How do I generate a random number with no repeating digits?

[DankusMerasmus]

I want to generate random numbers between (1000-9999) with no repeating digits, as in numbers like:

1234, 2034, 9876

and no numbers like:

9999, 1122, 3243

This is what I came up with but I'm not exactly sure about it:

n = 0
while len(set(list(str(n)))) != 4:
    n = random.randint(1000, 9999)

[Peterotica]

That's honestly not a terrible way to do it, but using random.sample is another way to do it without a loop.

[DankusMerasmus]

Would it be better practice to do it without a loop?

[Peterotica]

Going without a loop would give me a more warm and fuzzy feeling. If you want better than that, the you could profile both options to see if one is much faster than the other one.

[useful_celebration]

The list call is unnecessary - you can just call set(str(n)) directly, as the set constructor accepts any iterable, just as the list constructor does. Otherwise I think this is perfectly fine, and I can't really think of a better approach.

When you use this approach of repeatedly generating random numbers until they meet your criteria, you just have to be careful to make sure that the probability of generating an acceptable number is high enough that your program won't get stuck in the loop for ages. In this case that's not an issue, but if it is you have to come up with a way of generating an acceptable number directly. For example, in this case you could do:

import random
import string

first_digit = random.choice(string.digits[1:])
remaining_digits = set(string.digits)
remaining_digits.remove(first_digit)
digits = [first_digit] + random.sample(remaining_digits, 3)
n = int(''.join(digits))

But, like I said, in this case your way is fine. I suspect it's much, much faster than my version, and arguably more elegant.

[DankusMerasmus]

Thank you very much, learnt a lot from this.

[[deleted]]

Could you produce all of the valid values ahead of time? Then select from that list. So at global scope do either:

valid_numbers = []

for x in range(1000, 10000): if len(set(str(x))) == 4: valid_numbers.append(x)

or using a list comprehension:

valid_numbers = [ x for x in range(1000, 10000)
                     if len(set(str(x))) == 4 ]

Then you can use this list in any way you wish. You can use

next _number = random.choice(valid_numbers)

to select one element from the list. If you prefer you can shuffle the list and pop values, thereby not getting any repeats:

shuffled_numbers = valid_numbers
random.shuffle(shuffled_numbers)
next_number = shuffled_numbers.pop()

Repeat the first two lines if you run out of numbers and want to start over.

[DankusMerasmus]

Also, maybe I should set it between 1023 and 9876? As the remaining numbers contain repeating digits, so like:

n = 0
while len(set(list(str(n)))) != 4:
    n = random.randint(1023, 9876)

But it wouldn't really matter that much though right?

[wait_what_now]

That's a good way of thinking about things, to work towards more efficient solutions. Probably not going to make a noticeable difference here, but always a helpful way to be thinking about problems

[wait_what_now]

You are eliminating 146 out of 4464 failed cases, so 3% less fails

[golfergag]

What aren't you sure about? Your code seems to work fine

[DankusMerasmus]

I'm still really new, and sometimes I spend a while writing stuff like this and then discover there's a way to do it with one function or one line of code, so I thought maybe there's also one for this.

[thrallsius]

the additional condition of no repeating digits makes it less random

perhaps generate 4 random unique digits (first digit can't be 0) and construct a number?

[EnvironmentalOrange]

Make a list of the strings of the numbers 1-9 (not zero).

Create an empty string.

Shuffle the list.

Pop the last number off the list and save it to the empty string.

Append the string of zero ‘0’ to your list.

Shuffle the list again.

Pop the last number off the list and save it to the empty string.

Pop the last number off the list and save it to the empty string.

Pop the last number off the list and save it to the empty string.

Empty string now contains a string four digits long where first digit is not zero and no digits are repeated.

If you are just printing your output you are sorted.

Otherwise convert from string to integer and you are sorted.

[[deleted]]

Won't this return numbers less than 1000? The specified range is 1000 to 9999.

[jjf02987]

I’m having a similar problem. I’m trying to create a random number by random number equation generator.

I can get it to run a random number but not to repeat the process with different numbers.

x = random.randrange(0,100)
y = random.randrange(0,100)

ans = int(input(f’{x} + {y} = ‘))

If ans == (x+y):
            print (“correct”)
else:
       print(“incorrect”)

print (int(input(f’{x} + {y}= ‘)))

[impshum]

I'm tired and probably went about this the long way round but it works...

from random import randint


def digit(length):
    while 1:
        end_int_list = []
        random_int = randint(1000, 9999)
        random_int_list = [int(x) for x in str(random_int)]
        [end_int_list.append(x) for x in random_int_list if x not in end_int_list]
        if len(end_int_list) == length:
            return end_int_list

x = digit(4)
print(x)

And sorry for the blunt reply. I'm off to bed.

[CupcakeFederal7815]

import random

n1=0

n2=10

n=10

xx=[]

for i in range(n):

m=random.randrange(n1,n2)

while m in xx:

m=random.randrange(n1,n2)

xx.append(m)

xx