在 Python 中模拟滚动 2 个骰子
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simulating rolling 2 dice in Python
提问by A.Boland
I have been asked to simulate rolling two fair dice with sides 1-6. So the possible outcomes are 2-12.
我被要求模拟掷两个面为 1-6 的公平骰子。所以可能的结果是2-12。
my code is as follows:
我的代码如下：
def dice(n):
x=random.randint(1,6)
y=random.randint(1,6)
for i in range(n):
z=x+y
return z
My problem is that this is only returning the outcome of rolling the dice 1 time, so the outcome is always 2-12. I want it to return the sum of rolling the dice (n) times.
我的问题是这只是返回掷骰子 1 次的结果，所以结果总是 2-12。我希望它返回掷骰子 (n) 次的总和。
Does anyone have any suggestions for me?
有人对我有什么建议吗？
回答by Martijn Pieters
Roll the dice in the loop:
在循环中掷骰子：
def dice(n):
total = 0
for i in range(n):
total += random.randint(1, 6)
return total
The +=augmented assignment operator basically comes down to the same thing as total = total + random.randint(1, 6)when summing integers (it is slightly more complicated than that, but that only matters for mutable objects like lists).
该+=增强赋值运算符基本上可以归结为同样的事情，total = total + random.randint(1, 6)总结整数（这是略多于复杂，但只有事项可变对象如列表）时。
You could even use a generator expressionand the sum()function:
def dice(n):
return sum(random.randint(1, 6) for _ in range(n))
This basically does the same thing as the forloop in the first example; loop ntimes, summing up that many random numbers between 1 and 6 inclusive.
这基本上for与第一个示例中的循环执行相同的操作；循环n时间，总结了 1 到 6 之间的许多随机数。
If instead of rolling ntimes, you need to produce nresults of 2 dice rolls, you still need to roll in the loop, and you need to add the results to a list:
如果不是滚动n次数，你需要产生n2个骰子滚动的结果，你仍然需要在循环中滚动，你需要将结果添加到列表中：
def dice(n):
rolls = []
for i in range(n):
two_dice = random.randint(1, 6) + random.randint(1, 6)
rolls.append(two_dice)
return rolls
This too can be written out more compactly, with a list comprehension:
这也可以用列表理解更紧凑地写出来：
def dice(n):
return [random.randint(1, 6) + random.randint(1, 6) for _ in range(n)]
You could also use random.choice()from a list of generated sums; these are automatically weighted correctly; this basically pre-computes the 36 possible dice values (11 unique), each with equal probability:
您还可以使用random.choice()生成的总和列表；这些会自动正确加权；这基本上预先计算了 36 个可能的骰子值（11 个唯一值），每个骰子的概率相等：
from itertools import product
two_dice_sums = [a + b for a, b in product(range(1, 7), repeat=2)]
def dice(n):
return [random.choice(two_dice_sums) for _ in range(n)]
Either way, you'll get a list with nresults:
无论哪种方式，您都会得到一个包含n结果的列表：
>>> dice(5)
[10, 11, 6, 11, 4]
>>> dice(10)
[3, 7, 10, 3, 6, 6, 6, 11, 9, 3]
You could pass the list to the print()function to have these printed on one line, or on separate lines:
您可以将列表传递给print()函数以将这些打印在一行或单独的行上：
>>> print(*dice(5))
3 7 8 6 4
>>> print(*dice(5), sep='\n')
7
8
7
8
6
