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I want to generate a list of random integer within a range:

from random import randint lower = 0 upper = 20 length = 10 random_array = [randint(lower, upper) for _ in range(length)]

I don't think it is elegant and I come up with another solution:

random_array = [randint(lower, upper)] * length

But it generate a list of identical integers.

It might be that randint is executed before list.

I try putting off randint with lambda:

random_array = [(lambda: randint(lower, upper))()] * length

but fail.

So I want to know if it's possible to delay the evaluation of items (randint in this case) and then have it triggered once the list is assembled?

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    How is this a duplicate? The OP does not require the numbers to be unique. Commented May 16, 2016 at 15:23
  • @Selcuk No, they don't. Commented May 16, 2016 at 15:24
  • The linked duplicate does not match the asker's specific issue. This may be more useful: stackoverflow.com/questions/240178/… Commented May 16, 2016 at 15:24
  • So you find [(lambda: randint(lower, upper))()] * length more elegant than [randint(lower, upper) for _ in range(length)] ? Commented May 16, 2016 at 15:25
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    I think the actual question is "how do I defer the actual generation of numbers without precomputing the whole list"? The real answer is the use of a generator function. Commented May 16, 2016 at 15:26

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from random import randint lower = 0 upper = 20 length = 10 random_array = list() for _ in range(length): random_array.append(randint(lower, upper)) print random_array
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random_array = map(lambda _: randint(lower, upper), range(length))
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