How can I generate random even numbers in my program?

Answer 1

import random
number_even = (random.randint(0,100000)) * 2
number_odd = (random.randint(0,100000)) * 2 +1

Answer 2

import random as r
rn = r.choice(range(0, 101, 2))
print(rn)

Answer 3

Generate a random integer that you times by two :P
( the range should be the half of your need ^^ )

Answer 4

Simple logic:

1. Generate random number
2. Multiply by 2

Answer 5

@ Mahdi Mohammadi:

There's no third 'step' parameter in the method 'randint()' ^^ ( and in Python 3.x 'print' isn't no more a statement, but a function, so it require parenthesis )

Answer 6

Fill in the blanks to create a list of numbers multiplied by 10 in the range of 5 to 9.

a = [x*10

for

x

in

range(

5

, 9)]

Answer 7

import random
random.randint.__doc__

"Return random integer in range [a,b], including both end point"

So, to generate one random number in a range:

import random
n=random.randint(100,200)

Next, to 'selectively' print many random numbers, you need to loop, and have a test condition to decide to print the generated number or not, at each loop ^^

Answer 8

import random
lower_bound = -100
upper_bound = 101
step = 2
num = random.choice(range(lower_bound, upper_bound, step ))
print(num)

if you don't want negative number than change lower_bound to 0
if you want multiple numbers than create for loop and put last two lines in it

Answer 9

import random
for i in range(10):
num = random.randint(0,40)
if num%2 == 0:
print(num)
else:
pass

Answer 10

import random
def my_generator():
while True:
yield random.randrange(0,10000,2)
a=my_generator()
print(next(a))
#every time you run this code,it generates an even number in range 0-10000(you can easily change start and stop in rangerandom.randrange(start,stop,step),here step=2 to generate even numbers)

Answer 11

import random

# random even integers from 0 to 100
# or any other range you choose to enter
print(random.choice(range(0,101,2)))

Answer 12

If you're ever dealing with evens/odds, you need to know modulus. The formula you're working with would be x%2=0 where x is the number you're testing. To filter only evens your code would be

if (x % 2 = 0)
output x; (depending on what language you use your output command will be different)

Answer 13

randrange method of random module will do it
for example:

import random
for i in range(20):
print(random.randrange(0,100,2))

Answer 14

it's very simple
import random
number_even = (random.randint(0,100000)) * 2
number_odd = (random.randint(0,100000)) * 2 +1
use this code.

Answer 15

import random
num = random.randint (0,10000)
if(num%2 == 1)
num+=1

This would be the simplest way. You use modulus 2 to check if the number is odd, if it is, you add 1. Others are telling you to multiply by 2 and this isn't wrong, but you probably are also wanting to be able to keep your number within a range and have the numbers fluctuate more to seem more random. Give it a go :)

Answer 16

from random import randint
def rnd()
num = randint(0,100) # lower bound, upper bound
if num % 2==0:
return num
else:
return(rnd())

hopefully this works out well for you :)

Answer 17

The clearest and most understandable way is to test the output number using modulus within an if statement. Something like:

import randint from random
oddList = ()
num = randint(1,100)
if num%2 == 0:
oddList.append.(num)

should get you quite a long list of even numbers.

Answer 18

Import random from library and generate a random number.
Like --
import random as r
rnum=r.randint(range(1,10))

Answer 19

Thanks VIMALRAJ K

Answer 20

Fill in the blanks to create a list of numbers multiplied by 10 in the range of 5 to 9.

a = [x*10

for

x

in

range( 5, 9 ) ]