How do I generate random numbers in 3 lines - Linux Shell Script - linux

Question

I would like to write a code that can generate 3 rows of 6 random numbers spaced out, which shuffle after a given time (0.5 seconds), and no new rows are created, basically 6 random numbers keep generating in 3 rows.
The code I have so far is:
echo " "
echo " "
echo " "
for i in {1..5};
do
for i in {1..1};
do
echo -ne " $(($RANDOM % 100)) $(($RANDOM % 100)) $(($RANDOM % 100)) $(($RANDOM % 100)) $(($RANDOM % 100)) $(($RANDOM % 100))\r"
done
sleep 0.5
done
However, when I try to add the second and third row to this, it doesn't seem to work the way I want it. A sample output could look like:
45 88 85 90 44 22
90 56 34 55 32 45
58 99 42 10 48 98
and between these numbers, new ones will generate, keeping only 6 columns and 3 rows. I have tried making matrix too but it didn't work for me.

Answer 1

I don't know if you have it finish, but continuing on from the comment, I would fill an indexed array with random values between 1-100, e.g.
#!/bin/bash
for ((i = 0; i < 18; i++)); do ## fill array with random values
a[i]=$(($RANDOM % 100 + 1))
done
What you would then want is a function you could call, passing the number of values in each row (so you can output a '\n' after those digits print) and then the array values as arguments to the function to read into a local array within the function (of course, you can just use the original array without worrying about passing elements as arguments, but I prefer using local values within function to preserve values in other scopes unchanged. For that your print function could be something like:
## function to print array in shuffled order, using tput for cursor control
prnarray() {
local n=$1
local al=( ${#:2} )
local c=0
for i in $(shuf -i 0-$((${#al[#]} - 1))); do
[ "$c" -gt '0' -a $((c % n)) -eq '0' ] && printf "\n"
printf " %3d" "${al[i]}"
((c++))
done
printf "\n"
tput cuu 6 ## tput is used for cursor control to move back to top
}
Then you really don't need much else bu a loop to print the array, sleep for some period of time and then call prnarray again to overlay the output with a new shuffle. e.g.
tput sc ## save cursor position
## call prnarray 3 times with 5 sec. delay between displays
declare -i c=0
while (( c < 3 )); do
prnarray 3 ${a[#]}
((c++))
sleep 5
done
tput rc ## restore cursor position
Example Use/Output
The array will print in the same spot every 5 seconds with the same elements shuffled to different positions within the array.
$ sh randomshuf.sh
33 30 34
86 98 48
94 89 80
50 57 34
11 45 57
80 42 22
Give it a shot and let me know if you have any questions.
Note: to make it 3x6 change the lines:
tput cuu 3
and
prnarray 6 ${a[#]}
With those changed your output would resemble:
$ sh randomshuf.sh
85 9 45 14 18 16
6 59 43 19 29 58
7 89 18 72 29 29

Answer 2

I would recommend you to avoid using the shell for this. The shell is great for automating system administration tasks - e.g interact with files, directories, shell commands -, but it is also great to keep people away from learning a 'real' and most powerful programming language.
python, ruby or perl can help you out.
if all you have is a hammer, everything looks like a nail.
e.g ruby
def print_random_numbers(num)
random_numbers = []
num.times do |n|
random_numbers << rand(100)
end
puts random_numbers.join(' ')
puts
end
while true
3.times do
print_random_numbers(6)
end
sleep 0.5
end

Answer 3

I'm not sure whether I really understand your problem, but:
This gets you 6 numbers taken at random from the range 1-100
numbers=$(shuf -i 1-100 -n 6)
echo $numbers
The numbers are selected without repetition. If you want repetition, use -r.
This gives you a permutation of the numbers drawn before:
echo $numbers | tr ' ' '\n' | shuf | xargs echo