使用bash生成排列

Question

是否有可能编写一个bash脚本,可以从文件中读取每一行并为每个行生成排列(不重复)？使用awk/perl很好.

File
----
ab
abc


Output
------
ab
ba
abc
acb
bac
bca
cab
cba

Answer 1

我知道我比赛有点晚了,但为什么不支持扩张？

例如:

echo {a..z}{0..9}

输出:

a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 b0 b1 b2 b3 b4 b5 b6 b7 b8 b9 c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 d0 d1 d2 d3 d4 d5 d6 d7 d8 d9 e0 e1 e2 e3 e4 e5 e6 e7 e8 e9 f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 g0 g1 g2 g3 g4 g5 g6 g7 g8 g9 h0 h1 h2 h3 h4 h5 h6 h7 h8 h9 i0 i1 i2 i3 i4 i5 i6 i7 i8 i9 j0 j1 j2 j3 j4 j5 j6 j7 j8 j9 k0 k1 k2 k3 k4 k5 k6 k7 k8 k9 l0 l1 l2 l3 l4 l5 l6 l7 l8 l9 m0 m1 m2 m3 m4 m5 m6 m7 m8 m9 n0 n1 n2 n3 n4 n5 n6 n7 n8 n9 o0 o1 o2 o3 o4 o5 o6 o7 o8 o9 p0 p1 p2 p3 p4 p5 p6 p7 p8 p9 q0 q1 q2 q3 q4 q5 q6 q7 q8 q9 r0 r1 r2 r3 r4 r5 r6 r7 r8 r9 s0 s1 s2 s3 s4 s5 s6 s7 s8 s9 t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 u0 u1 u2 u3 u4 u5 u6 u7 u8 u9 v0 v1 v2 v3 v4 v5 v6 v7 v8 v9 w0 w1 w2 w3 w4 w5 w6 w7 w8 w9 x0 x1 x2 x3 x4 x5 x6 x7 x8 x9 y0 y1 y2 y3 y4 y5 y6 y7 y8 y9 z0 z1 z2 z3 z4 z5 z6 z7 z8 z9

另一个有用的例子

for X in {a..z}{a..z}{0..9}{0..9}{0..9}
    do echo $X;
done

Answer 2

纯粹的bash(使用local,更快,但不能使用下面的awk或下面的Python击败其他答案):

perm() {
  local items="$1"
  local out="$2"
  local i
  [[ "$items" == "" ]] && echo "$out" && return
  for (( i=0; i<${#items}; i++ )) ; do
    perm "${items:0:i}${items:i+1}" "$out${items:i:1}"
  done
  }
while read line ; do perm $line ; done < File

纯bash(使用子shell,慢得多):

perm() {
  items="$1"
  out="$2"
  [[ "$items" == "" ]] && echo "$out" && return
  for (( i=0; i<${#items}; i++ )) ; do
    ( perm "${items:0:i}${items:i+1}" "$out${items:i:1}" )
  done
  }
while read line ; do perm $line ; done < File

既然提问者提到Perl很好,我认为Python 2.6 +/3.X也很好:

python -c "from itertools import permutations as p ; print('\n'.join([''.join(item) for line in open('File') for item in p(line[:-1])]))"

对于Python 2.5 +/3.X:

#!/usr/bin/python2.5

# http://stackoverflow.com/questions/104420/how-to-generate-all-permutations-of-a-list-in-python/104436#104436
def all_perms(str):
    if len(str) <=1:
        yield str
    else:
        for perm in all_perms(str[1:]):
            for i in range(len(perm)+1):
                #nb str[0:1] works in both string and list contexts
                yield perm[:i] + str[0:1] + perm[i:]

print('\n'.join([''.join(item) for line in open('File') for item in all_perms(line[:-1])]))

在我的计算机上使用更大的测试文件:

First Python code
  Python 2.6:     0.038s
  Python 3.1:     0.052s
Second Python code
  Python 2.5/2.6: 0.055s
  Python 3.1:     0.072s
awk:              0.332s
Bash (local):     2.058s
Bash (subshell): 22+s

Answer 3

使用awk的更快版本

function permute(s, st,     i, j, n, tmp) {
    n = split(s, item,//)
    if (st > n) {  print s; return }
    for (i=st; i<=n; i++) {
        if (i != st) {
         tmp = item[st]; item[st] = item[i]; item[i] = tmp
         nextstr = item[1]
         for (j=2; j<=n; j++) nextstr = nextstr delim item[j]
        }else {
          nextstr = s
        }
       permute(nextstr, st+1)
       n = split(s, item, //)
   }
}
{ permute($0,1) }

用法:

$ awk -f permute.awk file

Answer 4

使用crunchutil,和bash:

while read a; do crunch 0 0 -p "$a"; done 2> /dev/null < File

输出:

ab
ba
abc
acb
bac
bca
cab
cba

这里的教程https://pentestlab.blog/2012/07/12/creating-wordlists-with-crunch/

Answer 5

有关排列示例，请参见《Perl Cookbook》。它们是基于字/数字的，但是在上面的示例中一个简单的split()/ join()就足够了。