how to create random vector?

hello,

I would like to know how I create a vector with random and integers 25 digit using round and randn?

Thanks.

Answer 1

[Karim]

RandomDouble = rand(25,1) % i.e. pick 25 random values between 0 and 1

RandomDouble = 25×1

0.2261 0.1115 0.6008 0.0399 0.7629 0.1976 0.1520 0.7056 0.1548 0.8280

RandomIntegers = randi(1000,25,1) % i.e. pick 25 integers between 1 and 1000

RandomIntegers = 25×1

543 552 914 309 791 211 947 842 821 615

[idan]

Thanks.

Answer 2

[Voss]

A vector of 5 random 19-digit numbers (change n_digits to 25 to get 25-digit numbers):

n_numbers = 5;
n_digits = 19;

digits = [randi(9,n_numbers,1) randi(10,n_numbers,n_digits-1)-1];

numbers = sum(digits.*10.^(n_digits-1:-1:0),2);

format long
disp(numbers);

   1.0e+18 *

   5.975211767544413
   3.355820591482017
   2.308621702816259
   6.853048767863059
   7.175091024278929

sprintf('%19d\n',numbers)

ans = 
    '5975211767544413184
     3355820591482017280
     2308621702816258560
     6853048767863058432
     7175091024278929408
     '

[Steven Lord]

Those numbers exceed flintmax so you're not generating the numbers you think you are. There are some (many) 19-digit numbers that your code cannot generate because they can't be represented in double precision.

rng default % for reproducibility
n_numbers = 1; % Let's just make 1 to start
n_digits = 19;
digits = [randi(9,n_numbers,1) randi(10,n_numbers,n_digits-1)-1]

digits = 1×19

8 9 1 9 6 0 2 5 9 9 1 9 9 4 8 1 4 9 7

That second randi call could be simplified a bit (to eliminate the subtraction) by specifying the first input as a vector.

randi([0 9], 1, 3)

ans = 1×3

9 6 0

Anyway, let's look at the number you created.

theNumber = polyval(digits, 10) % Another way to turn digits into a number

theNumber = 8.9196e+18

fromDouble = sprintf('%19d', theNumber)

fromDouble = '8919602599199481856'

Is it larger than flintmax? Yes.

isLargerThanFlintmax = theNumber > flintmax

isLargerThanFlintmax = logical
   1

Let's convert the numeric digits into the corresponding characters and compare.

fromDigits = sprintf('%s', char(digits + '0'))

fromDigits = '8919602599199481497'

Do they match?

[fromDouble; fromDigits]

ans = 2×19 char array
    '8919602599199481856'
    '8919602599199481497'

Nope. Not all numbers are representable in double precision when the magnitude is in the vicinity of theNumber. The distance from theNumber to the next representable number is:

distance = eps(theNumber)

distance = 1024

So adding something less than that distance to theNumber doesn't "get over the hump" to the next largest number.

isequal(theNumber, theNumber + 400)

ans = logical
   1