Python page count down print (len(elem_href1-(number)))

if you looking for the len of the strings and sorting it out, try this.

driver.get('https://stackoverflow.com/questions')elements = [x.get_attribute("href") for x in driver.find_elements_by_css_selector("#questions .question-hyperlink")]print(len(elements))numbers = sorted([len(e) for e in elements], reverse=True)print(numbers)

update

def page_counter():  for x in range(1000):      yield xcount = page_counter()driver.get('https://stackoverflow.com/questions')elements = [x.get_attribute("href") for x in driver.find_elements_by_css_selector("#questions .question-hyperlink")]print(len(elements)) links = dict((next(count) + 1, e) for e in elements)for key, value in links.items():   driver.get(value)   print(f'At Page: {key}')

2nd update

import operatorlinks = dict((next(count) + 1, e) for e in elements)desc_links = sorted(links.items(), key=operator.itemgetter(1))for link in desc_links:    driver.get(link[1])    print(f'At Page: {link[0]}')

For me it seems that len(link) has nothing related to shuffle. It looks like it just prints length of the string of current link.

If you want to see a number of 'pages left to go' you can just do:

for i, e in enumerate(elem_href1):    print(len(elem_href1) - i)

or

left_to_go = len(elem_href1)print(left_to_go)for e in elem_href1:    left_to_go -= 1    print(left_to_go)