如何在Bash中生成一个范围内的随机数?


当前回答

我喜欢这个技巧:

echo ${RANDOM:0:1} # random number between 1 and 9
echo ${RANDOM:0:2} # random number between 1 and 99

...

其他回答

还可以使用shuf(可在coreutils中使用)。

shuf -i 1-100000 -n 1

没有模块化的纯Bash随机数

lowerRange=10   # inclusive
upperRange=20   # exclusive

randomNumber=$(( RANDOM * ( upperRange - lowerRange) / 32767 + lowerRange ))

是什么:

perl -e 'print int rand 10, "\n"; '

0到9之间的随机数。

echo $((RANDOM%10))

一个bash函数,使用perl生成n位的随机数。指定数字数或n个0的模板。

rand() {
  perl -E '$ARGV[0]||=""; $ARGV[0]=int($ARGV[0])||length($ARGV[0]); say join "", int(rand(9)+1)*($ARGV[0]?1:0), map { int(rand(10)) } (0..($ARGV[0]||0)-2)' $1
}

用法:

$ rand 3
381
$ rand 000
728

调用rand n的演示,n在0到15之间:

$ for n in {0..15}; do printf "%02d: %s\n" $n $(rand $n); done
00: 0
01: 3
02: 98
03: 139
04: 1712
05: 49296
06: 426697
07: 2431421
08: 82727795
09: 445682186
10: 6368501779
11: 51029574113
12: 602518591108
13: 5839716875073
14: 87572173490132
15: 546889624135868

演示调用rand n,对于n,一个长度在0到15之间的0模板

$ for n in {0..15}; do printf "%15s :%02d: %s\n" $(printf "%0${n}d" 0) $n $(rand $(printf "%0${n}d" 0)); done
              0 :00: 0
              0 :01: 0
             00 :02: 70
            000 :03: 201
           0000 :04: 9751
          00000 :05: 62237
         000000 :06: 262860
        0000000 :07: 1365194
       00000000 :08: 83953419
      000000000 :09: 838521776
     0000000000 :10: 2355011586
    00000000000 :11: 95040136057
   000000000000 :12: 511889225898
  0000000000000 :13: 7441263049018
 00000000000000 :14: 11895209107156
000000000000000 :15: 863219624761093