[SOLVED] How to get random number in a range

Reply to [SOLVED] How to get random number in a range on Mon, 19 Jun 2017 13:55:35 GMT

TravisT — Mon, 19 Jun 2017 13:55:35 GMT

You should look at urandom. It gives you a random number within the byte size you pass into the function. This gives you a good value to scale with. I cannot remember how "good" this is for random numbers, especially for cryptography, but probably good enough for many cases.

You could do something simple, don't judge, quickly typing this without testing and to prove a point. This is not best practice.

the_range = 41-40
result = uos.urandom(1) / 256 * the_range

https://docs.pycom.io/pycom_esp32/library/uos.html#uos.urandom

Reply to [SOLVED] How to get random number in a range on Mon, 19 Jun 2017 15:03:47 GMT

amartinez — Mon, 19 Jun 2017 15:03:47 GMT

@TravisT said in How to get random number in a range:

result = uos.urandom(1) / 256 * the_range

Thanks for your reply, it gives me this error:

TypeError: unsupported types for truediv: 'bytes', 'int'

Reply to [SOLVED] How to get random number in a range on Mon, 19 Jun 2017 18:20:28 GMT

robert-hh — Mon, 19 Jun 2017 18:20:28 GMT

@amartinez you could code as follows:

import  uos
result = (uos.urandom(1)[0] / 256) + 40

Which will give you a number between 40.0 and 40.99609375. Since you started with a byte, there are only 256 different values.

Reply to [SOLVED] How to get random number in a range on Mon, 19 Jun 2017 17:42:02 GMT

livius — Mon, 19 Jun 2017 17:42:02 GMT

@amartinez
based on my old post
https://forum.pycom.io/topic/1110/random-number-generator/2

def Random():
   r = crypto.getrandbits(32)
   return ((r[0]<<24)+(r[1]<<16)+(r[2]<<8)+r[3])/4294967295.0

def RandomRange(rfrom, rto):
   return Random()*(rto-rfrom)+rfrom

Reply to [SOLVED] How to get random number in a range on Mon, 19 Jun 2017 18:52:25 GMT

amartinez — Mon, 19 Jun 2017 18:52:25 GMT

@robert-hh @livius thank you so much!