Question

我必须为我的计算机类做一个骰子滚动程序。我选择用Python编写我的。但是，当我执行它时，它会打印1000次数字列表。这是代码。任何帮助将不胜感激。

`

import random
def rollDie():
    one = 0
    two = 0
    three = 0
    four = 0
    five = 0
    six = 0
    seven = 0
    eight = 0
    nine = 0
    ten = 0 
    eleven = 0
    twelve = 0
    for i in range(0, 1000):
        roll = int(random.randint(1,6)) + (random.randint(1,6))
        if roll == 2:
            two = two+1
        elif roll == 3:
            three = three+1
        elif roll == 4:
            four = four+1
        elif roll == 5:
            five = five+1
        elif roll == 6:
            six = six+1
        elif roll == 7:
           seven = six+1
        elif roll == 8:
           eight = eight+1 
        elif roll == 9:
           nine = nine+1 
        elif roll == 10:
           ten = ten+1
        elif roll == 11:
           eleven = eleven+1
        elif roll == 12:
           twelve = twelve+1
    print "2:%s" % two
    print "3:%s" % three
    print "4:%s" % four
    print "5:%s" % five
    print "6:%s" % six
    print "7:%s" % seven
    print "8:%s" % eight
    print "9:%s" % nine
    print "10:%s" % ten
    print "11:%s" % eleven
    print "12:%s" % twelve
rollDie()

`

谢谢！

Answer 1

我的猜测是你在文本编辑器中的缩进是关闭的; print语句包含在for循环中。你也有一点问题，你打电话给seven = six+1。以下是您编辑过的代码，请尝试将此粘贴到您的文件中：

import random
def rollDie():
    one = 0
    two = 0
    three = 0
    four = 0
    five = 0
    six = 0
    seven = 0
    eight = 0
    nine = 0
    ten = 0 
    eleven = 0
    twelve = 0
    for i in range(0, 1000):
        roll = int(random.randint(1,6)) + (random.randint(1,6))
        if roll == 2:
            two = two+1
        elif roll == 3:
            three = three+1
        elif roll == 4:
            four = four+1
        elif roll == 5:
            five = five+1
        elif roll == 6:
            six = six+1
        elif roll == 7:
           seven = seven+1
        elif roll == 8:
           eight = eight+1 
        elif roll == 9:
           nine = nine+1 
        elif roll == 10:
           ten = ten+1
        elif roll == 11:
           eleven = eleven+1
        elif roll == 12:
           twelve = twelve+1
    print "2:%s" % two
    print "3:%s" % three
    print "4:%s" % four
    print "5:%s" % five
    print "6:%s" % six
    print "7:%s" % seven
    print "8:%s" % eight
    print "9:%s" % nine
    print "10:%s" % ten
    print "11:%s" % eleven
    print "12:%s" % twelve
rollDie()

此外，您的代码效率极低;请尝试使用以下代码：

import random
rolls = {}
for k in range(2, 13):
        rolls[k] = 0

for k in range(1000):
        roll = random.randint(1, 6)+random.randint(1, 6)
        rolls[roll]+=1

for k in rolls:
        print '%d occurred %d times!' %(k, rolls[k])

运行如下：

bash-3.2$
2 occurred 38 times!
3 occurred 51 times!
4 occurred 90 times!
5 occurred 99 times!
6 occurred 136 times!
7 occurred 173 times!
8 occurred 151 times!
9 occurred 90 times!
10 occurred 92 times!
11 occurred 58 times!
12 occurred 22 times!
bash-3.2$

如果我们计划时间，这是你的时间：

This ran in 0.06672 seconds!

这是我的：

This ran in 0.004228 seconds!

当然，这可能只是我的电脑给出了如此大的差异:)。如果您想知道我如何计时，我使用了time.time() - previous time.time()。

Answer 2

没有变量，我们可以用相同的方式实现掷骰子的相同功能：

import random
    from collections import Counter,OrderedDict


    def ran(i):
        roll = int(random.randint(1,6)) + (random.randint(1,6))
        return roll

    for i,j in OrderedDict(Counter(map(ran,range(1,1000)))).items():
        print i,":", j

说明：首先我们将rand int的所有值存储在dict.map链接中。然后usinf counter v可以找到所有频率然后odered dict用于根据键排序，即2号的第一个等等。

骰子滚动计划的问题

2 个答案: