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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

This is what the function is doing: it gets two arrays of tuples with 5 items each - hand = [H1,H2,H3,H4,H5] and board = [B1,B2,B3,B4,B5]

What I need to do is check all arrays formed by 2 items from hand and 3 items from board, like combination = [Hn,Hm,Bi,Bj,Bk] (100 combinations in total)

Then I need to compare each one of the combinations against a dictionary to get the combination rank, and then return the best array (best rank) and the rank itself:

def check_hand(hand, board, dictionary_A, dictionary_B):



 best_hand = []

 first_int = True



 for h1 in range (0, 4):

  for h2 in range (h1+1, 5):

   for b1 in range (0, 3):

    for b2 in range (b1+1, 4):

     for b3 in range (b2+1, 5):

      hand_check = []

      hand_check.append(hand[m1])

      hand_check.append(hand[m2])

      hand_check.append(board[b1])

      hand_check.append(board[b2])

      hand_check.append(board[b3])

      hand_check = sort(hand_check) #Custom sort for my array of objects

      hand_ranks = "".join([str(hand_check[0].rank),str(hand_check[1].rank),str(hand_check[2].rank),str(hand_check[3].rank),str(hand_check[4].rank)])



      if (hand_check[0].suit == hand_check[1].suit and hand_check[1].suit == hand_check[2].suit and hand_check[2].suit == hand_check[3].suit and hand_check[3].suit == hand_check[4].suit):

       control = [dictionary_A[hand_ranks][0],dictionary_A[hand_ranks][1]]

      else:

       control = [dictionary_B[hand_ranks][0],dictionary_B[hand_ranks][1]]



      if first_int:

       best_hand = hand_check

       rank = control

       first_int = False

      elif (int(control[0]) > int(rank[0])):

       rank = control

       best_hand = hand_check       

      elif (int(control[0]) == int(rank[0])):

       if (int(control[1]) > int(rank[1])):       

        rank = control

        best_hand = hand_check      



 return best_hand, rank[0]

I need to run this check for 2 million different hands and interact over 1000 times for every hand (Ideally I would run it for at least 100000 times for every hand, for a more statistically accurate result).
Any ideas on how to make it more efficient?

Examples:
I added some "prints" on the way for better understanding:

