I want count the same elements of two lists. Lists can have duplicate elements, so I can't convert this to sets and use & operator.
a=[2,2,1,1]
b=[1,1,3,3]
set(a) & set(b) work
a & b don't work
It is possible to do it withoud set and dictonary?
In Python 3.x (and Python 2.7, when it's released), you can use collections.Counter for this:
>>> from collections import Counter
>>> list((Counter([2,2,1,1]) & Counter([1,3,3,1])).elements())
[1, 1]
Here's an alternative using collections.defaultdict (available in Python 2.5 and later). It has the nice property that the order of the result is deterministic (it essentially corresponds to the order of the second list).
from collections import defaultdict
def list_intersection(list1, list2):
bag = defaultdict(int)
for elt in list1:
bag[elt] += 1
result = []
for elt in list2:
if elt in bag:
# remove elt from bag, making sure
# that bag counts are kept positive
if bag[elt] == 1:
del bag[elt]
else:
bag[elt] -= 1
result.append(elt)
return result
For both these solutions, the number of occurrences of any given element x in the output list is the minimum of the numbers of occurrences of x in the two input lists. It's not clear from your question whether this is the behavior that you want.
if bag[elt]: adds an item elt: 0 to the bag when elt in bag is false. When (as the OP hinted) there are few duplicates, it may be better to do if elt in bag and bag[elt]: otherwise bag may become bloated with useless 0 values.
bag[elt] -= 1 with an if not bag[elt]: del bag[elt]. Then the if test could just be if elt in bag:, which reads more nicely. I'll edit the answer.Using sets is the most efficient, but you could always do r = [i for i in l1 if i in l2].
[1, 1] if l1 == [1, 1] and l2 == [1], but it gives [1] if l1 == [1] and l2 == [1, 1]. Is that the behavior that you want?SilentGhost, Mark Dickinson and Lo'oris are right, Thanks very much for report this problem - I need common part of lists, so for:
a=[1,1,1,2]
b=[1,1,3,3]
result should be [1,1]
Sorry for comment in not suitable place - I have registered today.
I modified yours solutions:
def count_common(l1,l2):
l2_copy=list(l2)
counter=0
for i in l1:
if i in l2_copy:
counter+=1
l2_copy.remove(i)
return counter
l1=[1,1,1]
l2=[1,2]
print count_common(l1,l2)
1
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[1, 2, 1]and[1, 3, 2]?[1, 2, 2]and[1, 1, 2]?[1, 2]in both cases