Python - checking if the string has at least 3 numbers in it

Question

Im having some problems with the code below. I'm using PyCharm. The program is supposed to get an input for password of the user and check if it has at least 3 upper case chars and at least 3 numbers. The second task is what im having troubles with.

import sys


def num(s):
regex = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
for k in regex:
    if regex == s:
        print(y)
        y = y + 1
return y


passw = input("Enter password")
passw2 = list(passw)
lenght = (len(passw2))
if lenght < 8:
print("Password needs to be at least 8 chars long")
else:
    i = 0
    for x in passw2:
        if x.isupper():
            i += 1
if i < 3:
    print("You need at least 3 upper cased chars in your password")
    sys.exit()
else:
    numpassw2 = num(passw2)
    if numpassw2<3:
        print("At least 3 numbers needs to be given")
    else:
        print("OK,lets continue")

It gets stuck at calling out the num() function and gives following errors:

Traceback (most recent call last):
File "H:/szkola/python/projects/password/passwchecker.py", line 27, in <module>
numpassw2 = num(passw2)
File "H:/szkola/python/projects/password/passwchecker.py", line 10, in num
return y
UnboundLocalError: local variable 'y' referenced before assignment

Hi there welcome to SO. Please check over the [formatting](https://meta.stackexchange.com/questions/22186/how-do-i-format-my-code-blocks) in [your question](https://stackoverflow.com/posts/53571189/edit) and ensure that it accurately reflects your real code. — TrebledJ, Dec 01 '18 at 13:25
Some of your indents are off in the code you show here. This cannot run. — Jongware, Dec 01 '18 at 13:36

Alif Jahan · Accepted Answer · 2018-12-01T13:42:18.377

1

In your function num(s) declare a variable named y. Because of its missing in your function and it is raising an error.

def num(s):
    y = 0 # DECLARE y BEFORE CALL IT.
    regex = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
    for k in regex:
        for c in s:
            if c == k:
                print(y)
                y = y + 1
    return y

edited Dec 01 '18 at 13:42

answered Dec 01 '18 at 13:26

Alif Jahan

793
1
7
20

It made the function work, hovewer it doesnt return a correct statement. If i do PASSword123 for example, it still returns "At least 3 numbers needs to be given" – TheXsarix pl Dec 01 '18 at 13:35
Awesome. Works now. – TheXsarix pl Dec 01 '18 at 13:45

Patrick Artner · Answer 2 · 2018-12-01T13:48:55.690

Your problem is the result of false scoping - the variable you want to use is not known. Learn more here: Short Description of the Scoping Rules?

You can use sum() over booleans - each True sums as 1. There are pre-build function for checking islower, isupper, isdigit - you can those like this:

def check_pw(p):
    # count how much of which type is in the password
    num_digit = sum(c.isdigit() for c in p) # True == 1, False == 0
    num_upper = sum(c.isupper() for c in p)
    num_lower = sum(c.islower() for c in p)

    # check if all categories are at least 3 times in it 
    ok =  all(x >= 3 for x in [num_digit, num_lower, num_upper])

    # do some output / error handling 
    print("Numbers: {}".format(num_digit), 
          "UPPERS:  {}".format(num_upper),
          "lowers:  {}".format(num_lower))
    print(p, " is Ok: ", ok) 

    # return if ok or not - if not the calling code has to ask/check a new password
    return ok

for p in [ "no", "NO", "StillNo", "ABCdef12no", "YESyes123"]: 
    check_pw(p)  # not using the return value here

Output:

Numbers: 0 UPPERS:  0 lowers:  2
no  is Ok:  False

Numbers: 0 UPPERS:  2 lowers:  0
NO  is Ok:  False

Numbers: 0 UPPERS:  2 lowers:  5
StillNo  is Ok:  False

Numbers: 2 UPPERS:  3 lowers:  5
ABCdef12no  is Ok:  False

Numbers: 3 UPPERS:  3 lowers:  3
YESyes123  is Ok:  True

Doku:

samuq · Answer 3 · 2018-12-01T13:49:08.327

0

Try to change the code to something like this:

import sys


def num(s):
    y = 0
    regex = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
    for k in regex:
        for l in s:
            if regex == l:
                print(y)
                y += 1
    return y


passw = input("Enter password")
passw2 = list(passw)
lenght = (len(passw2))
if lenght < 8:
    print("Password needs to be at least 8 chars long")
else:
    i = 0
    for x in passw2:
        if x.isupper():
            i += 1
    if i < 3:
        print("You need at least 3 upper cased chars in your 
             password")
        sys.exit()
    else:
        numpassw2 = num(passw2)
        if numpassw2<3:
            print("At least 3 numbers needs to be given")
        else:
            print("OK,lets continue")

edited Dec 01 '18 at 13:49

answered Dec 01 '18 at 13:32

samuq

306
6
16

@TheXsarix pl check my answer – samuq Dec 01 '18 at 13:37
the else are off the place – TheXsarix pl Dec 01 '18 at 13:39
@TheXsarixpl Please upvote also my answer; the main problem was fixed first in my answer – samuq Dec 01 '18 at 13:51

score 0 · Answer 4 · answered Dec 01 '18 at 13:59

Here's an other option:

MIN_LENGTH = 8
MIN_UPPER = 3
MIN_DIGIT = 3


def check_pass(p):
    if not p or len(p) < MIN_LENGTH:
        return False # fast-exit, validate length 
    count_upper = count_digit = 0
    pass_upper = pass_digit = False
    for char in p: # iterate over the characters
        if char.isupper():
            count_upper += 1 # increment the counter and check if we've reached the limit
            pass_upper = count_upper >= MIN_UPPER
        elif char.isdigit():
            count_digit += 1 # increment the counter and check if we've reached the limit
            pass_digit = count_digit >= MIN_DIGIT
        else: # for any other character we have nothing to do
            continue 
        # we got here because the caracter was a digit or an upper, meaning that 1 of the counters has changed and maybe the "pass" flags also 
        if pass_upper and pass_digit:
            # we've passed the checks, so fast-exit we don't need to iterate anymore
            return True
    # if we've got here, it means that the password didn't passed the checks 
    return False

I think it's better as it has fewer iterations and has "fast-exits"

Python - checking if the string has at least 3 numbers in it

4 Answers4