Hand: Q♥K♥Q♠K♠3♦︎
Board: 2♣2♥J♦︎6♥4♦︎

2♣2♥J♦︎Q♥K♥ - 2♣2♥6♥Q♥K♥ - 2♣2♥4♦︎Q♥K♥ - 2♣6♥J♦︎Q♥K♥ - 2♣4♦︎J♦︎Q♥K♥ -
2♣4♦︎6♥Q♥K♥ - 2♥6♥J♦︎Q♥K♥ - 2♥4♦︎J♦︎Q♥K♥ - 2♥4♦︎6♥Q♥K♥ - 4♦︎6♥J♦︎Q♥K♥ -
2♣2♥J♦︎Q♥Q♠ - 2♣2♥6♥Q♥Q♠ - 2♣2♥4♦︎Q♥Q♠ - 2♣6♥J♦︎Q♥Q♠ - 2♣4♦︎J♦︎Q♥Q♠ -
2♣4♦︎6♥Q♥Q♠ - 2♥6♥J♦︎Q♥Q♠ - 2♥4♦︎J♦︎Q♥Q♠ - 2♥4♦︎6♥Q♥Q♠ - 4♦︎6♥J♦︎Q♥Q♠ -
2♣2♥J♦︎Q♥K♠ - 2♣2♥6♥Q♥K♠ - 2♣2♥4♦︎Q♥K♠ - 2♣6♥J♦︎Q♥K♠ - 2♣4♦︎J♦︎Q♥K♠ -
2♣4♦︎6♥Q♥K♠ - 2♥6♥J♦︎Q♥K♠ - 2♥4♦︎J♦︎Q♥K♠ - 2♥4♦︎6♥Q♥K♠ - 4♦︎6♥J♦︎Q♥K♠ -
2♣2♥3♦︎J♦︎Q♥ - 2♣2♥3♦︎6♥Q♥ - 2♣2♥3♦︎4♦︎Q♥ - 2♣3♦︎6♥J♦︎Q♥ - 2♣3♦︎4♦︎J♦︎Q♥ -
2♣3♦︎4♦︎6♥Q♥ - 2♥3♦︎6♥J♦︎Q♥ - 2♥3♦︎4♦︎J♦︎Q♥ - 2♥3♦︎4♦︎6♥Q♥ - 3♦︎4♦︎6♥J♦︎Q♥ -
2♣2♥J♦︎Q♠K♥ - 2♣2♥6♥Q♠K♥ - 2♣2♥4♦︎Q♠K♥ - 2♣6♥J♦︎Q♠K♥ - 2♣4♦︎J♦︎Q♠K♥ -
2♣4♦︎6♥Q♠K♥ - 2♥6♥J♦︎Q♠K♥ - 2♥4♦︎J♦︎Q♠K♥ - 2♥4♦︎6♥Q♠K♥ - 4♦︎6♥J♦︎Q♠K♥ -
2♣2♥J♦︎K♥K♠ - 2♣2♥6♥K♥K♠ - 2♣2♥4♦︎K♥K♠ - 2♣6♥J♦︎K♥K♠ - 2♣4♦︎J♦︎K♥K♠ -
2♣4♦︎6♥K♥K♠ - 2♥6♥J♦︎K♥K♠ - 2♥4♦︎J♦︎K♥K♠ - 2♥4♦︎6♥K♥K♠ - 4♦︎6♥J♦︎K♥K♠ -
2♣2♥3♦︎J♦︎K♥ - 2♣2♥3♦︎6♥K♥ - 2♣2♥3♦︎4♦︎K♥ - 2♣3♦︎6♥J♦︎K♥ - 2♣3♦︎4♦︎J♦︎K♥ -
2♣3♦︎4♦︎6♥K♥ - 2♥3♦︎6♥J♦︎K♥ - 2♥3♦︎4♦︎J♦︎K♥ - 2♥3♦︎4♦︎6♥K♥ - 3♦︎4♦︎6♥J♦︎K♥ -
2♣2♥J♦︎Q♠K♠ - 2♣2♥6♥Q♠K♠ - 2♣2♥4♦︎Q♠K♠ - 2♣6♥J♦︎Q♠K♠ - 2♣4♦︎J♦︎Q♠K♠ -
2♣4♦︎6♥Q♠K♠ - 2♥6♥J♦︎Q♠K♠ - 2♥4♦︎J♦︎Q♠K♠ - 2♥4♦︎6♥Q♠K♠ - 4♦︎6♥J♦︎Q♠K♠ -
2♣2♥3♦︎J♦︎Q♠ - 2♣2♥3♦︎6♥Q♠ - 2♣2♥3♦︎4♦︎Q♠ - 2♣3♦︎6♥J♦︎Q♠ - 2♣3♦︎4♦︎J♦︎Q♠ -
2♣3♦︎4♦︎6♥Q♠ - 2♥3♦︎6♥J♦︎Q♠ - 2♥3♦︎4♦︎J♦︎Q♠ - 2♥3♦︎4♦︎6♥Q♠ - 3♦︎4♦︎6♥J♦︎Q♠ -
2♣2♥3♦︎J♦︎K♠ - 2♣2♥3♦︎6♥K♠ - 2♣2♥3♦︎4♦︎K♠ - 2♣3♦︎6♥J♦︎K♠ - 2♣3♦︎4♦︎J♦︎K♠ -
2♣3♦︎4♦︎6♥K♠ - 2♥3♦︎6♥J♦︎K♠ - 2♥3♦︎4♦︎J♦︎K♠ - 2♥3♦︎4♦︎6♥K♠ - 3♦︎4♦︎6♥J♦︎K♠ -

Best Hand: 2♣2♥J♦︎K♥K♠ Rank: 3

Just wondering if there is a way to run the piece of code below more efficiently. I just started to get acquainted with parallel program and think this might be an answer? But I have no idea how to do it using imap or processes.

