Python 使用定义的最大长度的随机空间拉伸字符串

详细解释：
我编写了一个程序，用python模拟一个简单的Caesar编码算法。在加密的第一步之后，我们计算新字符串中字母的总和，然后“我想用随机空间拉伸原始字符串的大小！”它将被发送到目的地，并具有该长度，在删除空格后，它将破译字符串2次以获得原始字符串

简短解释：
只需完成stretch_str（text，length）函数即可生成字符串，如下所示：

text = "Hi I'm mark."
print(stretch_str(text, 23))
# The output can be :
# "H  i   I'  m m a  rk  ."
# or
# "H i  I ' m  m a r k   ."
# or any thing else like these...

以及守则：

import string


def caesar(plaintext, shift):
    alphabet = string.ascii_lowercase
    shifted_alphabet = alphabet[shift:] + alphabet[:shift]
    table = str.maketrans(alphabet, shifted_alphabet)
    return plaintext.translate(table)


def stretch_str(text, length):
    # I want to complete this function.
    # Stretch this text with random spaces.
    # The last character is dot, always.
    pass


# Sender section :
in_str = "some example sentence."
ciphered = caesar(in_str,-3)
total_sum = 0
for ch in ciphered[:-1].replace(" ", ""):
    total_sum += string.ascii_lowercase.index(ch)
ciphered_more = caesar(ciphered,total_sum%26)
msg = stretch_str(ciphered_more, total_sum)

# Receiver section :
number = len(msg)
ciphered_more = msg.replace(" ", "")
ciphered = caesar(ciphered_more, -(number%26))
original_str = caesar(ciphered, 3)
print(original_str)
# Printing the "someexamplesentence."

---

Tnx，你的答案看起来棒极了。但我会等待别人的回答，最终我会选择最好的。但你是我的首要任务。我现在不能投票支持你。但我会的。又是Tnx。

import random


def stretch_str(text, length):
    text = list(text)
    while len(text) < length:
        text.insert(random.randint(1, len(text) - 1), " ")
    return "".join(text)


text = "Hi I'm mark."
print(f"[{stretch_str(text, 23)}]")