Random integer with random.randint without certain number

Question

I am trying to generate a random number.

import random
print(random.randint(1,10))

Would I be able to generate a number between 1-10 but without ever generating x?

For example x = 3

Asked By: TNTDoctor

||

Source

Answer 1

Three approaches:

Loop until it’s valid:

while (val := random.randint(1, 10)) == 3:  # Python 3.8+, pre-3.8 is more verbose
    pass
print(val)

Generate from a smaller range, then increment all numbers equal to or higher than the excluded value:
```
val = random.randint(1, 9)
if val >= 3:
    val += 1
print(val)
```

Create a sequence of the numbers you allow, and use random.choice (probably the best solution if you’re doing this many times; make the sequence once, then reuse it after):

# Done once (must convert back to sequence from set, choice only works with sequences)
values = tuple(set(range(1, 11)) - {3})

# Done as often as you like
print(random.choice(values))

Answered By: ShadowRanger

Answer 2

If you use choice you can provide a list of acceptable values:

random.choice([[1,2,4,5,6,7,8,9]])

Answered By: Flow

Answer 3

Onliner, using range and sets:

x = 3
print(random.choice(list(set(range(1,11))-{x})))
# list(set(range(1,11))-{x})) -> [1, 2, 4, 5, 6, 7, 8, 9, 10]

Answered By: Maurice Meyer

Random integer with random.randint without certain number

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