Python - 尝试制作随机字符串生成器

Question

我正在尝试根据您提供的长度制作一个字符串生成器，它从 2 个数组中获取字母，一个为 Maj 提供，一个为 Min 提供，所以这是我的代码，但它通常返回“b”或错误

from random import randint
def randomstr(stringsize):
    Alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
    Alphabet2 = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
    i = stringsize+1
    LocalRanDom = ""
    StringGen = []
    while i < stringsize+1:
        i = i-1
    MajorMin = randint(1,2)
    print(Alphabet[1])
    if MajorMin == 1:
        LocalRanDom = randint(1,26)
        StringGen.append(Alphabet[LocalRanDom])
    if MajorMin == 2:
        LocalRanDom = randint(1,26)
        StringGen.append(Alphabet2[LocalRanDom])
    return StringGen 

randomstr(3)

Answer 1

这不是 pythonic 大声笑见克里斯的评论

首先，我不知道 i 是什么？ 你只需要i = stringsize+1然后

while i < stringsize+1:
    i = i-1

在那之后，你永远不会使用它。

如果您想要一个随机字符串生成器，您可以执行以下操作。

from random import random


def gen(length):
    string = ''
    for _ in range(length):
        shift = int(random()*26) # if you want cap letters, feel free to customzie
        asci = shift + 97
        string += chr(asci)
    return string

print(gen(3))

Answer 2

您的代码大多无法修复，但我会仔细阅读并解释我遇到的所有问题。 你也没有在你的问题中正确缩进你的代码，所以我在这方面做了一些假设。

解决代码问题

字母表生成

Alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]`
Alphabet2 = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]`

这些可以更简洁地表示为：

lowercase_letters = list(string.ascii_lowercase)
uppercase_letters = list(string.ascii_uppercase)

迭代逻辑

您当前的实现根本不会迭代，因为您分配i = stringsize+1然后创建一个 while 循环，条件i < stringsize+1 -这在第一次评估条件时永远不会为真。

正确的 Pythonic 方法是使用这样的 for 循环：

for i in range(stringsize):
    ...

字符串连接

Python 中的字符串在技术上是列表，但通过将单个字符附加到列表来构造字符串并不是很愉快。

一种方法是设置StringGen = ''然后在 for 循环中使用StringGen += c添加字符。 但是，这效率不高。 我将在这篇文章的底部提供一个解决方案来演示一个不涉及循环内串联的实现。

在条件逻辑中滥用整数

编码：

MajorMin = randint(1,2)
if MajorMin == 1:
    ...
if MajorMin == 2:
    ...

使用这个等效逻辑可以更清楚：

use_uppercase_letter = random.choice([True, False])
if use_uppercase_letter:
    ...
else:
    ...

替代实现

改进方法的变体

这是基于此处的点的randomstr的不同实现：

import string
import random


def randomstr(stringsize):
    lowercase_letters = list(string.ascii_lowercase)
    uppercase_letters = list(string.ascii_uppercase)

    def generate_letters(n):
        for i in range(n):
            use_uppercase_letter = random.choice([True, False])
            if use_uppercase_letter:
                yield random.choice(lowercase_letters)
            else:
                yield random.choice(uppercase_letters)

    return ''.join(c for c in generate_letters(stringsize))


print(randomstr(10))

我最好的破解方法

这是一个更简洁的实现，如果您需要，我会提供它，但它与您原来的方法有很大的不同。

import string
import random


def randomstr(stringsize):
    letters = list(string.ascii_lowercase + string.ascii_uppercase)
    return ''.join(random.choice(letters) for _ in range(stringsize))


print(randomstr(10))

示例运行

这些是您通过上述任一实现获得的输出示例。

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