Question

在编写这段代码时，我发现自己做了很多重复的事情，并且想知道是否有更简单，更简单或更短的方法来完成这样的重复性任务。

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

你去......

string.Empty

Answer 2

不是为每个随机输出获取命名变量，而是可以将每个可能值作为键的字典。这将缩短您的代码并使其可扩展到任何随机范围

float x= 15;
void *y;

y = &x;

printf("\tValue of X is : %f \n", x);
printf("\tValue of y is : %d", (int)*(float *)y);

示例输出：

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

Answer 3

首先，当你发现自己使用大量适合相同目的的变量时（例如零，一，二等数），你几乎肯定需要一个数组（Python术语中的list）。

import random

nums = [0] * 7 # number of digits from zero to six

for i in range(1000001):
    r = random.randint(0, 6) # you can get rid of this variable 
                             # and use random.randint(0, 6) as an array index
    nums[r] += 1 # each member of a list represents a number of some digit's occurrences

print(''.join("{} -> {}\n".format(a, b) for a, b in zip(range(7), nums)))

如果你想要一些非常简短，快速和强大的东西：

import random, collections

print(''.join("{} -> {}\n".format(a, b) for a, b in collections.Counter(random.randint(0, 6) for _ in range(100)).items()))

选中this，了解如何使用collections.Counter计算列表项的出现次数。

这段代码很快，并且在使用生成器时不会浪费内存。

Answer 4

此代码可能有所帮助，计数器字典将数字作为键，数字出现为值。

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

Answer 5

>>> from random import randint
>>> randomnumber = [0,0,0,0,0,0,0]
>>> i = 0
>>> while i < 100:
...  random = (randint(0,6))
...  randomnumber[random] +=1
...  i = i +1
... 
>>> randomnumber
[18, 15, 10, 8, 16, 19, 14]

Answer 6

在您的特殊情况下，您可以

from random import randint
results=[0]*7 # same as [0,0,0,0,0,0,0]

i=0
while i < 1000000:
    n = randint(0,6)
    results[n]+=1
    i+=1

for i in range(len(results)):
    print(i, results[i])

Answer 7

这些答案都可行，但是他们将花费很长时间非常循环，而实际上你根本不需要使用循环。这里的代码占用了一段时间，而不是需要循环超过一百万次：

from random import random

nums = [random() for _ in range(7)]
nums = [int((x / sum(nums)) * 1000000) for x in nums]

不可否认，这种方法总共有点1000000，但它的更多，更多更快，你也可以添加一些随机的，使它实际上是1000000。

有关详细信息，请参阅：

Getting N random numbers that the sum is M

Random numbers that add to 100: Matlab（有关此生成的统计分布的信息）

在N次范围内的随机数选择中获取整数的出现次数

7 个答案: