The Proposed Solution in PEP202 Python
>>> print [i for i in range(10)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> print [i for i in range(20) if i%2 == 0]
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18]
>>> nums = [1, 2, 3, 4]
>>> fruit = ["Apples", "Peaches", "Pears", "Bananas"]
>>> print [(i, f) for i in nums for f in fruit]
[(1, 'Apples'), (1, 'Peaches'), (1, 'Pears'), (1, 'Bananas'),
(2, 'Apples'), (2, 'Peaches'), (2, 'Pears'), (2, 'Bananas'),
(3, 'Apples'), (3, 'Peaches'), (3, 'Pears'), (3, 'Bananas'),
(4, 'Apples'), (4, 'Peaches'), (4, 'Pears'), (4, 'Bananas')]
>>> print [(i, f) for i in nums for f in fruit if f[0] == "P"]
[(1, 'Peaches'), (1, 'Pears'),
(2, 'Peaches'), (2, 'Pears'),
(3, 'Peaches'), (3, 'Pears'),
(4, 'Peaches'), (4, 'Pears')]
>>> print [(i, f) for i in nums for f in fruit if f[0] == "P" if i%2 == 1]
[(1, 'Peaches'), (1, 'Pears'), (3, 'Peaches'), (3, 'Pears')]
>>> print [i for i in zip(nums, fruit) if i[0]%2==0]
[(2, 'Peaches'), (4, 'Bananas')]