Bash searching for words in file with same characters [duplicate] - string

Question

is it possible to write a bash script that can read in each line from a file and generate permutations (without repetition) for each? Using awk / perl is fine.
File
----
ab
abc
Output
------
ab
ba
abc
acb
bac
bca
cab
cba

Answer 1

I know I am a little late to the game but why not brace expansion?
For example:
echo {a..z}{0..9}
Outputs:
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
Another useful example:
for X in {a..z}{a..z}{0..9}{0..9}{0..9}
do echo $X;
done

Answer 2

Pure bash (using local, faster, but can't beat the other answer using awk below, or the Python below):
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
Pure bash (using subshell, much slower):
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
Since asker mentioned Perl is fine, I think Python 2.6+/3.X is fine, too:
python -c "from itertools import permutations as p ; print('\n'.join([''.join(item) for line in open('File') for item in p(line[:-1])]))"
For 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])]))
On my computer using a bigger test file:
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

Using the crunch util, and bash:
while read a; do crunch 0 0 -p "$a"; done 2> /dev/null < File
Output:
ab
ba
abc
acb
bac
bca
cab
cba
Tutorial here https://pentestlab.blog/2012/07/12/creating-wordlists-with-crunch/

Answer 4

A faster version using awk
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) }
usage:
$ awk -f permute.awk file

Answer 5

See the Perl Cookbook for permutation examples. They're word/number oriented but a simple split()/join() on your above example will suffice.

Answer 6

Bash word-list/dictionary/permutation generator:
The following Bash code generates 3 character permutation over 0-9, a-z, A-Z. It gives you (10+26+26)^3 = 238,328 words in output.
It's not very scalable as you can see you need to increase the number of for loop to increase characters in combination. It would be much faster to write such thing in assembly or C using recursion to increase speed. The Bash code is only for demonstration.
P.S.
You can populate $list variable with list=$(cat input.txt)
#!/bin/bash
list=`echo {0..9} {a..z} {A..Z}`
for c1 in $list
do
for c2 in $list
do
for c3 in $list
do
echo $c1$c2$c3
done
done
done
SAMPLE OUTPUT:
000
001
002
003
004
005
...
...
...
ZZU
ZZV
ZZW
ZZX
ZZY
ZZZ
[babil#quad[13:27:37][~]> wc -l t.out
238328 t.out

Answer 7

$ ruby -ne '$_.chomp.chars.to_a.permutation{|x| puts x.join}' file # ver 1.9.1

Answer 8

Because you can never have enogh cryptic Bash-oneliners:
while read s;do p="$(echo "$s"|sed -e 's/./&,/g' -e 's/,$//')";eval "printf "%s\\\\n" "$(eval 'echo "$(printf "{'"$p"'}%.0s" {0..'"$((${#s}-1))"'})"')"|grep '\(.\)\1*.*\1' -v";echo;done <f
It's pretty fast - at least on my machine here:
$ time while read s;do p="$(echo "$s"|sed -e 's/./&,/g' -e 's/,$//')";eval "printf "%s\\\\n" "$(eval 'echo "$(printf "{'"$p"'}%.0s" {0..'"$((${#s}-1))"'})"')"|grep '\(.\)\1*.*\1' -v";echo;done <f >/dev/null
real 0m0.021s
user 0m0.000s
sys 0m0.004s
But be aware that this one will eat a lot of memory when you go beyond 8 characters...

Answer 9

file named input:
a
b
c
d
If you want the output:
a b
a c
a d
b b
b c
b d
c c
c d
d d
You can try the following bash script:
lines=$(wc -l input | awk '{print $1}')
for ((i=1 ; i<=$lines ; i++)); do
x=$(sed -n ''$i' p' input)
sed -n ''$i',$ p' input > tmp
for j in $(cat tmp) ; do
echo $x $j
done
done

Answer 10

Hows about this one
lines="a b c"
for i in $lines; do
echo $i >tmp
for j in $lines ; do
echo $i $j
done
done
it will print
a a
a b
a c
b a
b b
b c
c a
c b
c c

Answer 11

Just a 4-lines-bash joke - permutation of 4 letters/names:
while read line
do
[ $(sed -r "s/(.)/\1\n/g" <<<$line | sort | uniq | wc -l) -eq 5 ] && echo $line
done <<<$(echo -e {A..D}{A..D}{A..D}{A..D}"\n") | sed -r "s/A/Adams /;s/B/Barth /; s/C/Cecil /; s/D/Devon /;"
Adams Barth Cecil Devon
Adams Barth Devon Cecil
...
I like Bash! :-)