please help me, i want to write a function that takes a list as an argument and returns a new list of elements that appear more than once from the original list. I tried writing it this way but it doesn't give me the answer
def list(n):
l = [ ]
for s in n:
if n.count(s) > 1:
l.append(s)
return l
return None
You're close. You need to remove the return statement from the for loop. As it is, you return unconditionally after the first element.
def list(n):
l = []
for s in n:
if n.count(s) > 1:
l.append(s)
return l
Second, I highly recommend that you don't use list as the name of this function as then you shadow the builtin list function which is very useful.
L as variable name is not the best for readability.
You can use filter() function to do that.
This is not the most fast for CPU but laconic and pythonic enough.
my_list = [1, 2, 2, 3, 4, 4, 5, 5, 6, 7]
print filter(lambda x: my_list.count(x) > 1, my_list)
Also you can use a list comprehension as 6502 mentioned:
my_list = [1, 2, 2, 3, 4, 4, 5, 5, 6, 7]
print [x for x in my_list if my_list.count(x) > 1]
set(my_list) instead of my_list as last argument of filter?
set() will return you a collection of unique values.filter+lambda can be substituted by shorter and more pythonic list comprehensions: [x for x in set(L) if L.count(x) > 1]set() constructor will iterate through the list to create a set so we'll get one more hidden loop?count?def list_duplicates(seq):
seen = set()
seen_add = seen.add
#adds all elements it doesn't know yet to seen and all other to seen_twice
seen_twice = set( x for x in seq if x in seen or seen_add(x) )
# turn the set into a list (as requested)
return list( seen_twice )
a = [1,2,3,2,1,5,6,5,5,5]
list_duplicates(a)
This will print
[1,2,5]
I think solution of Eugene Naydenov is good, but I some enhance it:
In [1]: my_list = [1, 2, 2, 3, 4, 4, 5, 5, 6, 7];
In [2]: set_list = lambda(lst) : list(set(i for i in filter(lambda x: my_list.count(x) > 1, lst))) #set_list is function
In [3]: set_list(my_list)
Out[3]: [2, 4, 5]
