How to find the difference between 3 lists that may have duplicate numbers

Question

I have 3 lists and I want to find the difference between 1st / 2nd and 2nd / 3rd and print them.

Here is my code:

n1 = [1,1,1,2,3] n2 = [1,1,1,2] # Here the 3 is not found ("3" is not present in n1 at all) n3 = [1,1,2] # here 1 is not found (there are fewer "1"s in n3 than in n2) for x in n1: if x not in n2: print x for m in n2: if m not in n3: print m 

but I get only 3 as output.

How to conclude 1 and 3? I tried using set , but it printed 3 again.

Answer 1

Since you seem to care about how many times the item was found in both lists, you need to either remove matching items from the list that you are comparing:

 comp = n2[:] # make a copy for x in n1: if x not in comp: print x else: comp.remove(x) # output: 3 

or use collections.Counter

 from collections import Counter print Counter(n1) - Counter(n2) # output: Counter({3: 1}) 

which tells you which elements in n1 not in n2 , or they can be found more often in n1 than in n2 .

So for example:

 >>> Counter([1,2,2,2,3]) - Counter([1,2]) Counter({2: 2, 3: 1}) 

Answer 2

you can use set to find the difference between the list.

 print set(n1).difference(set(n2))