I want to have a deck of cards that looks like this:

['H1', 'H2', 'H3', 'H4', 'H5', 'H6'...]

For that I wrote the following code:

    import itertools
    kind = 'HCDS'
    value = ['1','2','3','4','5','6','7','8','9','10','J','Q','K','A']
    temp = list(itertools.product(kind,value))

    deck = []
    for i in range(0, len(temp)): 
        (a,b) = temp[i]

I was wondering if it could be done easier without the use of this temp variable or the for loop.

  • 2
    \$\begingroup\$ One (off-topic) suggestion - just to ease parsing and keystrokes - use "T" for 10... (eg. '9', 'T', 'Q', 'K', 'A' ]) \$\endgroup\$ Nov 26, 2016 at 20:40
  • \$\begingroup\$ Since this is code review: the proper way to do this would be to have a list of (immutable) objects with suit and value properties.. \$\endgroup\$ Nov 27, 2016 at 2:15
  • 1
    \$\begingroup\$ @BlueRaja-DannyPflughoeft you may want to add that as an answer, so it's not lost in the comments for the next person who sees this question. I had the same thought when I was initially looking at this question. \$\endgroup\$ Nov 27, 2016 at 5:00

2 Answers 2


You could just use str.join in a list comprehension and create deck directly:

import itertools 
kind = 'HCDS' 
value = ['2','3','4','5','6','7','8','9','10','J','Q','K','A']
deck = ["".join(card) for card in itertools.product(kind,value)]

Note that a regular French card set starts at 2, there is no 1.


You can unpack the tuple directly in the list comprehension and use + to join the cards.

In [12]: kind = 'HCDS'
In [13]: value = ['1','2','3','4','5','6','7','8','9','10','J','Q','K','A']
In [14]: deck = [k + v for k, v in itertools.product(kind, value)]
In [15]: deck[:4], deck[-4:]
Out[15]: (['H1', 'H2', 'H3', 'H4'], ['SJ', 'SQ', 'SK', 'SA'])

Also, instead of using itertools.product, you could use a double-for-loop list comprehension:

In [16]: deck = [k + v for k in kind for v in value]
In [17]: deck[:4], deck[-4:]
Out[17]: (['H1', 'H2', 'H3', 'H4'], ['SJ', 'SQ', 'SK', 'SA'])

Not that it matters much in this example, but for just joining two strings, + is quite a bit faster than join, and the double-for-loop also seems to be a bit faster then itertools.product:

In [21]: %timeit ["".join(card) for card in itertools.product(kind,value)]
10000 loops, best of 3: 80.4 us per loop
In [22]: %timeit [k + v for k, v in itertools.product(kind, value)]
10000 loops, best of 3: 42.1 us per loop
In [23]: %timeit [k + v for k in kind for v in value]
10000 loops, best of 3: 35.5 us per loop

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