intersection of two lists python
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> list(set(a) & set(b))
[1, 3, 5]intersection of two lists python
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> list(set(a) & set(b))
[1, 3, 5]intersection in list
def intersection(lst1, lst2):
lst3 = [value for value in lst1 if value in lst2]
return lst3
# Driver Code
lst1 = [4, 9, 1, 17, 11, 26, 28, 54, 69]
lst2 = [9, 9, 74, 21, 45, 11, 63, 28, 26]
print(intersection(lst1, lst2))intersection of lists in python
import numpy as np
recent_coding_books = np.intersect1d(recent_books,coding_books)intersect in python list
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> list(set(a) & set(b))
[1, 3, 5]list intersection python
#get name1 and name2 as a list input
name1 =list("".join(str(x)for x in input("Enter name1").replace(" ","")))
name2 =list("".join(str(x)for x in input("Enter name2").replace(" ","")))
#check using list comprehension if x in name1 is in name2
#this will return multiple instances of the same character from name1 that matches with the name2
common = [x for x in name1 if x in name2]
#create a set out of the output so as to have only unique values of the repeated characters
unique = set(common)
#thus the above set will have common unrepeated characters from both names
#create a variable and initialize it to zero
d=0
#run a loop that checks the minimum occurrence of
#the character from the set in name1 & name2
#Minimum because for ex: a might exist thrice in name1, but only twice in name2
#we will need to take only 2 common occurrences from the name2
#thus finding the minimum occurrence of the character from both names
for x in unique:
d = d + min(name1.count(x),name2.count(x))
#multiplying by two, because if one character from name 1 matches with one,
#character from name2, then it makes two in total
difference = (len(name1) + len(name2)) - d*2
print(difference)Copyright © 2021 Codeinu
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