances in any places
ie.
import re
r = re.findall('\\bcat\\b', "cat catalog cattering")
print(len(r), 'cat(s)')
Gives "1 cat(s)", not "3 cat(s)"
If you need word only in some tags then you use first response.css(), response.xpath(), etc.
EDIT:
Example which shows how to use
re.findall(pattern, response.text)
but it can find text inside tag too.
It also shows how to use
response.css('body').re(pattern)
It counts 'view', '\\bviews\\b' and '\d+ views' on Stackoverflow and display first three elements
You can run it without creating project.
import scrapy
import re
class MySpider(scrapy.Spider):
name = 'myspider'
start_urls = ['https://stackoverflow.com/']
def parse(self, response):
print('url:', response.url)
for pattern in ['view', '\\bviews\\b', '\d+ views']:
print('>>> pattern:', pattern)
result = re.findall(pattern, response.text)
print('>>> re:', len(result), result[0:3])
result = response.css('body').re(pattern)
print('>>> response.re:', len(result), result[0:3])
# --- it runs without project and saves in `output.csv` ---
from scrapy.crawler import CrawlerProcess
c = CrawlerProcess({'USER_AGENT': 'Mozilla/5.0'})
c.crawl(MySpider)
c.start()