You could do the following:
from fake_useragent import UserAgent
import random
import re
def grp(pat, txt):
r = re.search(pat, txt)
return r.group(0) if r else '&'
ua = UserAgent()
browsers = {
'chrome': r'Chrome/[^ ]+',
'safari': r'AppleWebKit/[^ ]+',
'opera': r'Opera\s.+$',
'firefox': r'Firefox/.+$',
'internetexplorer': r'Trident/[^;]+',
}
for k, v in browsers.items():
print(sorted(ua.data_browsers[k], key=lambda a: grp(v, a))[-1])
The output of the script is:
Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/41.0.2228.0 Safari/537.36
Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.75.14 (KHTML, like Gecko) Version/7.0.3 Safari/7046A194A
Opera/9.80 (Windows NT 6.1; Opera Tablet/15165; U; en) Presto/2.8.149 Version/11.1
Mozilla/5.0 (Windows NT 6.1; WOW64; rv:40.0) Gecko/20100101 Firefox/40.1
Mozilla/5.0 (compatible, MSIE 11, Windows NT 6.3; Trident/7.0; rv:11.0) like Gecko
Please note that the grp
function was blatantly stolen from this answer
If you instead are only looking for a single browser, like you asked, this would choose randomly while honoring the probabilities as lined out in the project's readme
browser = random.choice(ua.data_randomize)
print(sorted(ua.data_browsers[browser], key=lambda a: grp(browsers[browser], a))[-1])