edited 30 secs ago

200_success

131k17157422

asked 2 hours ago

Vandré Dreer Bonaite

New contributor

1

$begingroup$
Welcome to Code Review. Is this real working Python code? (Introducing a comment with // is not syntactically valid.) What's mao? Also, please provide some example inputs and outputs so that we can better understand what this code accomplishes.
$endgroup$
– 200_success
2 hours ago

$begingroup$
The comment is not in the original code. It is a working code. What I am trying to accomplish is to find a better way to interact 100 times between hand and board. I updated the original post with examples.
$endgroup$
– Vandré Dreer Bonaite
1 hour ago

$begingroup$
In your example, 2s5h3sTc looks like an incomplete combination. Also, you still haven't explained what mao[m1] is.
$endgroup$
– 200_success
1 hour ago

$begingroup$
Yes, it was incomplete. And mao = hand (also updated that).
$endgroup$
– Vandré Dreer Bonaite
27 mins ago

add a comment |

This is what the function is doing: it gets two arrays of tuples with 5 items each - hand = [H1,H2,H3,H4,H5] and board = [B1,B2,B3,B4,B5]

What I need to do is check all arrays formed by 2 items from hand and 3 items from board, like combination = [Hn,Hm,Bi,Bj,Bk] (100 combinations in total)

Then I need to compare each one of the combinations against a dictionary to get the combination rank, and then return the best array (best rank) and the rank itself:

def check_hand(hand, board, dictionary_A, dictionary_B):



 best_hand = []

 first_int = True



 for h1 in range (0, 4):

  for h2 in range (h1+1, 5):

   for b1 in range (0, 3):

    for b2 in range (b1+1, 4):

     for b3 in range (b2+1, 5):

      hand_check = []

      hand_check.append(hand[m1])

      hand_check.append(hand[m2])

      hand_check.append(board[b1])

      hand_check.append(board[b2])

      hand_check.append(board[b3])

      hand_check = sort(hand_check) #Custom sort for my array of objects

      hand_ranks = "".join([str(hand_check[0].rank),str(hand_check[1].rank),str(hand_check[2].rank),str(hand_check[3].rank),str(hand_check[4].rank)])



      if (hand_check[0].suit == hand_check[1].suit and hand_check[1].suit == hand_check[2].suit and hand_check[2].suit == hand_check[3].suit and hand_check[3].suit == hand_check[4].suit):

       control = [dictionary_A[hand_ranks][0],dictionary_A[hand_ranks][1]]

      else:

       control = [dictionary_B[hand_ranks][0],dictionary_B[hand_ranks][1]]



      if first_int:

       best_hand = hand_check

       rank = control

       first_int = False

      elif (int(control[0]) > int(rank[0])):

       rank = control

       best_hand = hand_check       

      elif (int(control[0]) == int(rank[0])):

       if (int(control[1]) > int(rank[1])):       

        rank = control

        best_hand = hand_check      



 return best_hand, rank[0]

Examples:
I added some "prints" on the way for better understanding:

Hand: Q♥K♥Q♠K♠3♦︎
Board: 2♣2♥J♦︎6♥4♦︎

Best Hand: 2♣2♥J♦︎K♥K♠ Rank: 3

edited 30 secs ago

200_success

131k17157422

asked 2 hours ago

Vandré Dreer Bonaite

New contributor

1

$begingroup$
Welcome to Code Review. Is this real working Python code? (Introducing a comment with // is not syntactically valid.) What's mao? Also, please provide some example inputs and outputs so that we can better understand what this code accomplishes.
$endgroup$
– 200_success
2 hours ago

$begingroup$
The comment is not in the original code. It is a working code. What I am trying to accomplish is to find a better way to interact 100 times between hand and board. I updated the original post with examples.
$endgroup$
– Vandré Dreer Bonaite
1 hour ago

$begingroup$
In your example, 2s5h3sTc looks like an incomplete combination. Also, you still haven't explained what mao[m1] is.
$endgroup$
– 200_success
1 hour ago

