I'm using C# and I need to generate a random 10 digit number. So far, I've only had luck finding examples indicating min maximum value. How would i go about generating a random number that is 10 digits, which can begin with 0, (initially, I was hoping for random.Next(1000000000,9999999999) but I doubt this is what I want).
My code looks like this right now:
[WebMethod]
public string GenerateNumber()
{
Random random = new Random();
return random.Next(?);
}
**Update ended up doing like so,
[WebMethod]
public string GenerateNumber()
{
Random random = new Random();
string r = "";
int i;
for (i = 1; i < 11; i++)
{
r += random.Next(0, 9).ToString();
}
return r;
}
Use this to create random digits with any specified length
public string RandomDigits(int length)
{
var random = new Random();
string s = string.Empty;
for (int i = 0; i < length; i++)
s = String.Concat(s, random.Next(10).ToString());
return s;
}
try (though not absolutely exact)
Random R = new Random();
return ((long)R.Next (0, 100000 ) * (long)R.Next (0, 100000 )).ToString ().PadLeft (10, '0');
To get the any digit number without any loop, use Random.Next with the appropriate limits [100...00, 9999...99].
private static readonly Random _rdm = new Random();
private string PinGenerator(int digits)
{
if (digits <= 1) return "";
var _min = (int)Math.Pow(10, digits - 1);
var _max = (int)Math.Pow(10, digits) - 1;
return _rdm.Next(_min, _max).ToString();
}
This function calculated the lower and the upper bounds of the nth digits number.
To generate the 10 digit number use it like this:
PinGenerator(10)
If you want ten digits but you allow beginning with a 0 then it sounds like you want to generate a string, not a long integer.
Generate a 10-character string in which each character is randomly selected from '0'..'9'.
private void button1_Click(object sender, EventArgs e)
{
Random rand = new Random();
long randnum2 = (long)(rand.NextDouble() * 9000000000) + 1000000000;
MessageBox.Show(randnum2.ToString());
}
I came up with this method because I dont want to use the Random method :
public static string generate_Digits(int length)
{
var rndDigits = new System.Text.StringBuilder().Insert(0, "0123456789", length).ToString().ToCharArray();
return string.Join("", rndDigits.OrderBy(o => Guid.NewGuid()).Take(length));
}
hope this helps.
// ten digits
public string CreateRandomNumber
{
get
{
//returns 10 digit random number (Ticks returns 16 digit unique number, substring it to 10)
return DateTime.UtcNow.Ticks.ToString().Substring(8);
}
}
(1000000000,9999999999) is not random - you're mandating that it cannot begin with a 1, so you've already cut your target base by 10%.
Random is a double, so if you want a integer, multiply it by 1,000,000,000, then drop the figures after the decimal place.
private static Random random = new Random((int)DateTime.Now.Ticks);//thanks to McAden
public long LongBetween(long maxValue, long minValue)
{
return (long)Math.Round(random.NextDouble() * (maxValue - minValue - 1)) + minValue;
}
I tried to write a fast one:
private int GetNDigitsRandomNumber(int digits)
{
var min = 1;
for (int i = 0; i < digits-1; i++)
{
min *= 10;
}
var max = min * 10;
return _rnd.Next(min, max);
}
Here is a simple solution using format string:
string r = $"{random.Next(100000):00000}{random.Next(100000):00000}";
Random 10 digit number (with possible leading zeros) is produced as union of two random 5 digit numbers. Format string "00000" means leading zeros will be appended if number is shorter than 5 digits (e.g. 1 will be formatted as "00001").
For more about the "0" custom format specifier please see the documentation.
Random random = new Random();
string randomNumber = string.Join(string.Empty, Enumerable.Range(0, 10).Select(number => random.Next(0, 9).ToString()));
To generate a random 10 digit number in C#
Random RndNum = new Random();
int RnNum = RndNum.Next(1000000000,9999999999);
Related
I'm very new to coding and I just can't wrap my head around Loops/Arrays/Randoms. I understand the concept but when it comes to applying it, I'm just lost.
Here I'm trying to generate 100 random numbers between 1-1000 and it has to output the maximum value. Here's my code so far:
Random rnd = new Random();
int nums = rnd.Next(0, 1001);
for (int i = 1; i <= 100; i++)
{
}
Console.WriteLine(nums);
Console.ReadLine();
It's only giving me one number. :(
I'd greatly appreciate any help!
Thanks!
You can accumulate your random generated number to the array and then by using Max function of the array you can find the maximum value
class Program
{
public static void Main(string[] args)
{
Random rnd = new Random();
int[] intArr = new int[100];
for (int i = 0; i < intArr.Length; i++)
{
int num = rnd.Next(1, 1000);
intArr[i] = num;
Console.WriteLine(num);
}
Console.WriteLine();
int maxNum = intArr.Max();
Console.WriteLine("The max num is:" + maxNum);
Console.ReadLine();
}
}
Click to watch online demo
You need to call rnd.Next() inside loop.
Random rnd = new Random();
for (int i = 1; i <= 100; i++)
{
int nums = rnd.Next(0, 1001);
Console.WriteLine(nums);
}
Console.ReadLine();
A good approach would be initializing a variable that stores your max. Then generate a random number within your iterative block and if it is greater than your max, set it as the new max.
Random r = new Random();
int max = 0; //declare our max variable
for(int i = 0; i < 100; i++)
{
int rand = r.Next(0, 1001);
if(rand > max) //if the new random value is greater than our max, set max = rand
max = rand;
}
Console.WriteLine(max); //Output the maximum value
Console.ReadLine();
If you want to output every random value and then output the max out of all the values generated, simply modify the code above by outputting rand within your loop as well.
Hope this helps!
I am not sure, are you asking like this?
Random random = new Random();
int[] nums = new int[100];
// when for loop ends, nums are full of 100 numbers
for (int i = 0; i < nums.Length; i++)
{
int newNum = random.Next(1, 1000);
// show every number
Console.WriteLine(newNum);
nums[i] = newNum;
}
// get the max number
var maxNum = nums.Max();
Console.WriteLine(maxNum);
If you want to see the code for Loops/Arrays/Randoms all working together you can use the below with the comments walking through what each line is doing (Working .NET Fiddle Example)
public static void Main()
{
// Pass in what range we want our randomly generated numbers to be in
// In your case, between 1 - 1000 and we want to create 100 of them.
//(See GenerateRandomNumbers())
var random = GenerateRandomNumbers(1, 1000, 100);
//Take our newly returned randomly created numbers and
//pass them to our GetMaxNumber method so it can find the Max number
//See (GetMaxNumber())
var result = GetMaxNumber(random);
//We now have our max number; print it to the Console.
Console.WriteLine("Max: " + result);
}
public static int GetMaxNumber(params int[] inputs)
{
//Create a variable that will store the largest number we find in our array
int max = inputs[0];
//Iterate (loop) through all of the 100 values in our array that we passed in
//Here we define "input" which will hold the value for each value in inputs as we check
//if the value of input is greater than our current value of max. If it is greater than our
//current value of max, then we need to update max to now be equal to the value of our input.
//Note: it will do this comparison 100 times beginning with the first value in the inputs array
foreach (var input in inputs)
{
if (input > max)
{
//input's value is greater than the current value of max; update max so that it is equal to the current value of input.
max = input;
}
//no more code; return to top of foreach loop and set input to the next value in inputs
}
//When we get here, it means our foreach loop has completed going through and comparing all 100 values of inputs to see which value is the largest.
