Common Elements Between Two Lists With No Duplicates

Problem is this, take two lists, say for example these two: a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89] b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13] And write a program that ret

Solution 1:

1. You are appending a list containing i to c, so i not in c will always return True. You should append i on its own: c.append(i)

Or

2. Simply use sets (if order is not important):

   a = [1, 1, 2, 2, 3, 5, 8, 13, 21, 34, 55, 89]
   b = [1, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
   c = set(a) & set(b)  #  & calculates the intersection.print(c)
   #  {1, 2, 3, 5, 8, 13}

EDIT As @Ev. Kounis suggested in the comment, you will gain some speed by using c = set(a).intersection(b).

Solution 2:

using list comprehensions, l think it had be short and simple if it was implemented as following

a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
[i for i in a and b if i in a and b]

result:

[1, 2, 3, 5, 8, 13]

Solution 3:

The below code would work:

newlist = []
for x in b:
    if x in a:
        if x in newlist:
            print("duplicate")
        else:
            newlist.append(x)

for y in newlist:
    print(y)   

Solution 4:

Using intuition of sets, You could do something like this...

filtered_arr = list(set(b)-set(a))

First you convert 2 arrays into sets, then take the substitute of it convert the result into the list again.

Solution 5:

c = []
for items in a:
    for numbers in b:
        if items == numbers:
            if items in c:
                Noneelse:
                c.append(items)
print(c)