I have a list, which may/may not contain duplicate elements. Given another list/set of Elements, I need a list of all unique elements present in that list.
Input:
input_list = ['android', 'ios', 'android', '', 'none', 'android', 'junk_os']
os_list = ['android', 'ios', 'windows']
Output:
output = ['android', 'ios']
What will be most pythonic and efficient way of doing this ? Length of given list may be around 10, while Os_list is fixed to 3. But this line will be iterated over 10000 times.
You could use sets to do this
set(os_list).intersection(input_list)
Edit Since the os_list is fixed you may as well have that stored as a set:
os_list = {'android', 'ios', 'windows'}
# then it's a little less work to do each time in the loop
output = os_list.intersection(input_list)
Profiling
# me
os_set = {'android', 'ios', 'windows'}
%timeit os_set.intersection(input_list)
# 1000000 loops, best of 3: 323 ns per loop
# vks
os_list = ['android', 'ios', 'windows']
%timeit [i for i in os_list if i in input_list]
# 1000000 loops, best of 3: 550 ns per loop
Using Padraic Cunningham's method you can avoid the function lookup and scrape a little more performance out of it. As a bonus it ends up looking like a function name that makes sense.
os_set = {'android', 'ios', 'windows'}
unique_valid_devices = os_set.intersection
%timeit output_list = unique_valid_devices(input_list)
1000000 loops, best of 3: 290 ns per loop
si = os_set.intersection and si(os_list) should be faster againList = ['android', 'ios', 'android', '', 'none', 'android']
OS_list = ['android', 'ios', 'windows']
y=set(List)
print [i for i in OS_list if i in y]
You can use set here which is O(1).
The fastest though is (for smaller List having upto 600-700 elements)
List = ['android', 'ios', 'android', '', 'none', 'android']
OS_list = ['android', 'ios', 'windows']
[i for i in OS_list if i in List]
Time check:
s1="""List = ['android', 'ios', 'android', '', 'none', 'android']
OS_list = ['android', 'ios', 'windows']
[i for i in OS_list if i in List]"""
s2="""List = ['android', 'ios', 'android', '', 'none', 'android']
OS_list = ['android', 'ios', 'windows']
set(List).intersection(OS_list)"""
print timeit.timeit(stmt=s1,number=1000)
print timeit.timeit(stmt=s2,number=1000)
Output:
0.000895947527903
0.00130528204537
if i in y there if you search in list its O(n) but if y is a set its O(1)