//now return this value to Main()
return max;
}
public static int[] GenerateRandomNumbers(int beginRange, int endRange, int maxNumbers)
{
// Instantiate random number generator
Random rnd = new Random();
//Generate and display
int[] intArr = new int[maxNumbers];
//Generate 100 numbers with numbers between 1 and 1000
for (int i = 0; i < intArr.Length; i++)
{
int num = rnd.Next(beginRange, endRange);
intArr[i] = num;
}
return intArr;
}
I am trying to do string manipulation. Here's my C# code :
static void Main(string[] args)
{
string input;
string output;
int length;
Console.WriteLine("input = ");
input = Console.ReadLine();
length = input.Length;
if ((input != "") || (length != 0))
{
Random randem = new Random();
int i = -1; //because I do not want the first number to be replaced by the random number
char[] characters = input.ToCharArray();
while (i < length)
{
int num = randem.Next(0, 9);
char num1 = Convert.ToChar(num);
i = i + 2; //so that every next character will be replaced by random number.. :D
characters[i] = num1; //*error* here
}
output = new string(characters);
Console.WriteLine(output);
}
For example:
User input : "i_love_to_eat_fish"
Desired output : "i2l4v1_9o5e8t7f8s2"
notice that the only unchanged character in
the char[] characters is : "i l v _ o e t f s". (desired output from the program)
I've already tried using this code, but still,
keep getting error at characters[i] = num1;
Am I on the right track?
I'm guessing the error you get is IndexOutOfRangeException this is because of the i = i + 2;. The while makes sure that i is less than length, but then adding 2 could result in it being more. Just add a check that it isn't beyond the length.
i = i + 2;
if(i < length)
characters[i] = num1;
Or just change to a for loop.
Random randem = new Random();
char[] characters = input.ToCharArray();
for(int i = 1; i < length; i += 2)
{
int num = randem.Next(1, 10); // max value is exclusive
char num1 = num.ToString()[0];
characters[i] = num1;
}
output = new string(characters);
Console.WriteLine(output);
Also as Shar1er80 points out you're currently converting the digit to the char that has the same ASCII value, and not the the actual characters that represent the digit. The digits 0-9 are represented by the the values 48-57. You can change the call to Random.Next to be:
int num = randem.Next(48, 58); // The upper bound is exclusive, not inclusive
char num1 = (char)num;
Or as Shar1er80 does it
int num = randem.Next(0,10) // Assumming you want digits 0-9
char num1 = num.ToString[0];
Also note that the max value for Random.Next is exclusive, so if you want to include the possibility of using a 9 you have to use an upper bound that is 1 greater than the greatest value you want.
Whenever you reach i = 17 you add 2 to i . That makes i = 19 with length of input equal to 18 that causes out of range exception.
The error you are getting is IndexOutOfTheRangeException, which explains everthing in itself.
It means that index you are feeding to array in the loop is going beyond its length-1 (as arrays have 0-based indexing)
So when you do i+2, you need to check if i+2 is not exceeding i.length-1 at any point of time; which does in your loop.
In general just check if you are supplying indexes between 0 and Array.Length-1
its because you start at index -1, and characters doesn't contain an index of -1.
EDIT: Sorry no the corrct answer is it must be while(i < length - 2)
Change this line
char num1 = Convert.ToChar(num);
To
char num1 = num.ToString()[0];
Then... Put
characters[i] = num1;
In an if block
if (i < length)
characters[i] = num1;
I want to generate an example of a valid input by Regex pattern. I'm programming with C# .Net . Like this:
//this emthod doesn't exists, its an example of funcionality that I want.
Regex.GenerateInputExample("^[0-9]{15}$");
So, this example gives-me a possible value, like 000000000000000. How to do this?
