CODEWITHSUNDEEP
python
mrnovice
mrnovice

Reputation: 253

How to check if a specific digit is in an integer

I want to check if, for example, the digit '2' is in 4059304593. My aim is to then check if there are any of the digits 1-9 not in my integer. This is what I have tried:

for i in xrange(10):
    for j in xrange(100):
        num = str(i^j)
        one_count = 0
        two_count = 0
        for k in xrange(len(num)):
            if num[k] == 1:
                one_count += 1
            if num[k] == 2:
                two_count += 1

Then my "counts" would go all the way down to nine_count, and if any of the counts are 0, then that digit isn't in 'num'. From what I've read on these sites, my script would be inefficient - can someone point out a better way?

Upvotes: 3

Views: 13606

Answers (6)

Synonian_raj
Synonian_raj

Reputation: 346

number = 4059304593
digit = 2
def hasDigit(number, digit):
    while number > 0:
        d = number % 10
        if d == digit:
            return True
        number = number // 10
    return False

Upvotes: 0

Jean-François Fabre
Jean-François Fabre

Reputation: 140307

This "digit" thing calls for a string approach, not a numeric one (reminds me of some Project Euler puzzles).

I'd create a set out of the digits of your number first (removing duplicates at the same time)

s = set(str(4059304593))

then to check for a digit:

print('2' in s)

(note that in for a set is performant)

then, to check whether s contains all the 013456789 digits:

print(s.issuperset("013456789"))

(if this must be done multiple times, it may be worth to create a set with the string, issuperset will work faster)

Upvotes: 9

Miguel
Miguel

Reputation: 1907

Another way is using - with sets:

set('0123456789') - set(str(4059304593))

Result are all digits that aren't in your integer:

{'2', '7', '1', '6', '8'}

Upvotes: 1

L3viathan
L3viathan

Reputation: 27331

Since you just want to find out which digits aren't in your number:

def not_in_number(n):
    return {*range(10)} - {*map(int, {*str(n)})}

Usage:

>>> not_in_number(4059304593)
{1, 2, 6, 7, 8}

This takes the set of digits ({*range(10)}) and substracts from it the set of digits of your number ({*map(int, {*str(n)})}), created by mapping the set of digit characters to integers. If you find the {*...} notation confusing, you can always use set(...) instead, which will also work for Python 2.7+:

def not_in_number(n):
    return set(range(10)) - set(map(int, set(str(n))))

Upvotes: 1

Tony Tannous
Tony Tannous

Reputation: 14863

L = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
j = 0
nL = [0, 0, 0, 0, 0, 0, 0, 0, 0]
n = 1004 #number
while n:
    i = n%10
    nL[i] = 1
    n //= 10

OUT

nL = [1, 1, 0, 0, 1, 0, 0, 0, 0, 0]

Explanation: if nL[i] is 1, then the i-th digit is in n

Upvotes: 1

Eric Duminil
Eric Duminil

Reputation: 54303

You could convert your number to a string, then to a set to get the unique digits.

You just have to iterate over digits in 0-9 to find the ones not present in your original number :

>>> set(map(int,str(4059304593)))
set([0, 9, 3, 4, 5])
>>> digits = _
>>> [i for i in range(10) if i not in digits]
[1, 2, 6, 7, 8]

Upvotes: 2

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