$begingroup$
Yes, it was incomplete. And mao = hand (also updated that).
$endgroup$
– Vandré Dreer Bonaite
27 mins ago

add a comment |

This is what the function is doing: it gets two arrays of tuples with 5 items each - hand = [H1,H2,H3,H4,H5] and board = [B1,B2,B3,B4,B5]

What I need to do is check all arrays formed by 2 items from hand and 3 items from board, like combination = [Hn,Hm,Bi,Bj,Bk] (100 combinations in total)

Then I need to compare each one of the combinations against a dictionary to get the combination rank, and then return the best array (best rank) and the rank itself:

def check_hand(hand, board, dictionary_A, dictionary_B):



 best_hand = []

 first_int = True



 for h1 in range (0, 4):

  for h2 in range (h1+1, 5):

   for b1 in range (0, 3):

    for b2 in range (b1+1, 4):

     for b3 in range (b2+1, 5):

      hand_check = []

      hand_check.append(hand[m1])

      hand_check.append(hand[m2])

      hand_check.append(board[b1])

      hand_check.append(board[b2])

      hand_check.append(board[b3])

      hand_check = sort(hand_check) #Custom sort for my array of objects

      hand_ranks = "".join([str(hand_check[0].rank),str(hand_check[1].rank),str(hand_check[2].rank),str(hand_check[3].rank),str(hand_check[4].rank)])



      if (hand_check[0].suit == hand_check[1].suit and hand_check[1].suit == hand_check[2].suit and hand_check[2].suit == hand_check[3].suit and hand_check[3].suit == hand_check[4].suit):

       control = [dictionary_A[hand_ranks][0],dictionary_A[hand_ranks][1]]

      else:

       control = [dictionary_B[hand_ranks][0],dictionary_B[hand_ranks][1]]



      if first_int:

       best_hand = hand_check

       rank = control

       first_int = False

      elif (int(control[0]) > int(rank[0])):

       rank = control

       best_hand = hand_check       

      elif (int(control[0]) == int(rank[0])):

       if (int(control[1]) > int(rank[1])):       

        rank = control

        best_hand = hand_check      



 return best_hand, rank[0]

Examples:
I added some "prints" on the way for better understanding:

Hand: Q♥K♥Q♠K♠3♦︎
Board: 2♣2♥J♦︎6♥4♦︎

Best Hand: 2♣2♥J♦︎K♥K♠ Rank: 3

edited 30 secs ago

200_success

131k17157422

asked 2 hours ago

Vandré Dreer Bonaite

New contributor

This is what the function is doing: it gets two arrays of tuples with 5 items each - hand = [H1,H2,H3,H4,H5] and board = [B1,B2,B3,B4,B5]

What I need to do is check all arrays formed by 2 items from hand and 3 items from board, like combination = [Hn,Hm,Bi,Bj,Bk] (100 combinations in total)

Then I need to compare each one of the combinations against a dictionary to get the combination rank, and then return the best array (best rank) and the rank itself:

def check_hand(hand, board, dictionary_A, dictionary_B):



 best_hand = []

 first_int = True



 for h1 in range (0, 4):

  for h2 in range (h1+1, 5):

   for b1 in range (0, 3):

    for b2 in range (b1+1, 4):

     for b3 in range (b2+1, 5):

      hand_check = []

      hand_check.append(hand[m1])

      hand_check.append(hand[m2])

      hand_check.append(board[b1])

      hand_check.append(board[b2])

      hand_check.append(board[b3])

      hand_check = sort(hand_check) #Custom sort for my array of objects

      hand_ranks = "".join([str(hand_check[0].rank),str(hand_check[1].rank),str(hand_check[2].rank),str(hand_check[3].rank),str(hand_check[4].rank)])



      if (hand_check[0].suit == hand_check[1].suit and hand_check[1].suit == hand_check[2].suit and hand_check[2].suit == hand_check[3].suit and hand_check[3].suit == hand_check[4].suit):

       control = [dictionary_A[hand_ranks][0],dictionary_A[hand_ranks][1]]

      else:

       control = [dictionary_B[hand_ranks][0],dictionary_B[hand_ranks][1]]



      if first_int:

       best_hand = hand_check

       rank = control

       first_int = False

      elif (int(control[0]) > int(rank[0])):

       rank = control

       best_hand = hand_check       

      elif (int(control[0]) == int(rank[0])):

       if (int(control[1]) > int(rank[1])):       

        rank = control

        best_hand = hand_check      



 return best_hand, rank[0]

