Question

我必须编写一个python代码，它应该采用两个序列（s1，s2）为python中给定数量的试验生成序列2（s2）的随机字符串，并返回每个字符串（s1固定，s2更改）审判。例如：

input seq: [aabcc,aabcc]
Output seq for 4 trials:

Trial1:[aabcc, aabcc]
Trial2:[aabcc, aaabc]
Trial3:[aabcc, aaaaa]
Trial4:[aabcc, ccaab]

就像从给定输入序列中为给定数量的试验生成随机序列。有人可以帮我使用python中的基本for和while循环来编码吗？（即没有内置功能）。

Answer 1

import random

# if you can't use random.choice, this is a one-for-one substitute
def random_choice(seq):
    """
    Return a random element from a non-empty sequence
    """
    return seq[int(random.random() * len(seq))]

def random_string_from_sample(s, length=None):
    """
    Return a random string based on the letter-distribution in s
    """
    if length is None:
        length = len(s)
    return ''.join(random_choice(s) for _ in range(length))

def main():
    s0, s1 = "aabcc", "aabcc"
    trials = 4
    for i in range(1, trials+1):
        print("Trial{}:[{}, {}]".format(i, s0, random_string_from_sample(s1)))

if __name__=="__main__":
    main()

产生

Trial1:[aabcc, bbaca]
Trial2:[aabcc, cbaac]
Trial3:[aabcc, cacac]
Trial4:[aabcc, caacc]

Answer 2

此解决方案提取唯一字符，因此结果不会针对输入中频率较高的字符进行加权。如果您不想要这种行为，只需删除要设置的转换。

from random import randint

s = "aabbcc"

chars = list(set(s))
nchars = len(chars)
trials = 4 

for i in range(trials):
    rng_sample = ''.join([chars[randint(0,nchars-1)] for _ in range(len(s))])
    print rng_sample

Answer 3

S1显然很容易返回。 s2可以变成一个列表并随机播放：

s2 ="aabbcc"
import random
h = list(s2)

random.shuffle(h)

newString = ''.join(h)
print (newString)

生成字符串的随机序列

3 个答案: