Opposite of set.intersection in python?
PythonSetPython Problem Overview
In Python you can use a.intersection(b) to find the items common to both sets.
Is there a way to do the disjoint opposite version of this? Items that are not common to both a and b; the unique items in a unioned with the unique items in b?
Python Solutions
Solution 1 - Python
You are looking for the symmetric difference; all elements that appear only in set a or in set b, but not both:
a.symmetric_difference(b)
From the set.symmetric_difference() method documentation:
> Return a new set with elements in either the set or other but not both.
You can use the ^ operator too, if both a and b are sets:
a ^ b
while set.symmetric_difference() takes any iterable for the other argument.
The output is the equivalent of (a | b) - (a & b), the union of both sets minus the intersection of both sets.
Solution 2 - Python
a={1,2,4,5,6}
b={5,6,4,9}
c=(a^b)&b
print(c) # you got {9}
Solution 3 - Python
The best way is a list comprehension.
a = [ 1,2,3,4]
b = [ 8,7,9,2,1]
c = [ element for element in a if element not in b]
d = [ element for element in b if element not in a]
print(c)
# output is [ 3,4]
print(d)
# output is [8,7,9]
You can join both lists
Solution 4 - Python
Try this code for (set(a) - intersection(a&b))
a = [1,2,3,4,5,6]
b = [2,3]
for i in b:
if i in a:
a.remove(i)
print(a)
the output is [1,4,5,6]
I hope, it will work
Solution 5 - Python
e, f are two list you want to check disjoint
a = [1,2,3,4]
b = [8,7,9,2,1]
c = []
def loop_to_check(e,f):
for i in range(len(e)):
if e[i] not in f:
c.append(e[i])
loop_to_check(a,b)
loop_to_check(b,a)
print(c)
## output is [3,4,8,7,9]
This loops around to list and returns the disjoint list