How to remove duplicates in my python script?

I have used a regex search to filter down some results from a text file (searching for ".js") which has given me roughly around 16 results some of which are duplicates. I want to remove duplicates from that output and print either onto the console or redirect it into a file. I have attempted the use of sets and dictionary.fromkeys with no success! Here is what I have at the moment, thank you in advance:

#!/usr/bin/python

import re
import sys

pattern = re.compile("[^/]*\.js")

for i, line in enumerate(open('access_log.txt')):
    for match in re.findall(pattern, line):
        x = str(match)
        print x

Using sets to eliminate duplicates:

#!/usr/bin/python

import re

pattern = re.compile("[^/]*\.js")

matches = set()
with open('access_log.txt') as f:
    for line in f:
        for match in re.findall(pattern, line):
            #x = str(match) # or just use match
            if match not in in matches:
                print match
                matches.add(match)

But I question your regex:

You are doing a findall on each line, which suggests that each line might have multiple "hits", such as:

file1.js file2.js file3.js

But in your regex:

[^/]*\.js

[^/]* is doing a greedy match and would return only one match, namely the complete line.

If you made the match non-greedy, i.e. [^/]*?, then you would get 3 matches:

'file1.js'
' file2.js'
' file3.js'

But that highlights another potential problem. Do you really want those spaces in the second and third matches for these particular cases? Perhaps in the case of /abc/ def.js you would keep the leading blank that follows /abc/.

So I would suggest:

#!/usr/bin/python

import re

pattern = re.compile("""
    (?x)            # verbose mode
    (?:             # first alternative:
        (?<=/)      # positive lookbehind assertion: preceded by '/'
        [^/]*?      # matches non-greedily 0 or more non-'/'
    |               # second alternative
        (?<!/)      # negative lookbehind assertion: not preceded by '/'
        [^/\s]*?    # matches non-greedily 0 or more non-'/' or non-whitespace
    )
    \.js            # matches '.js'
    """)

matches = set()
with open('access_log.txt') as f:
    for line in f:
        for match in pattern.findall(line):
            if match not in matches:
                print match
                matches.add(match)

If the filename cannot have any whitespace, then just use:

pattern = re.compile("[^\s/]*?\.js")