CODEWITHSUNDEEP
pythonloopsnested-loops
Ruud14
Ruud14

Reputation: 127

How can I do this, but with a changeable number of dice?

How can I do this, but with a changeable number of dice? Which means that all_dices can have any length.

 for first_dice_number in all_dices[0]:
    for second_dice_number in all_dices[1]:
        for thrird_dice_number in all_dices[2]:
            result = first_dice_number+second_dice_number+thrird_dice_number
            results.append(result)

For clarification, all_dices looks like: 

[[1,2,3,4,5,6],[1,2,3,4,5,6],[1,2,3,4,5,6]]

This most likely is a noob question but I just can't figure it out.

The problem is that I don't want to put a 100 for loops inside each other.

Upvotes: 2

Views: 63

Answers (3)

John Coleman
John Coleman

Reputation: 52008

The itertools solution is probably the best in Python, but another way to handle dice is to think about enumerating possible rolls of n dice as counting from 0 to 6^n -1 in base 6 (but using "digits" 1 to 6 rather than 0 to 5):

def decode(dice_code, num_dice):
    dice = []
    for _ in range(num_dice):
        dice_code,r = divmod(dice_code,6)
        dice.append(1+r)
    return dice

For example, decode(0,5) is [1,1,1,1,1] and decode(7775,5) is [6,6,6,6,6]. Just loop over range(6**n), decoding as needed.

Upvotes: 0

Sam Washington
Sam Washington

Reputation: 670

Recursion is made for those things, put one loop in a function, and keep calling yourself until you do not have dices

all_dices=[range(1,6),range(1,6),range(1,6),range(1,6)]

def flip(dicenum,totaluntilhere):
  if dicenum>len(all_dices)
    results.append(totaluntilhere+all_dices[dicenum]
  else
  for dice in all_dices[dicenum+1]:
    flip(dicenum+1,totaluntilhere+dice)

flip(-1,0)

I am not so good in python, but that's the idea

Upvotes: 0

chepner
chepner

Reputation: 532023

You want a (Cartesian) product of the sets of values.

from itertools import product

results = [sum(numbers) for numbers in product(*all_dices)]

For example:

>>> list(product([1,2,3], [1,2,3,4]))
[(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4)]

Upvotes: 5

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