The current answers are good, but do not talk about how they are just syntactic sugar to some pattern that we are so used to.

Let's start with an example, say we have 10 numbers, and we want a subset of those that are greater than, say, 5.

>>> numbers = [12, 34, 1, 4, 4, 67, 37, 9, 0, 81]

For the above task, the below approaches below are totally identical to one another, and go from most verbose to concise, readable and pythonic:

Approach 1

result = []
for index in range(len(numbers)):
    if numbers[index] > 5:
        result.append(numbers[index])
print result  #Prints [12, 34, 67, 37, 9, 81]

Approach 2 (Slightly cleaner, for-in loops)

result = []
for number in numbers:
    if number > 5:
        result.append(number)
print result  #Prints [12, 34, 67, 37, 9, 81]

Approach 3 (Enter List Comprehension)

result = [number for number in numbers if number > 5]

or more generally:

[function(number) for number in numbers if condition(number)]

where:

• function(x) takes an x and transforms it into something useful (like for instance: x*x)
• if condition(x) returns any False-y value (False, None, empty string, empty list, etc ..) then the current iteration will be skipped (think continue). If the function return a non-False-y value then the current value makes it to the final resultant array (and goes through the transformation step above).

To understand the syntax in a slightly different manner, look at the Bonus section below.

For further information, follow the tutorial all other answers have linked: List Comprehension

Bonus

(Slightly un-pythonic, but putting it here for sake of completeness)

The example above can be written as:

result = filter(lambda x: x > 5, numbers)

The general expression above can be written as:

result = map(function, filter(condition, numbers)) #result is a list in Py2

If you want the effect of a nested for loop, use:

import itertools
for i, j in itertools.product(range(x), range(y)):
    # Stuff...

If you just want to loop simultaneously, use:

for i, j in zip(range(x), range(y)):
    # Stuff...

Note that if x and y are not the same length, zip will truncate to the shortest list. As @abarnert pointed out, if you don't want to truncate to the shortest list, you could use itertools.zip_longest.

UPDATE

Based on the request for "a function that will read lists "t1" and "t2" and return all elements that are identical", I don't think the OP wants zip or product. I think they want a set:

def equal_elements(t1, t2):
    return list(set(t1).intersection(set(t2)))
    # You could also do
    # return list(set(t1) & set(t2))

The intersection method of a set will return all the elements common to it and another set (Note that if your lists contains other lists, you might want to convert the inner lists to tuples first so that they are hashable; otherwise the call to set will fail.). The list function then turns the set back into a list.

UPDATE 2

OR, the OP might want elements that are identical in the same position in the lists. In this case, zip would be most appropriate, and the fact that it truncates to the shortest list is what you would want (since it is impossible for there to be the same element at index 9 when one of the lists is only 5 elements long). If that is what you want, go with this:

def equal_elements(t1, t2):
    return [x for x, y in zip(t1, t2) if x == y]

This will return a list containing only the elements that are the same and in the same position in the lists.

Try using this:

for k in range(1,c+1,2):