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 returns a list that contains only the elements that are common between the lists (without duplicates). Make sure your program works on two lists of different sizes.
Here's my code:
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 = []
for i in a:
if i in b and i not in c:
c.append([i])
print(c)
My output is still giving me duplicates despite the 'i not in c' statement. why is this? I'm sure its blatantly obvious, I just cant see it!
i
to c
, so i not in c
will always return True
. You should append i
on its own: c.append(i)
Or
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)
.
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]
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)
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.
c = [] for items in a: for numbers in b: if items == numbers: if items in c: None else: c.append(items) print(c)
a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
result = []
[(result.append(i)) for i in a if i in b and i not in result]
print(result) # [1, 2, 3, 5, 8, 13]
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