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

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])

Thanks for any suggestions :D

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

can be:

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

if caps >= 1:

can be:

if caps:

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]))

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.

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.

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)
...

and then

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