Python, min among first objects in lists

Question

I have a very particular problem, I have a deck of cards. The deck is a list of lists of tuples, each internal list is a suit with card tuples (suit, value), sorted smallest to largest(2-Ace). I would like to find the smallest card in the deck. So basically I want to take the first object from each suit, and find the smallest. Short of an horridly ugly for loop, what is the most pythonic way to do this?

Answer 1

min takes a key function. You can use this to get the first element to use it as the comparison:

min(my_list, key=lambda x: x[0])

Answer 2

min(suit[0] for suit in deck, key=lambda suit,value: value)

Since you noted that the suits are sorted, you can simply get the first item of each suit list to get the smallest value of that suit; then compare those first values using min() with the key argument specified to look at the value portion of each.

Answer 3

Lets assume that the deck list is created as below:

TheHearts = TheSpades = TheDiamonds = TheClubs = [2,3,4,5,6,7,8,9,10,"j","q","k","a"]
TheDeck = [TheHearts, TheSpades, TheDiamonds, TheClubs]

the following line can provide you the first object of the first list:

print TheDeck[0][0]

or even if you make a dictionary, then you can use the names instead of numbers:

TheHearts = TheSpades = TheDiamonds = TheClubs = [2,3,4,5,6,7,8,9,10,"j","q","k","a"]
TheDeck = {'TheHearts':TheHearts, 'TheSpades':TheSpades, 'TheDiamonds':TheDiamonds, 'TheClubs':TheClubs}
print TheDeck['TheHearts'][0]