可以缩短或改进此代码吗? [英] can this code be shortened or improved?

问题描述

可以缩短/改善吗?我正在尝试在python中进行密码检查.

Can this be shortened/improved? I'm trying to make a password checker in python.

是否可以将if放入for循环中?如果可以，怎么办?

Could the if's be put into a for loop? And if so, how?

pw = input("Enter password to test: ")

caps = sum(1 for c in pw if c.isupper())
lower = sum(1 for c in pw if c.islower())
nums = sum(1 for c in pw if c.isnumeric())

scr = ['weak', 'medium', 'strong']
r = [caps, lower, nums]


if len(pw) < 6:
    print("too short") 
elif len(pw) > 12:
    print("too long")

if caps >= 1:
    if lower >= 1:
        if nums >= 1:
            print(scr[2])
        elif nums < 1:
            print("your password is " + scr[1])
    elif lower < 1:
        print("your password strength is " + scr[0])
elif caps < 1:
    print("your password strength is " + scr[1])

感谢您的任何建议:D

Thanks for any suggestions :D

推荐答案

caps = sum(1 for c in pw if c.isupper())

可以是:

caps = sum(c.isupper() for c in pw)

---

if caps >= 1:

可以是:

if caps:

---

最显着的改进:可以完全移除底部的if/elif块

---

The most significant improvement: The bottom if/elif block can be completely removed by doing

i_strength = sum(map(bool,[caps,lower,nums])) - 1 #or sum(map(bool,r)) - 1
print('your password is {}'.format(scr[i_strength]))

说明:map(bool,[caps,lower,nums])累加每个caps,lower,nums不为零的次数.将它们与sum相加会为您提供强度"，您已经方便地将其放入了列表中，可以通过索引进行访问.

Explanation: map(bool,[caps,lower,nums]) accumulates how many times each of caps,lower,nums is non-zero. Adding them up with sum gives you your "strength", which you've conveniently already put into a list, which can be accessed by index.

所有这些改进都利用了python中虚假"的概念，在布尔上下文中也称为对象的值.通常，空值和零值是False，对布尔值求和就等于加一和零，因此就可以了.

All of these improvements leverage the concept of "falsiness" in python, otherwise known as an object's value in a boolean context. Generally empty and zero things are False, and summing booleans is equivalent to adding ones and zeroes, so there you go.

当然，除了检查它们是否为非零之外，似乎您没有在做带有高/低/数字计数的事情.所以清理只是

Of course, it doesn't seem that you're doing anything with the counts of upper/lower/nums other than checking if they're nonzero. So a cleanup would just be

caps = any(c.isupper() for c in pw)
...

然后

i_strength = sum([caps,lower,nums]) -1

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