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)
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