Generate random integer between two ranges

Question

I want to get a random number between (65, 90) and (97, 122) range. I can select random number in range using

import random
random.randint(65,95)

but ˋrandint` take the only integer so how can I do?? any idea

Thanks in advance

you could generate 2 ranges, and then a `random.choice([first range rand, second range rand])` — depperm, Jul 31 '17 at 12:01
Are you sure you're not really after `random.choice(string.ascii_letters)` ? — Jon Clements, Jul 31 '17 at 12:02
what John means is that the code is probably used to generate letters, ASCII codes are probably not that interesting. — Jean-François Fabre, Jul 31 '17 at 12:08
@Kallz maybe [edit] your question: what do you want to do with this number. Because Jon Clements comment is a very good suggestion. — Jean-François Fabre, Jul 31 '17 at 12:10

score 1 · Answer 1 · answered Jul 31 '17 at 12:09

1

Alternatively:

def random():
  arr1 = np.random.randint(65,90)
  arr2 = np.random.randint(97,122)
  out = np.stack((arr1,arr2))
  out = np.random.choice(out)
  return out

answered Jul 31 '17 at 12:09

Andreas Look

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score 1 · Accepted Answer · answered Jul 31 '17 at 12:10

1

def random_of_ranges(*ranges):
  all_ranges = sum(ranges, [])
  return random.choice(all_ranges)

print(random_of_ranges(range(65, 90), range(97, 122)))

This is a very general way of handling this. You can pass any number of ranges (or subset of them). But it is rather costly, especially for very large sets (like range(1, 100000000)). For your usecase (and maybe several others in the vicinity) it would be sufficient, though.

answered Jul 31 '17 at 12:10

Alfe

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Why would you build `all_ranges` here? `random.choice(random.choice(ranges))` does the same without rebuilding a list each time... – Jon Clements Jul 31 '17 at 12:23
2

It does the same if all ranges are of equal size (which is the case in the Q above). In the general case this need not be true, and `choice(choice([ range(1, 100), range(200, 202) ]))` will return 200 and 201 way more often than e. g. 50 or 37. But of course, building the list just once and then use it in all calls with this input would be a good optimization. – Alfe Jul 31 '17 at 12:25
Good point - well made :) – Jon Clements Jul 31 '17 at 12:26
I have a feeling it's going to be slower, but...: `heapq.nlargest(1, itertools.chain(ranges), key=lambda L: random.random())` :p – Jon Clements Jul 31 '17 at 12:29
Why using `nlargest` instead of `max`? – Alfe Jul 31 '17 at 13:20
1

Just to make my non-serious suggestion more obviously non-serious? – Jon Clements Jul 31 '17 at 13:22

Nicola Coretti · Answer 3 · 2017-08-01T12:00:38.680

-1

Make your own?

import random

def my_rand():
  numbers = list()
  numbers.append(random.randint(65,95))
  numbers.append(random.randint(97,122))
  return numbers[random.randint(0,1)]

edited Aug 01 '17 at 12:00

answered Jul 31 '17 at 12:04

Nicola Coretti

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That's not what is probably wanted if the ranges are of different size. – Alfe Jul 31 '17 at 12:06
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plus the code has a syntax error. – Jean-François Fabre Jul 31 '17 at 12:07
@Jean-FrançoisFabre fixed, thanks for taking the time to complain instead of fixing it ;) – Nicola Coretti Aug 01 '17 at 12:01
@Alfe assumption – Nicola Coretti Aug 01 '17 at 12:02
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@Nicoretti No. Not assumption. Comment. _You_ assumed something about the wanted distribution and implemented a function based on your assumption (without any declaration of the assumption). I _commented_ that this might _probably_ not be what PO wants. Without any more information from PO that statement about the probability is not an assumption, it's a fact. – Alfe Aug 01 '17 at 16:01
@Nicoretti I would have fixed it if it wasn't already downvoted. I figured it wasn't worth the trouble. – Jean-François Fabre Aug 01 '17 at 18:01

Generate random integer between two ranges

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