Import random

Question

#range(6), random.randint(1, 49), : whats else it needs to stop printing a repeat nums? in range of 6

Answer 1

Mohamed, why the trailing `else` clause at the end of the `while` block? It's doing essentially nothing.

Anyway, kenji, what about this?

from random import randint

seen = set()
while len(seen) < 6:
seen.add(randint(1, 49))
print(seen)

Answer 2

from random import randint

a = []
for i in range(6):
x = randint(1, 49)
if x not in a:
a. append(x)
print(a)

Answer 3

from random import randint

a = []
counter = 0
while counter < 6:
x = randint(1, 49)
if x not in a:
a. append(x)
counter +=1
else:
pass
print(a)

Answer 4

thx a lot DZ, now I know

Answer 5

you are welcome :)

Answer 6

I knew that, I just wanted to do that for him to see if he will notice that or not hhhh that's all.

Answer 7

Use a set comprehension, sets can't have repeated values, i.e.,

from random import randint

uniques = {randint(1, 49) for _ in range(6)}

Answer 8

DZ seems The best ans, but the output is sometimes 4 nums or 5 nums only in range(6), can't get it?

Answer 9

Ah lol I see, I ruined it.

Answer 10

import random
for i in range (6):
print ( random.randint(1, 49))

#output, 3 , 15 , 30, 3, 16, 47
#how can it not output a num two times, e.g '3'?

Answer 11

import random
arr=[]
f=0
while True :
a=random.randint(1,49)
if a not in arr :
print(a)
arr.append(a)
f+=1
if f==6 :
break

Answer 12

@ kenji

i posted an answer before...
its seems perfect ....
it always prints 6 different numbers

Answer 13

hah got me! I just a newbie, wondering the random function should have some kind of methods to do all the work not repeated the nums, would it be without a loops.

Answer 14

Edged yours more clean, thx a lot!

Answer 15

#how about this
import random
num = list (range (1, 50))
picks = random.sample (num, k=6)
print (picks)

Answer 16

exactly what u want
be more specific please