Question

我想帮助找到更好的方法来制作骰子计划的概率。

这是我的所有代码：

import random,collections

probabiltyOfDice = {}
rolls = 100

#Counters
counter1 = 0
counter2 = 0
counter3 = 0
counter4 = 0
counter5 = 0
counter6 = 0

for x in range(rolls):
    output = random.randint(1,6)
    if output == 1:
        counter1 += 1
        probabiltyOfDice[1] = counter1
    elif output == 2:
        counter2 += 1
        probabiltyOfDice[2] = counter2
    elif output == 3:
        counter3 += 1
        probabiltyOfDice[3] = counter3
    elif output == 4:
        counter4 += 1
        probabiltyOfDice[4] = counter4
    elif output == 5:
        counter5 += 1
        probabiltyOfDice[5] = counter5
    elif output == 6:
        counter6 += 1
        probabiltyOfDice[6] = counter6

ordered = collections.OrderedDict(sorted(probabiltyOfDice.items()))
print(ordered)

我的方法有效，如果常量滚动数设置为100，则输出为

OrderedDict([(1, 15), (2, 16), (3, 14), (4, 22), (5, 21), (6, 12)])

OrderedDict显示骰子编号以及该编号出现的次数。我能够通过使用6个计数器变量并使用一组if-elif语句来解决这个问题。

我正在使用集合库来排序字典以进行显示。但是，如果有人能告诉我更好的方法来解决这个问题，我希望如此。

显然，对于这种情况，使用6个变量和许多if语句可能是一种冗长的方法。

Answer 1

我们可以通过使用列表来简化代码。这里的列表包含六个元素，最初都设置为零：

import random, collections

rolls = 100

#Counters
counters = [0] * 6

for x in range(rolls):
    output = random.randint(1,6)
    counters[output-1] += 1

ordered = collections.OrderedDict(enumerate(counters, 1))
print(ordered)

对于跑步，它会在我的机器上打印：

>>> print(ordered)
OrderedDict([(0, 12), (1, 11), (2, 17), (3, 31), (4, 14), (5, 15)])

我们可以通过生成0和5之间的数字（包括两者）来轻松提高代码的效率：

rolls = 100

#Counters
counters = [0] * 6

for x in range(rolls):
    output = random.randint(0, 5)
    counters[output] += 1

ordered = collections.OrderedDict(enumerate(counters, 1))
print(ordered)

Answer 2

因此，python有一个非常方便的字典子类，名为Counter，实质上，它在计算计数方面非常快。我精心设计了以下内容：

final WebElement Elementform1 = driver.findElement(By.xpath(k.toString()));

Actions action = new Actions(driver);
action.contextClick(Elementform1).build().perform();
Robot robot = null;
try {
    robot = new Robot();
} catch (AWTException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
robot.keyPress(KeyEvent.VK_DOWN);
//robot.keyRelease(KeyEvent.VK_DOWN);
try {
    Thread.sleep(2000);
} catch (InterruptedException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
robot.keyPress(KeyEvent.VK_DOWN);
try {
    Thread.sleep(2000);
} catch (InterruptedException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
robot.keyPress(KeyEvent.VK_DOWN);
try {
    Thread.sleep(2000);
} catch (InterruptedException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
robot.keyPress(KeyEvent.VK_DOWN);
try {
    Thread.sleep(2000);
} catch (InterruptedException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
robot.keyPress(KeyEvent.VK_ENTER);
try {
    Thread.sleep(2000);
} catch (InterruptedException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
robot.keyRelease(KeyEvent.VK_DOWN);
try {
    Thread.sleep(2000);
} catch (InterruptedException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
robot.keyPress(KeyEvent.VK_ENTER);
try {
    Thread.sleep(2000);
} catch (InterruptedException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

现在，问题在于它只能在合理的大范围内工作。因此，为了避免创建中间列表，我们可以具有以下内容：随机导入来自馆藏进口柜台

import random
from collections import Counter

prob = Counter([random.randint(1,6) for _ in range(10000)])

print(prob)
>> Counter({1: 1650, 2: 1750, 3: 1679, 4: 1546, 5: 1686, 6: 1689})

编辑：添加第三个版本，基本上是第一个版本，放弃了更方便的生成器的中间列表，如@Mark Dickinson所示：

prob = Counter()

for _ in range(10000000):
   prob[random.randint(1,6)] += 1

print(prob)
>> Counter({1: 13084052,
     2: 13087518,
     3: 13090177,
     4: 13086193,
     5: 13081778,
     6: 13085341})

计算可能性的更好方法 - Python

2 个答案: