CODEWITHSUNDEEP
pythonlistelementcontain
Nikolay
Nikolay

Reputation: 90

Checking elements in the list

I've spend few hours solving this problem but program doesn't work (syntax error). Cheking an answer for similar question didn't help. What's wrong with the code below? I want to chek if list (password) contains at least one digit, as well as containing one uppercase letter and one lowercase letter in it. Please, provide me with the simplest way, I'm a beginner ...

def checkio(password):    
    array = list(password)
    #for letter in array:
    if len(array) < 10 or len(array) > 64:
        return False
    if (any(l.isdigit() for l in array) and (any(l.isupper( for l in array) and (any(l.islower for l in array):
        return True
    else:
        return False

Upvotes: 2

Views: 115

Answers (4)

vikramls
vikramls

Reputation: 1822

When you do any(l.isdigit() for l in array), you are creating a generator. The generator has to be "consumed" for the value to be correct.

In this case, you are better served by using a list instead. Also the call to array = list(password) is unnecessary, since strings are iterable in python. Here's how the code should look:

def checkio(password):    
    if len(password) < 10 or len(password) > 64:
        return False
    if any([c.isdigit() for c in password]) and any([c.islower() for c in password]) and any([c.isupper() for c in password]):
        return True
    else:
        return False

In this version, the any() function is called on temporary lists created using the [c.func() for c in password].

Upvotes: 1

user4179775
user4179775

Reputation:

You can do it this way, You are missing some parenthesis, another thing is, you said at least one digit, the length should be <1. You don't need also to convert to a list, you can iterate strings

def checkio(password):    
    if len(password) < 1 or len(password) > 64:
        return False
    if (any(x.isdigit() for x in password)) and (any(l.isupper for l in password)) and (any(l.islower for l in password)):
        return True
    else:
        return False

print checkio("StackO3f") #True

print checkio("S") #False

print checkio("sssss") #False

Upvotes: 1

TomDillinger
TomDillinger

Reputation: 194

Your parentheses are very wrong. try this.

def checkio(password):    
  array = list(password)
  #for letter in array:
  if len(array) < 10 or len(array) > 64:
      return False
  if ((any(l.isdigit() for l in array)) and (any(l.isupper() for l in array)) and ((any(l.islower() for l in array)))):
      return True
  else:
      return False

Upvotes: 2

mgilson
mgilson

Reputation: 310227

Sometimes these things are easiest to see if you format the code nicely. You're missing some parenthesis:

def checkio(password):
    if 10 < len(password) or len(password) > 64:
        return False

    return (any(l.isdigit() for l in password) and
            any(l.isupper() for l in password) and
            any(l.islower() for l in password)):

Note, you shouldn't have to construct a list from the password -- python strings are iterable and have a well defined length.

Upvotes: 1

Related Questions