- >>> A = (1,2,3)

- >>> B = (2,3,4,5)

- >>> [i for i in A if i in B]

- [2, 3]

- >>> [i for i in A if i not in B]

- [1]

Posted by Alger at February 18, 2016 - 6:40 AM

- set(A).intersection(B)

Posted by Ulysses at February 26, 2016 - 6:49 AM

Thank you

#Method:

a=[2,3,4,5]

b=[2,5,8]

tmp = [val for val in a if val in b]

print tmp

#[2, 5]

#Methods two

print list(set(a).intersection(set(b)))

Union of 2 to get two list

print list(set(a).union(set(b)))

3 to get two list difference sets

print list(set(b).difference(set(a))) # B are not found in the a.

#Method:

a=[2,3,4,5]

b=[2,5,8]

tmp = [val for val in a if val in b]

print tmp

#[2, 5]

#Methods two

print list(set(a).intersection(set(b)))

Union of 2 to get two list

print list(set(a).union(set(b)))

3 to get two list difference sets

print list(set(b).difference(set(a))) # B are not found in the a.

Posted by Lori at February 28, 2016 - 7:01 AM

All methods of set second parameter need not be set, Iterable as long as you can

Posted by Ulysses at March 06, 2016 - 7:27 AM

Thank you for your reply

Posted by Lori at March 10, 2016 - 8:16 AM

A = (1,2,3)

B = (2,3,4,5)

With what method can easily find out the A that exists in the B, or for those who don't exist in B

A in B

set(A) & set(B)

A is not in B

set(A) - set(B)

B = (2,3,4,5)

With what method can easily find out the A that exists in the B, or for those who don't exist in B

A in B

set(A) & set(B)

A is not in B

set(A) - set(B)

Posted by Rudolf at March 20, 2016 - 8:22 AM

A=(1,2,3)

N=[]

B=[2,3,4,5]

for i in A:

for j in B:

if i==j:

N.append(i)

B.remove(i)

break

print N,'common in A and B'

print B, 'only exist in B'

N=[]

B=[2,3,4,5]

for i in A:

for j in B:

if i==j:

N.append(i)

B.remove(i)

break

print N,'common in A and B'

print B, 'only exist in B'

Posted by Asa at March 23, 2016 - 8:56 AM

B = (2,3,4,5)

With what method can easily find out the A that exists in the B, or for those who don't exist in B

Thank you

Started by Lori at February 12, 2016 - 6:39 AM