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I want to generate a unique random integer for the variable SEED. This code is part of a bigger script that is going to run multiple times and output each time, so feeding SEED with a non-duplicate random integer is important. Also, I tried random.sample and such that return lists/sequence not an integer, but this is not the case here because of the following line using torch.

SEED = random.randint(1, 1000)
random.seed(SEED)
np.random.seed(SEED)
torch.manual_seed(SEED)
torch.cuda.manual_seed(SEED)
torch.backends.cudnn.deterministic = True
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    So what exactly is the problem with the snippet you shared? Commented Jun 25, 2021 at 18:42
  • I need it to generate a duplicate free random integer, but it does not! Commented Jun 25, 2021 at 18:48

3 Answers 3

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You can also use time.time():

import time
SEED = int(time.time())

This is technically unique since it'll be a different timestamp every time you run the code.

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4 Comments

even if I run it inside a loop that lasts around 3 hours?
I don't know exactly what is your use case, but why don't you initialize SEED outside the loop?
because if I do so it's going to run the loops with the same SEED. I need to run each iteration with a unique SEED.
What if I need to generate multiple unique seeds within 1 second? Your method will fail.
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Perhaps this is a little too brute-force, but randomly popping integers out of a list of integers seems to work:

min_seed = 1
max_seed = 1000
seed_list = [*range(min_seed, max_seed+1)]

n_repetitions = 300  # or whatever, as long as n_repetitions < len(seed_list)
for repetition in range(0, n_repetitions):
    seed = seed_list.pop(random.randint(min_seed,len(seed_list)))

Since the integers get popped out of the list as they're used as seeds, there won't be repeats.

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8 Comments

Lengthy but functional but that's all that matters! Thanks!
Why the list comprehension instead of just list(range(...))?
No real reason, just cranking out code. :^) I prefer the comprehension just because it conveys a bit more intention, at least to me - it reads as "I'm building a list" rather than "I want this other thing to be a list."
For what it's worth, timeit is telling me that list(range(...)) is considerably faster. For 10,000 repetitions of each, the comprehension takes 0.198 sec, and list(range(...)) 0.0855 sec. So there, a counterargument against my original way!
[range(a, b)] creates a list with length 1, but list(range(a, b)) creates a list with length b - 1 - a.
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As an ultimate solution cast UUID to long integer from here:

import uuid
uuid.uuid4().int & (1<<64)-1
9518405196747027403L
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3 Comments

how can I edit this to get up to 3 digits?
uuid.uuid4().int & (1<<10)-1
@MahanAghaZahedi while a UUID is guaranteed to be unique, a 3-digit piece of it is not. In fact any manipulation will lose the uniqueness guarantee.

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