Examples:
I added some "prints" on the way for better understanding:

Hand: Q♥K♥Q♠K♠3♦︎
Board: 2♣2♥J♦︎6♥4♦︎

Best Hand: 2♣2♥J♦︎K♥K♠ Rank: 3

python performance playing-cards

edited 30 secs ago

200_success

131k17157422

asked 2 hours ago

Vandré Dreer Bonaite

New contributor

edited 30 secs ago

200_success

131k17157422

asked 2 hours ago

Vandré Dreer Bonaite

New contributor

edited 30 secs ago

200_success

131k17157422

edited 30 secs ago

200_success

131k17157422

edited 30 secs ago

200_success

131k17157422

asked 2 hours ago

Vandré Dreer Bonaite

New contributor

asked 2 hours ago

Vandré Dreer Bonaite

asked 2 hours ago

Vandré Dreer Bonaite

New contributor

Vandré Dreer Bonaite is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

1

$begingroup$
Welcome to Code Review. Is this real working Python code? (Introducing a comment with // is not syntactically valid.) What's mao? Also, please provide some example inputs and outputs so that we can better understand what this code accomplishes.
$endgroup$
– 200_success
2 hours ago

$begingroup$
The comment is not in the original code. It is a working code. What I am trying to accomplish is to find a better way to interact 100 times between hand and board. I updated the original post with examples.
$endgroup$
– Vandré Dreer Bonaite
1 hour ago

$begingroup$
In your example, 2s5h3sTc looks like an incomplete combination. Also, you still haven't explained what mao[m1] is.
$endgroup$
– 200_success
1 hour ago

$begingroup$
Yes, it was incomplete. And mao = hand (also updated that).
$endgroup$
– Vandré Dreer Bonaite
27 mins ago

add a comment |

1

$begingroup$
Welcome to Code Review. Is this real working Python code? (Introducing a comment with // is not syntactically valid.) What's mao? Also, please provide some example inputs and outputs so that we can better understand what this code accomplishes.
$endgroup$
– 200_success
2 hours ago

$begingroup$
The comment is not in the original code. It is a working code. What I am trying to accomplish is to find a better way to interact 100 times between hand and board. I updated the original post with examples.
$endgroup$
– Vandré Dreer Bonaite
1 hour ago

$begingroup$
In your example, 2s5h3sTc looks like an incomplete combination. Also, you still haven't explained what mao[m1] is.
$endgroup$
– 200_success
1 hour ago

$begingroup$
Yes, it was incomplete. And mao = hand (also updated that).
$endgroup$
– Vandré Dreer Bonaite
27 mins ago

Welcome to Code Review. Is this real working Python code? (Introducing a comment with // is not syntactically valid.) What's mao? Also, please provide some example inputs and outputs so that we can better understand what this code accomplishes.

– 200_success
2 hours ago

The comment is not in the original code. It is a working code. What I am trying to accomplish is to find a better way to interact 100 times between hand and board. I updated the original post with examples.

– Vandré Dreer Bonaite
1 hour ago

In your example, 2s5h3sTc looks like an incomplete combination. Also, you still haven't explained what mao[m1] is.

– 200_success
1 hour ago

Yes, it was incomplete. And mao = hand (also updated that).

– Vandré Dreer Bonaite
27 mins ago

add a comment |

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Vandré Dreer Bonaite is a new contributor. Be nice, and check out our Code of Conduct.

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Vandré Dreer Bonaite is a new contributor. Be nice, and check out our Code of Conduct.

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