generate a list of random numbers

Similar Questions

• I'm trying to generate a table of random numbers 30. After each 5 readings a new vi must open and indicate to the user than 5 readings were made. and continue with the generation of the table again.

  because I don't have a sensor now, I am currently generating a table of random numbers 30. After each 5 readings a warning should be given to the user 5 readngs are completed. This cycle must be repeated. the size of the table is 30.

  Please help me, waiting for response as soon as possible.

  Once I have the transducer, I'll take 30 analog samples and then after each 5 smaples this wraning will be displayed din a new VI

  Use a while loop with a delay time representing your sampling interval.

  Use is equal to the count Terminal to see if 4, then 4th iteration = 5th sample.

  Use a box structure. The real deal will only run on the 4th iteration.

  In the case of true place a Subvi with your message of your choice in the front panel. Go to the properties of the VI window and set ' open the front panel when it is called.

  The condition to closing of attention is not given to your description.

  Consider that rather than usign a Subvi to do this, you can use the "dialog box one/two/three button" or "display message" live in the palette "user interface and dialogue."

  Please try it out and send your own VI. Do not provide us with a working solution.

  Kind regards

• The value of the seed of the generator of random numbers in Matlab node

  Hi all

  I need to use random variables inside a Matlab Labview Commsuite node. I generate with the randn() function. Is it possible to put the seed of the random number generator such as the random numbers are reproducible?

  Hi Steve,.

  You can use randn ('seed', seeds) to initialize the random number generator. The seed must be a positive integer value. For example, randn ('seed', 3). In addition, this seed is global within the entire application. This means that if the randn() function is called MathScript multi-node that could run in parallel, random results are not guaranteed to be the same after you change any part of your code. But you should always get the same results for repeated passages of the same code.

• How LabVIEW can generate random numbers according to any pdf or histogram?

  I would like to know if there is a VI that generates a set of random numbers according to any probability density, the probability mass function or the graph histogram (this without explicitly knowing the function).

  Thank you.

  Acoustic wrote:

  I would like to know if there is a VI that generates a set of random numbers according to any probability density, the probability mass function or the graph histogram (this without explicitly knowing the function).

  If you have a given arbitrary distribution, you can simulate random data of thesholding mapping in the integral standard service with a simple random number 0... 1.

  This old post of mine should show you the General procedure and more details. (Is there even an former example). Of course, make sure that the desired function is all positive, of course.

  

  

• Medium-sized random numbers

  So what I need to do, is to generate a series of random numbers. I've got that.

  However, the tricky part is that I need to average the random number and the previous 3 numbers random and exit on a waveform graph.

  Any help or pointers would be great. Thank you

  Simple as pie. He called a VI means Pt by PT. set your buffer to 4 and it will be an average intensity absorbed with the last 3.

  

• How can I generate multiple unique random numbers?

  Hello

  I am trying to generate multiple random numbers between a given set of numbers (1-52) and do not have the same number generated twice within this group. I can compare the numbers of last and next with the function to compare, but how would I go about comparison of all numbers generated without using a huge list of shift registers...

  Any help/ideas are welcome and appreciated.

  Jason

  Solved my problem. IM passing the random number through a registry to offset to each case and build a table every time. I then searches the table for the new random number. If the number is not found I get a value of-1, another thing is an index value of 0-everything. If a comparator greater than (-1) indicates that the same number is in the table and then I can raise this matter until the same number is not found.

  Kind regards

  Jason

• Random number generator. Even and odd numbers

  I have a problem.

  I have a generator of random numbers, it goes from 0 to 20, I need to have two indicators, we show you how many numbers are pairs, and how much isn't.

  I tried to table D use the decimated 1, but as these numbers are random, it doesn´t works for me. I m new with LabView and I Don t know if there is another way to make it work, I need help, especially with a few examples.

  Thank you very much.

  
  

• How to generate random numbers from 1 to 5

  How to generate random numbers from 1 to 5

  -1110340081

  Thank you I ended up

• How the random numbers will be generated

  The "dice" in LabVIEW function is ised to generate random numbers between 0 and 1. If I create a [100000] array with random numbers between 0 and 100, the appearance of 1 to 100 is the same (about 1000 times each), but the appearance of the 0 is only 500.

  So my question is, on the basis of which will be generated random numbers?

  Mitu salvation,

  It is not a problem of the RNG, it's a problem of your function rounded!

  "To U8" allows you to convert the entire random DBL. ToU8 rounds up to the next integer. So all the number of 0-0.5 will get rounded to 0, but all the numbers from 0.5 to 1.5 will get rounded to 1. If you have twice the range of a number rounded to 1 at the beach of rounding to zero - the same goes for your "end of range" with rounding to 100. 100 should be also less likelihood (in your VI) to appear...

  To get the same probability, you should (explicitly) roundpupils before converting in U8!

  BTW. You can also search the forum to get the same answer by searching for "random"...

• I want just a bunch of random numbers (about 120 of them) from the list, select 'all' and get the computer to rearrange in CNC

  I try both notepad and wordpad. I'm in Winows XP 2003.  I want just a bunch of random numbers (about 120 of them) from the list, select "all" and reorganize in order digital computer. I can't understand how I've done it before. Google says select the balls feature, but all that is put a point in front of a certain number. There is no arrow down to select. No A - Z. No 1,2,3. what Miss me? is the Notepad or wordpad not the place to do that?

  Use Excel.

• Generating random numbers with format 18XX88YYYYY

  I've been tryig to generate random numbers with
  format 18xx88yyyyy
  here
  18, followed by two random number then 88 followed by five random number. Here 18 and 88 are set to place one, two, five and sixth position.
  
  but I am unable to get the logic.
  
  So please help
  
  
  Thanx
  
  Achyot

  Try this:

  SELECT    '18'
         || ROUND (DBMS_RANDOM.VALUE (1, 100))
         || '88'
         || ROUND (DBMS_RANDOM.VALUE (1, 100000))
    FROM DUAL
  

  - - - - - - - - - - - - - - - - - - - - -
  Kamran Agayev a. (10g OCP)
  http://kamranagayev.WordPress.com

• Random numbers and generate funds

  Hello.

  I would like to create a very simple equation random addition.

  I have: 3 boxes of dynamic text (r1_txt, r2_txt, a1_txt)

  I want the first two random numbers and 'a1_txt' to display the sum of the first two random numbers.

  Everything is perfect to the point of adding the two numbers and drop the answer in a1_text. What should I do for my last line of code?

  Thanks for your help!

  random numbers
  1.
  function randomNumbers (min:Number, max: Number) {}
  var Results:Number = Math.floor (Math.random () * max) + min;
  return results;
  }

  2.
  new_mc.addEventListener (MouseEvent.CLICK, showRandomnumber);

  3.
  function showRandomnumber(event:MouseEvent):void {}
  r1_txt. Text = randomNumbers (1,100)
  r2_txt. Text = randomNumbers (1,100)
  a1_txt. Text = Number (r1_txt.text) + Number (r2_txt.text);
  }

  Your showRandomnumber() function has a few problems.  TextFields working with channels, so whatever it is assigned to a textfield must be a string value.  You should actually get errors for the first two lines.

  function showRandomnumber(event:MouseEvent):void {}
  r1_txt. Text = String (randomNumbers (1,100));
  r2_txt. Text = String (randomNumbers (1,100));
  a1_txt. Text = String (Number (r1_txt.text) + Number (r2_txt.text));
  }

  If you change which is not fix things, then look to your textfields to the problem

• 6 random numbers without repetition

  I'm doing a generator of 6 random numbers between 1 and 48. When the user clicks a button randomly 6 numbers appear in 6 different textFeilds. The problem is that I don't want the same number twice. How can I generate 6 different random numbers without using back and out of the service?

  import flash.events.Event;

  Stop();

  btn.addEventListener (MouseEvent.CLICK, random1);

  function random1 {(evt:Event)}

  display1. Text = "";

  Display2.text = "";

  Display3.text = "";

  display4. Text = "";

  display5. Text = "";

  display6. Text = "";

  var r1 = Math.floor (Math.random () *(1+48-1)) + 1;

  var r2 = Math.floor (Math.random () *(1+48-1)) + 1;

  var A3 = Math.floor (Math.random () *(1+48-1)) + 1;

  var r4 = Math.floor (Math.random () *(1+48-1)) + 1;

  var A5 = Math.floor (Math.random () *(1+48-1)) + 1;

  var A6 = Math.floor (Math.random () *(1+48-1)) + 1;

  If (r2 = r1 | r3 == r2 | r4 == r3 | r5 == r4 | r6 = r5) {}

  return;

  }

  var list: Array = new Array();

  Liste.push (R1, R2, R3, R4, R5, R6);

  Liste.sort (Array.Numeric);

  display1. Text = String (list [0]);

  Display2.Text = String (list [1]);

  Display3.Text = String (list [2]);

  display4. Text = String (list [3]);

  display5. Text = String (list [4]);

  display6. Text = String (list [5]);

  }

  Here is the code for the approach mentioned with your included textfields

  btn.addEventListener (MouseEvent.CLICK, random1);

  function random1 {(evt:Event)}

  var nums:Array = new Array();

  Complete the table of numbers

  for (var i: uint = 1; i<=48;>
  Nums.push (i);
  }

  Shuffle (NUMS);

  Take the first six of the mixed table

  var list: Array = nums.splice (0.6);

  assign values to the textfields

  for (var j: uint = 1; j<=6;>
  This ["Display" + String (j)] .text = String (list [j]);
  }
  }

  function shuffle(a:Array) {}
  var p:int;
  var t: *;
  var ivar:int;
  for (ivar =. Length-1; Ivar > = 0; Ivar-) {}
  p = Math.Floor ((Ivar+1) * Math.Random ());
  t = a [ivar];
  a [ivar] = a [p];
  a [p] = t;
  }
  }

• random numbers to make simple bingo game

  Hi people.
  
