如何在3行中生成随机数-Linux Shell脚本

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 would recommend you to avoid using the shell for this. The shell is great for automating system administration tasks - eg 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.

eg 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 2

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

Answer 3

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 , eg

#!/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. eg

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