Building on @Dex'ter answer@Dex'ter answer, I think that using a dictionary to store your results call for the wrong iteration patterns. I would store them in a list and grab them directly with a for
loop when I need them.
Building becomes something along the lines of:
vacancies = [{
'position': element.contents[0].strip(),
'link': 'http://rabota.ua{}'.format(element.get('href')),
} for element in href_tag]
Updating would be:
for vacancy, company in zip(vacancies, company_href):
vacancy.update({'company': company.get_text()})
And printing can be as simple as:
for vacancy in vacancies:
print(vacancy['company'], vacancy['position'], vacancy['link'], sep='\n')
Writting that, I think that using the whole building + updating technique is also a poor decision as you could build everything at once:
href_tag = content.find_all('a', class_='t')
company_href_tag = content.find_all("a", class_="rua-p-c-default")[2:]
vacancies = [{
'position': element.contents[0].strip(),
'link': 'http://rabota.ua{}'.format(element.get('href')),
'company': company.get_text(),
} for element, company in zip(href_tag, company_href_tag)]