  I do a simple game of Bingo in Director.
  
  1 generate a random number up to 75; Number = Random (75)
  2. check whether this number is on the list
  3. If it is not the number is valid and code executed to display "bingo ball.
  4 Add the number to the list
  and repeat.
  
  Is this valid? Are there problems with this approach, IE. as the numbers are called, the random number generated is more likely to be on the list already if will quickly take a? OR y at - it another way to do this?
  
  Thanks in advance
  

  The problem with your approach will be obvious when you deleted a bunch of numbers already. What you will see is the number of times you have to repeat step 2 will develop whenever you shoot a ball of bing as the number of balls drawn increases. How I would approach the opposite...

  1. create your list of numbers from 1 to 75, procedurally at the beginning:

  bingoList =]
  Repeat the witih j = 1 to 75
  bingoList.add (j)
  end repeat

  -Gives us a list containing the numbers from 1 to 75.

  2. now, all you have to do is to generate a random number based on the number of items in the list and then remove this from the list:

  listCount = bingoList.count
  bingoBallPosition = random (listCount)
  -get the value of ball bingo at the position in the list:
  bingoBall = bingoList [bingoBallPosition]
  -remove this bullet list number
  bingoList.deleteAt (bingoBallPosition)

  You'll have to do other things such as the definition of the bingoList as a global variable or property based on how you code your project...

  You may want to be present in a moviescript upwards if you are a novice programmer, you can call the managers of anywhere in your project... If you would probably create an newGame Manager who would set up your list (and other things)...

  -moviescript

  Global bingoList, bingoBall

  on newGame
  bingoList =]
  Repeat the witih j = 1 to 75
  bingoList.add (j)
  end repeat
  end newGame

  on getBingoBall
  listCount = bingoList.count
  bingoBallPosition = random (listCount)
  -get the value of ball bingo at the position in the list:
  bingoBall = bingoList [bingoBallPosition]
  -remove this bullet list number
  bingoList.deleteAt (bingoBall)
  -do other things here like the bingo ball display
  end getBingoBall

• Random numbers, not so much to chance

  I use a simple piece of code I found online to create a random number between two numbers

  new Random().nextInt(6-1)+1);
  

  I am trying to create random numbers for dice.

  Everything seems to work fine, but the numbers are not that random. I have a loop configuration to display the code above 50 times and it will make a loop over and over again the same number. Here is an example of the results

  1,1,1,1,1,1,4,4,4,4,2,2,2,2,2,1,1,1,1,1

  Therefore, it is quite difficult to create a type of yahtzee game when you always roll a yahtzee on the first roll lol.

  Is there something I'm missing here?

  I agree that the Random function generates only numbers random nickname and there are interesting ways of the algorithm of seeds, I believe that in this case, the problem is that Random is not used correctly.  I think that random is intended to be used as shown in the following fragment.  I also think that random is perfectly satisfactory for use in this case, so I would not personally recommend looking for another algorithm.

  Random myGenerator = new Random(System.currentTimeMillis());
  String thrown = ""; // Yes, this should be a StringBuffer....
  for ( int i = 0; i < 50; i++ ) {
      thrown = thrown + Integer.toString(myGenerator.nextInt(6)+1) + ", ";
  }
  System.out.println("Thrown: " + thrown);
  

  Here is the result of a few tracks:

  Lift: 5, 6, 2, 3, 1, 5, 3, 5, 4, 1, 5, 6, 3, 3, 6, 6, 5, 5, 3, 5, 1, 6, 6, 2, 1, 5, 5, 1, 6, 5, 5, 2, 5, 3, 6, 6, 5, 3, 5, 1, 2, 5, 1, 6, 2, 3, 1, 3, 2, 6,
  Lift: 4, 1, 4, 1, 5, 3, 5, 3, 2, 4, 3, 2, 5, 3, 4, 4, 3, 4, 2, 3, 2, 3, 1, 2, 2, 4, 5, 1, 5, 6, 2, 6, 6, 1, 1, 4, 5, 3, 4, 5, 6, 6, 6, 5, 6, 6, 2, 4, 5, 4,

  Edit: updated to fix the sample code fragment andgive output.