So, this problem would take some time to solve, since the functionality is not built in. I'll give a general way to solve it:
Using an ascii (or unicode) chart find out the character codes that correspond to the characters you are using for your regex (65 =A, 69 = D, etc)
Create a random function with those bounds. Multiple bounds would take a little more trickery (A-Z =26, 0-9 = 10, so a random number from 0- 35)
Random random = new Random();
int randomNumber = random.Next(65, 70); // this generates a random number including the bounds of 65-69)
char temp = (char)random;
Next you would take the randomly generated characters and add them together into a string.
int lowerBound = 65, upperBound =69;
int length = 6;
char temp;
int randomNumber;
string result= "";
Random rand = new Random();
for (int a = 0; a <= length; a++)
{
randomNumber = rand.Next(lowerBound, upperBound);
temp = (char)randomNumber;
result = result + temp;
} //result is the indirect regex generated string
Indirectly giving you a regex generated string.
The next step is parsing information out of a regex. I've provided a simple case below that will not work for every regex, due to regex complexity.
Regex bob = new Regex("[A-Z]");
int lowerBound = Convert.ToInt32(bob.ToString()[1]);
int upperBound = Convert.ToInt32(bob.ToString()[3]);
int length = 6; //length of the string to be generated
char temp;
int randomNumber;
string result= "";
Random rand = new Random();
for (int a = 0; a <= length; a++)
{
randomNumber = rand.Next(lowerBound, upperBound);
temp = (char)randomNumber;
result = result + temp;
}
( This process could be streamlined into class and utilized etc)
I'm looking for some succinct, modern C# code to generate a random double number between 1.41421 and 3.14159. where the number should be [0-9]{1}.[0-9]{5} format.
I'm thinking some solution that utilizes Enumerable.Range somehow may make this more succinct.
You can easily define a method that returns a random number between two values:
private static readonly Random random = new Random();
private static double RandomNumberBetween(double minValue, double maxValue)
{
var next = random.NextDouble();
return minValue + (next * (maxValue - minValue));
}
You can then call this method with your desired values:
RandomNumberBetween(1.41421, 3.14159)
Use something like this.
Random random = new Random()
int r = random.Next(141421, 314160); //+1 as end is excluded.
Double result = (Double)r / 100000.00;
Random r = new Random();
var number = r.Next(141421, 314160) / 100000M;
Also you can't force decimal number to match your pattern. E.g. if you have 1.5 number it will not match 1.50000 format. So, you should format result as string:
string formattedNumber = number.ToString("0.00000");
I used this. I hope this helps.
Random Rnd = new Random();
double RndNum = (double)(Rnd.Next(Convert.ToInt32(LatRandMin.Value), Convert.ToInt32(LatRandMax.Value)))/1000000;
Check out the following link for ready-made implementations that should help:
MathNet.Numerics, Random Numbers and Probability Distributions
The extensive distributions are especially of interest, built on top of the Random Number Generators (MersenneTwister, etc.) directly derived from System.Random, all providing handy extension methods (e.g. NextFullRangeInt32, NextFullRangeInt64, NextDecimal, etc.). You can, of course, just use the default SystemRandomSource, which is simply System.Random embellished with the extension methods.
Oh, and you can create your RNG instances as thread safe if you need it.
Very handy indeed!
here my solution, it's not pretty but it works well
Random rnd = new Random();
double num = Convert.ToDouble(rnd.Next(1, 15) + "." + rnd.Next(1, 100));
Console.WriteLine(num);
Console.ReadKey();
JMH BJHBJHHJJ const int SIZE = 1;
int nnOfTosses,
headCount = 0, tailCount = 0;
Random tossResult = new Random();
do
{
Console.Write("Enter integer number (>=2) coin tosses or 0 to Exit: ");
if (!int.TryParse(Console.ReadLine(), out nnOfTosses))
{
Console.Write("Invalid input");
}
else
{
//***//////////////////
// To assign a random number to each element
const int ROWS = 3;
double[] scores = new double[ROWS];
Random rn = new Random();
// Populate 2D array with random values
for (int row = 0; row < ROWS; row++)
{
scores[row] = rn.NextDouble();
}
//***//////////////////////////
for (int i = 0; i < nnOfTosses; i++)
{
double[] tossResult = new double[i];
tossResult[i]= tossResult.nextDouble();
}
Console.Write("Number of Coin Tosses = " + nnOfTosses);
Console.Write("Fraction of Heads = ");
Console.Write("Fraction of Tails = ");
Console.Write("Longest run is ");
}
} while (nnOfTosses != 0);
Console.ReadLine();
I want to make a method that takes a variable of type int or long and returns an array of ints or longs, with each array item being a group of 3 digits. For example:
int[] i = splitNumber(100000);
// Outputs { 100, 000 }
int[] j = splitNumber(12345);
// Outputs { 12, 345 }
int[] k = splitNumber(12345678);
// Outputs { 12, 345, 678 }
// Et cetera
I know how to get the last n digits of a number using the modulo operator, but I have no idea how to get the first n digits, which is the only way to make this method that I can think of. Help please!
Without converting to string:
int[] splitNumber(int value)
{
Stack<int> q = new Stack<int>();
do
{
q.Push(value%1000);
value /= 1000;
} while (value>0);
return q.ToArray();
}
This is simple integer arithmetic; first take the modulo to get the right-most decimals, then divide to throw away the decimals you already added. I used the Stack to avoid reversing a list.
Edit: Using log to get the length was suggested in the comments. It could make for slightly shorter code, but in my opinion it is not better code, because the intent is less clear when reading it. Also, it might be less performant due to the extra Math function calls. Anyways; here it is:
int[] splitNumber(int value)
{
int length = (int) (1 + Math.Log(value, 1000));
var result = from n in Enumerable.Range(1,length)
select ((int)(value / Math.Pow(1000,length-n))) % 1000;
return result.ToArray();
}
By converting into a string and then into int array
int number = 1000000;
string parts = number.ToString("N0", new NumberFormatInfo()
{
NumberGroupSizes = new[] { 3 },
NumberGroupSeparator = "."
});
By using Maths,
public static int[] splitNumberIntoGroupOfDigits(int number)
{
var numberOfDigits = Math.Floor(Math.Log10(number) + 1); // compute number of digits
var intArray = new int[Convert.ToInt32(numberOfDigits / 3)]; // we know the size of array
var lastIndex = intArray.Length -1; // start filling array from the end
while (number != 0)
{
var lastSet = number % 1000;
number = number / 1000;
if (lastSet == 0)
{
intArray[lastIndex] = 0; // set of zeros
--lastIndex;
}
else if (number == 0)
{
intArray[lastIndex] = lastSet; // this could be your last set
--lastIndex;
}
else
{
intArray[lastIndex] = lastSet;
--lastIndex;
}
}
return intArray;
}
Try converting it to string first and do the parsing then convert it back to number again
Convert to string
Get length
If length modulus 3 == 0
String substring it into ints every 3
else if
Find remainder such as one or two left over
Substring remainder off of front of string
Then substring by 3 for the rest
You can first find out how large the number is, then use division to get the first digits, and modulo to keep the rest:
int number = 12345678;
int len = 1;
int div = 1;
while (number >= div * 1000) {
len++;
div *= 1000;
}
int[] result = new int[len];
for (int i = 0; i < result.Length; i++) {
result[i] = number / div;
number %= div;
div /= 1000;
}
You can use this with the System.Linq namespace from .NET 3.5 and above:
int[] splitNumber(long value)
{
LinkedList<int> results = new LinkedList<int>();
do
{
int current = (int) (value % 1000);
results.AddFirst(current);
value /= 1000;
} while (value > 0);
return results.ToArray();// Extension method
}
I use LinkedList<int> to avoid having to Reverse a list before returning. You could also use Stack<int> for the same purpose, which would only require .NET 2.0:
int[] splitNumber(long value)
{
Stack<int> results = new Stack<int>();
do
{
int current = (int) (value % 1000);
results.Push(current);
value /= 1000;
} while (value > 0);
return results.ToArray();
}