Calling random method

[sp3cpk]

Hey guys! Can anyone help me with this? I've been stuck for a while..

So I'm trying to figure out how to make it so when my condition calls on a state that case will have two methods in it and one method is selected at random

Example:

if this is my condition

do that

case that:

method1();

method2();

So my question is how would I make it select method1() or method2() at random, a 50% chance?

Edited April 5, 2017 by sp3cpk

[Charlotte]

option == random(2)

if option == 1 , do method1

else

method2

[whipz]

6 minutes ago, Charlotte said:

option == random(2)

if option == 1 , do method1

else

method2

was just about to say that you can also turn it into a switch case

[sp3cpk]

19 minutes ago, Charlotte said:

option == random(2)

if option == 1 , do method1

else

method2

got it! Thanks a bunch

[Vilius]

1 hour ago, Charlotte said:

option == random(2)

if option == 1 , do method1

else

method2

Yeah and when it returns a 0 what are you going to do then?

The chance would be 1/3 instead of 1/2 to get 1 or 2

Correct code would be:

switch(random(2)){
  case 0:
    method1();
    break;
  case 1:
    method2();
    break;
}

 

Edited April 5, 2017 by Vilius

[Final]

Just remember the random can be a zero.

EDIT: Villius sniped me.

Edited April 5, 2017 by Final

[sp3cpk]

9 minutes ago, Final said:

Just remember the random can be a zero.

EDIT: Villius sniped me.

I implemented the code... however, i did option == random(1)

it always generates a 1?;o how come?

but when i leave it at 2 it'll generate a 0-2?

I tried a switch case too same issue ;o? @Vilius

Edited April 5, 2017 by sp3cpk

[Vilius]

33 minutes ago, sp3cpk said:

I implemented the code... however, i did option == random(1)

it always generates a 1?;o how come?

but when i leave it at 2 it'll generate a 0-2?

I tried a switch case too same issue ;o? @Vilius

Random doesn't mean that it scatters everything evenly, flipping a coin a 100 times and getting tails 99 times out of 100 is still random, flip was random and the result was random.

Changing it to the examples that I posted below might not do any difference.

If you want something different you can try:

if((Math.random()<0.5)? false: true)
	method1();
else
	method2();

 

result would be after 100 iterations (true being first method is called, false being method2)

true, true, false, true, false, true, false, true, false, true, false, false, false, false, false, false, false, true, false, true, true, false, false, false, true, true, false, true, false, true, false, true, true, true, true, false, true, false, false, false, true, true, true, false, false, false, true, true, true, true, true, true, false, false, true, true, true, false, false, true, false, false, true, true, true, true, true, false, true, false, false, true, true, false, false, false, true, false, true, false, true, false, false, false, true, false, true, false, false, false, true, false, false, false, true, true, true, false, true, false, 

If you want to use something else then do:

Random random = new Random();
switch(random.nextInt(2)){
  case 0:
    	method1();
    break;
  case 1:
    	method2();
    break;
}
//OR
Random random = new Random();
if(random.nextBoolean())
  method1();
else
  method2();

results after 100 iterations:

1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0

Edited April 5, 2017 by Vilius

[Lewis]

int method;

method = random(0,1);
 

if (method == 0 {

do w.e
} else if (method == 1) {

do w.e

}

[sp3cpk]

5 minutes ago, Vilius said:

Random doesn't mean that it scatters everything evenly, flipping a coin a 100 times and getting tails 99 times out of 100 is still random, flip was random and the result was random.

If you want something different you can try:

if((Math.random()<0.5)? false: true)
	method1();
else
	method2();

result would be after 100 iterations (true being first method is called, false being method2)

 

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If you want to use something else then do:

Random random = new Random();
switch(random.nextInt(2)){
  case 0:
    	method1();
    break;
  case 1:
    	method2();
    break;
}
    

results after 100 iterations:

 

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0
1
0
1
0
0
1
0
0
0
1
0
1
0
1
0
0
0
0
1
1
0
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Oh i totally understand! I tested it for a while :P and never got 2 like over 100 results...

I will give that a try thanks

EDIT: simply changing it to method = random(0,1); worked thanks @Lewis and @Vilius and everyone else

Edited April 5, 2017 by sp3cpk

[sp3cpk]

5 minutes ago, Lewis said:

 

Didn't mean to post

Edited April 5, 2017 by sp3cpk

[Vilius]

24 minutes ago, sp3cpk said:

Oh i totally understand! I tested it for a while :P and never got 2 like over 100 results...

I will give that a try thanks

EDIT: simply changing it to method = random(0,1); worked thanks @Lewis and @Vilius and everyone else

I edited my post, it was fucked up for some reason, the spoilers dont seem to work for me haha

Edited April 5, 2017 by Vilius

[Vilius]

I have to correct myself, my first example should work with random(2) instead of random(1), because random(int i) is generating a number exclusively, meaning it generates a number between 0-1 and never returns a 1, doing random(2) would generate 0-1 without including the number 2.

random(int min, int max) is inclusive meaning that it will include the numbers defined in the parameters. Example: random(1, 3) would generate 1, 2, 3.

Edited April 5, 2017 by Vilius

[Charlotte]

 Was not really thinking about that. 

I don't code

[Lewis]

47 minutes ago, Charlotte said:

 Was not really thinking about that. 

I don't code