Efficient User-Agent Regex to find Safari in Python

Question

To find out if user-agent pertains to Safari, one must look for the presence of Safari but not the presence of Chrome. I am also assuming that this needs to be case-insensitive.

I am trying to do this using regular expressions in Python without subsequently needing to traverse groups to match strings.

One way to solve this is :

r1 = re.compile ("Safari", re.I)
r2 = re.compile ("Chrome", re.I)

if len(r1.findall (userAgentString)) > 0 and len(r2.findall(userAgentString)) <=0):
    print "Found Safari"

I have also tried to attempt using

r = re.compile ("(?P<s>Safari)|(?P<c>Chrome)", re.I)
m = r.search (userAgentString)
if (m.group('s') and not m.group('c')):
    print "Found Safari"

This does not work because search will stop after finding the first instance of one of 'Chrome' or 'Safari' (probably obvious to the Regex-Gurus..).

I can get it to work slightly efficiently using the re.finditer() funciton as follows :

r = re.compile ("(?P<s>Safari)|(?P<c>Chrome)", re.I)
safari = chrome = False
for i in r.finditer (userAgentString):
    if i.group('s'):
        safari = True
    if i.group('c'):
        chrome = True
if safari and not chrome:
    print "Found Safari"

Is there a more efficient way to do this ? (Please note I am looking for efficiency not convenience). Thanks.

Sample User-agents :

Safari : "Mozilla/5.0 (iPad; CPU OS 6_0 like Mac OS X) AppleWebKit/536.26 (KHTML, like Gecko) Version/6.0 Mobile/10A5355d Safari/8536.25"

Chrome : "Mozilla/5.0 (Windows NT 6.2; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/32.0.1667.0 Safari/537.36"

For it's worth, I timed it and jwodder is on the mark with the efficiency of a simple 'lower()' and an 'in'. Came out to be about 10 times faster than a precompiled regex. Unless I did something wrong in setup/timeit..

    import timeit
    setup = '''
import re
r = re.compile ('(?P<m>MSIE)|(?P<c>Chrome)|(?P<s>Safari)', re.I)
def strictBrowser (userAgentString):
    c=s=m=False
    for f in r.finditer(userAgentString):
        if f.group('m'):
            m = True
        if f.group('c'):
            c = True
        if f.group('s'):
            s = True
    # msie or (safari but not chrome)
    # all chromes us will have safari in them..
    return m or (s and not c)
'''
    print timeit.timeit(
        'strictBrowser ("Mozilla/5.0 (iPad; CPU OS 6_0 like Mac OS X) AppleWebKit/536.26 (KHTML, like Gecko) Version/6.0 Mobile/10A5355d Safari/8536.2")',
        setup=setup, number=100000
        )
    setup = '''
def strictBrowser (userAgentString):
    userAgentString = userAgentString.lower()
    if (
        'msie' in userAgentString or
        ('safari' in userAgentString and 'chrome' not in userAgentString)
        ):
        return True
    return False
'''
    print timeit.timeit(
        'strictBrowser ("Mozilla/5.0 (iPad; CPU OS 6_0 like Mac OS X) AppleWebKit/536.26 (KHTML, like Gecko) Version/6.0 Mobile/10A5355d Safari/8536.2")',
        setup=setup, number=100000
        )

Output :
0.0778814506637
0.00664118263765

Can you include sample User-Agent strings, please?

Martijn Pieters
– Martijn Pieters

2013-11-10 01:18:16 +00:00
Commented Nov 10, 2013 at 1:18 — Martijn Pieters
– Martijn Pieters, Commented Nov 10, 2013 at 1:18

jwodder · Accepted Answer · 2013-11-10 01:20:07Z

4

Since you're testing whether certain fixed strings appear in a given string, it's probably easiest and most efficient to forgo regexes entirely:

if 'safari' in userAgentString.lower() and 'chrome' not in userAgentString.lower():
    print "Found Safari"

answered Nov 10, 2013 at 1:20

jwodder

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5 Comments

user1055761 Over a year ago

I am wondering what is the cost of a lower() on userAgentString ? In your case you are calling it two times. I am also wondering which is faster a lower on the entire userAgent string and then an 'in' 2 times or a regex. (Please note that I could compile the regex once at startup and call only the search() function each time on the user agent strings.. (sorry if I am missing something)

jwodder Over a year ago

@user1055761: Run some tests (say, with timeit) and find out.

steveha Over a year ago

The .lower() method function is pretty inexpensive, but it is pretty easy to use a variable to avoid evaluating it twice and that is what I would recommend.

Alan Moore Over a year ago

Why are you so worried about efficiency? There's not going to be a significant difference in performance whichever approach you take.

user1055761 Over a year ago

I ran tests with timeit and jwodder is on the mark ! Have added it to the main question.

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