Python – Paired Existence in Records

Sometimes, while working with Python records, we can have a problem in which we need to check for paired existence inside the record, or else if one doesn’t exist, other also should not. This kind of problem is common in domains such as Data Science and web development. Let’s discuss certain ways in which this task can be performed.

Input : 
test_list = [(‘Gfg’, ‘is’, ‘Best’), (‘Gfg’, ‘is’, ‘good’), (‘CS’, ‘is’, ‘good’)] 
pairs = (‘is’, ‘good’) 
Output : [(‘Gfg’, ‘is’, ‘good’), (‘CS’, ‘is’, ‘good’)]
Input : 
test_list = [(‘Gfg’, ‘is’, ‘Best’), (‘Gfg’, ‘is’, ‘good’), (‘CS’, ‘is’, ‘better’)] 
pairs = (‘better’, ‘good’) 
Output : [] 

Method #1 : Using generator expression 
This is brute force way in which this task can be performed. In this, we check for the existence/non-existence of both numbers and accept the result if, none or both are present.

Python3

# Python3 code to demonstrate working of
# Paired Existence in Records
# Using generator expression

# initializing list
test_list = [('Gfg', 'is', 'Best'),
             ('Gfg', 'is', 'good'),
             ('CS', 'is', 'good')]

# printing original list
print("The original list is : " + str(test_list))

# initializing Pairs
pairs = ('Gfg', 'Best')

# Paired Existence in Records
# Using generator expression
res = []
for sub in test_list:
    if ((pairs[0] in sub and pairs[1] in sub) or (
         pairs[0] not in sub and pairs[1] not in sub)):
        res.append(sub)

# printing result
print("The resultant records : " + str(res))

Output : 

The original list is : [(‘Gfg’, ‘is’, ‘Best’), (‘Gfg’, ‘is’, ‘good’), (‘CS’, ‘is’, ‘good’)] 
The resultant records : [(‘Gfg’, ‘is’, ‘Best’), (‘CS’, ‘is’, ‘good’)] 

Method #2 : Using XNOR
This is yet another way to solve this problem. In this, use the power of XOR operator to perform this task and negate the result.

Python3

# Python3 code to demonstrate working of
# Paired Existence in Records
# Using XNOR

# initializing list
test_list = [('Gfg', 'is', 'Best'),
             ('Gfg', 'is', 'good'),
             ('CS', 'is', 'good')]

# printing original list
print("The original list is : " + str(test_list))

# initializing Pairs
pairs = ('Gfg', 'Best')

# Paired Existence in Records
# Using XNOR
res = []
for sub in test_list:
    if (not ((pairs[0] in sub) ^ (pairs[1] in sub))):
        res.append(sub)

# printing result
print("The resultant records : " + str(res))

Output : 

The original list is : [('Gfg', 'is', 'Best'), ('Gfg', 'is', 'good'), ('CS', 'is', 'good')]
The resultant records : [('Gfg', 'is', 'Best'), ('CS', 'is', 'good')]