check if all the digits of the given number are different

Question

I want to create a 4-digit number between 1000 and 10000 with different digits, but I am a bit inexperienced as I am new to this stuff. Can you help me?

import random

number = random.choice(range(1000,10000))


print(number)

Asked By: hamsi

||

Source

Answer 1

You need to iterate through the digits of the number and check if that digit is in the number more than once.

def different_digits(num):
    for d in str(num):
        if str(num).count(d) > 1:
            return False
    return True

print(different_digits(1234)) # True
print(different_digits(1224)) # False

This can even be simplified with all:

def different_digits(num):
    return all(str(num).count(d) == 1 for d in str(num))

print(different_digits(1234)) # True
print(different_digits(1224)) # False

Now to get the 4-digit number just use a while loop:

x = random.randint(1000, 9999)
while not different_digits(x):
    x = random.randint(1000, 9999)

Answered By: The Thonnu

Answer 2

import random

a = 0
while a == 0:
    a, b, c, d = random.sample(range(10), 4)
k = 1000*a + 100*b + 10*c + d
print(k)

Answered By: progerg

Answer 3

Using a direct selection of the digits, without any trial and error:

import random

s = '123456789'
# select first digit
a = random.sample(s, 1)
# select last 3 digits
b = random.sample(list(set(s).difference(a))+['0'], 3)

out = int(''.join(a+b))

Example output: 6784

Answered By: mozway

check if all the digits of the given number are different

Question:

Answers: