Compare Two List, And Create Other Two Lists With Intersection And Difference
I have 2 lists A and B. In the B list I can have multiple elements from list A. For example: A = [1,3,5,7, 9, 12, 14] B = [1,2,3,3,7,9,7,3,14,14,1,3,2,5,5] I want to create: to
Solution 1:
#to create a list with ids that are in A and found in B (unique)
resultlist=list(set(A)&set(B))
print(list(set(A)&set(B)))
#to create a list of ids that are in A and have no corresponding in B (unique)print(list(set(A)-set(B)))
#the numbers in B, that don't have a corespondent in Aprint(list(set(B)-set(A)))
Solution 2:
Convert the list to set
and then perform set operations
>>>set_A = set(A)>>>set_B = set(B)>>>list(set_A & set_B)
[1, 3, 5, 7, 9, 14] # set intersection
>>>list(set_A - set_B) # set difference
[12]
>>>list(set_B - set_A)
[2]
Solution 3:
With python you can simply use the set type:
list(set(A) & set(B))
will return a list containing the element intersection between lists A
and B
.
list(set(A) - set(B))
Will return a list containing all the elements that are in A
and not in B
.
Vice versa:
list(set(B) - set(A))
Will return a list containing all the elements that are in B
and not in A
.
Solution 4:
you could use the 'a in L' functionality, which will return True if an element is in a List. e.g.
A = [1,3,5,7, 9, 12, 14]
B = [1,2,3,3,7,9,7,3,14,14,1,3,2,5,5]
common = []
uncommon = []
for a in A:
if a in B:
common.append(a)
else:
uncommon.append(a)
print(common)
print(uncommon)
this should give you a good hint on how to approach the other question. best
Solution 5:
Use set operations:
A = [1, 3, 5, 7, 9, 12, 14]
B = [1, 2, 3, 3, 7, 9, 7, 3, 14, 14, 1, 3, 2, 5, 5]
sa = set(A)
sb = set(B)
# intersectionl1 = list(sa & sb)
# [1, 2, 3, 5, 7, 9, 12, 14]# differencesl2 = list(sa - sb)
# [12]l3 = list(sb - sa)
# [2]
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