获取随机数选择中整数出现次数N次

Question

在写这段代码时，我发现自己在做很多重复性的事情，并且想知道是否有一种更简单、更简单或更短的方法来完成这样的重复性任务。在

以下是相关代码：

 from random import randint
    RandomNumber = Zeroes = Ones = Twos = Threes = Fours = Fives = Sixes = i = 0

    while i < 1000000:
        RandomNumber = (randint(0,6))
        if RandomNumber == 0:
            Zeroes = Zeroes + 1
        if RandomNumber == 1:
            Ones = Ones + 1
        if RandomNumber == 2:
            Twos = Twos + 1
        if RandomNumber == 3:
            Threes = Threes + 1
        if RandomNumber == 4:
            Fours = Fours + 1
        if RandomNumber == 5:
            Fives = Fives + 1
        if RandomNumber == 6:
            Sixes = Sixes + 1

        i = i + 1

Answer 1

给你。。。在

from random import randint
outcomes=[0]*7
for i in range(1000000):
    outcomes[randint(0,6)]+=1

Answer 2

您不必为每个随机输出获取命名变量，而是可以将字典中每个可能的值作为键。这将缩短您的代码，并使其可扩展到任何随机范围

from random import randint
randomMax = 6
randomList= {i:0 for i in range(0,randomMax+1)}
totalIterations = 10000
while totalIterations >0:
    randomList[randint(0,randomMax)]+=1
    totalIterations-=1

样本输出：

^{pr2}$

Answer 3

这段代码可能会有所帮助，计数器字典将数字作为键，数字出现作为值。在

from random import randint

counter = {}

while i < 1000000:
    RandomNumber = (randint(0,6))
    if RandomNumber in counter:
        counter[RandomNumber] += 1
    else:
        counter[RandomNumber] = 1

    i += 1