How to check if a number is 2 digit in python

I have created a function which checks if the number has any other digit other than 3 or 5, return false.

def isDecent(n):
    digits = list(map(int, str(n)))
    for digit in digits:
        if digit != 3 or digit != 5: return False
    return True

print(isDecent(5553334))

But unfortunately it's not working! Can I know the problem?

asked Mar 7, 2014 at 13:40

5

It's always true that digit != 3 OR digit != 5. So it will always return False.

Change the or to and.

Alternatively, check it as

if digit not in (3, 5):

In fact, we don't need to convert the characters to ints just to check equality, and you could just do:

def isDecent(n):
    digits = str(n)
    return all(digit in ("3", "5") for digit in digits)

Or even digit in "35"... but I think that's less readable.

Another option is to use sets:

def isDecent(n):
    return set(str(n)).issubset('35')

answered Mar 7, 2014 at 13:43

RemcoGerlichRemcoGerlich

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5

Since the why part has been already answered, let me show you how to do this efficiently

def isDecent(n):
    return all(char in '35' for char in str(n))

The sister version of the same

def isDecent(n):
    return not any(char not in '35' for char in str(n))

answered Mar 7, 2014 at 13:48

thefourtheyethefourtheye

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1

Your issue is that you are using an or instead of an and.

Just change your code to this :

def isDecent(n):
    digits = list(map(int, str(n)))
    for digit in digits:
        if digit != 3 and digit != 5: return False
    return True

print(isDecent(555333))

And then everything works !

answered Mar 7, 2014 at 13:43

I suggest avoiding the conversion to a string:

def isDecent(n):
    if n == 0:
        return False

    while n: 
        if n % 10 not in (3, 5, ): 
            return False 
        n //= 10

    return True 

answered Mar 7, 2014 at 13:48

tmr232tmr232

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2

You can use set:

d=str(55555333)
r=''.join(set(d))
if r in '353':
    print "Number contains 3 or 5"

answered Mar 7, 2014 at 13:50

venpavenpa

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4

Other's have shown the problem in your code, here's an alternative.

def isDecent(n):
     if not set(str(n)) - {'3','5','.'}:
             return False
     return True

It returns False on if str(n) contains anything other than '3' , '5' or '.'

answered Mar 7, 2014 at 13:48

GuyGuy

6055 silver badges21 bronze badges

Another solution:

def isDecent(n):
    s = set([ x for x in str(n)])
    if len(s) > 2:
        return False
    return [False,True]['3' in s and '5' in s]

for i in [5553334, 5553133, 111115, 33333555]:
    print i, isDecent(i)

Output:

5553334 False
5553133 False
111115 False
5553335 True

answered Mar 7, 2014 at 13:50

James SapamJames Sapam

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5

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