如何创建一个包含5个字符的随机字符串？ （1,2,3,4,5）并定义此字符的百分比

Question

如何创建一个包含5个字符的随机字符串？ （1,2,3,4,5）并定义这些字符的百分比。

我想在新的random_char

中定义％[/ 1“，”2“，”3“，”4“，”5“]中每个字符的％次数。

例如：

1. 从100开始是50％，应该是'1'
的50个字符
2. 从100开始是20％，应该是'2'
的20个字符
3. 是10％表格100应该是10个字符'3'
4. 从100开始是15％，应该是'4'
的15个字符
5. 从100开始是5％，应该是'5'
的5个字符

当然不超过100％。

import random
string = ["1","2","3","4","5"]

random_char=''
for i in range(0,100):
    random_char+=random.choice(string)

print(random_char)

Answer 1

我不确定我理解你的问题，但这是猜测：

import random

num_chars = 100
characters = "12345"
percentages = .50, .20, .10, .15, .05
counts = [int(percent*num_chars) for percent in percentages]
random_chars = ['' for _ in range(num_chars)]   # Initialize to empty chars.

for char, count in zip(characters, counts):
    for _ in range(count):
        while True:
            index = random.randrange(0, num_chars)  # Pick random spot.
            if not random_chars[index]:  # Empty?
                random_chars[index] = char
                break

random_chars = ''.join(random_chars)
print('random_chars:', random_chars)

# Verify result is correct.
if len(random_chars) != num_chars:
    print('Generated string is wrong length ({} instead of {})'.format(
        len(random_chars), num_chars))

# Check distribution of characters.
for char in characters:
    count = sum(1 for i in range(num_chars) if random_chars[i] == char)
    print('Character {!r} is in result: {} times'.format(char, count))

示例输出：

random_chars: 1422121211111141334154411511142221114211131212342242222332511241351121114314151431434111111111121111
Character '1' is in result: 50 times
Character '2' is in result: 20 times
Character '3' is in result: 10 times
Character '4' is in result: 15 times
Character '5' is in result: 5 times

但是，正如我在评论中所说，使用random.shuffle()会更有效率和直截了当：

num_chars = 100
characters = "12345"
percentages = 50, 20, 10, 15, 5
random_chars = [char*round(percent/100*num_chars)
                    for char, percent in zip(characters, percentages)]
random.shuffle(random_chars)
random_chars = ''.join(random_chars)
print('random_chars:', random_chars)

Answer 2

如果你总是想要一个100-char-long字符串中的固定比例，你可以简单地用你想要的字符串洗牌：

import random


s = list('1' * 50 + '2' * 20 + '3' * 10 + '4' * 15 + '5' * 5)
random.shuffle(s)
''.join(s)

这给你一些类似的东西：

  

'1112213241213121114254411211141541342111141411411152423322114133124141512212211521213131111111311214'

如果您希望它是随机的并且有任何长度，那么您可以使用choices代替。

options = '12345'
weights = [.5, .2, .1, .15, .05]
length = 10  # how many characters long

''.join(random.choices(options, weights, k=length))

结果如下：

  

'5514232551'

Answer 3

函数<add name="Owin" verb="" path="*" type="Microsoft.Owin.Host.SystemWeb.OwinHttpHandler, Microsoft.Owin.Host.SystemWeb"/> 是python 3.6中的新功能，但您可以使用random.choices轻松实现它：

random.shuffle

示例输出：

import random

def choices(options, percentages, length):
    result = []
    for (char, weight) in zip(options, percentages):
        n = int(weight / 100.0 * length)
        result.extend(char * n)
    random.shuffle(result)
    return ''.join(result)

def test(options, percentages, length):
    string = choices(options, percentages, length)
    print(string)
    for char in options:
        count = string.count(char)
        perc = count / length * 100.0
        print('Character {!r} is in result: {} times = {}%'.format(char, count, perc))
    print()

percentages = 50, 20, 10, 15, 5

test(['1', '2', '3', '4', '5'], percentages, 40)
test(['a', 'b', 'c', 'd', 'e'], percentages, 80)
test(['q', 'w', 'e', 'r', 't'], percentages, 100)