There any way to optimize these two functions ?
first function:
def searchList(list_, element):
for i in range (0,len(list_)):
if(list_[i] == element):
return True
return False
second function:
return_list=[]
for x in list_search:
if searchList(list_users,x)==False:
return_list.append(x)
Yes:
return_list = [x for x in list_search if x not in list_users]
The first function basically checks for membership, in which case you could use the in keyword. The second function can be reduced to a list comprehension to filter out elements from list_search list based on your condition.
in is O(n) for list. If you wanted to get really crazy you can write a binary search (if applicable) which would reduce this to logn!For first function
def searchList(list, element):
return element in list
You can make it in 1 line
searchList = lambda x,y: y in x
For 2nd, use a list comp like shown in the other answer
What you are doing with your two functions is building the complement as ozgur pointed out. Using sets is the most easy thing here
>>> set([2,2,2,3,3,4])- set([1,2,2,4,5])
set([3])
your list_search would be the first list and your list_users the second list.
The only difference is that your new user is only once in the result no matter how often it is in the list_search
Disclaimer: I assumed list_search has no duplicate elements. Otherwise, use this solution.
What you want is exactly the set complement of list_users in list_search.
As an alternative approach, you can use sets to get the difference between two lists and I think it should be much more performant than the naive look up which takes 0(n^2).
>>> list_search = [1, 2, 3, 4]
>>> list_users = [4, 5, 1, 6]
>>> print set(list_search).difference(list_users)
>>> {2, 3}
list_usersto asetand check for item simply using theinoperator.