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
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