This question already has answers here:
How to zero pad a sequence of integers in bash so that all have the same width?
(15 answers)
Closed 6 years ago.
I have a folder with several files that are named like this:
file.001.txt.gz, file.002.txt.gz, ... , file.150.txt.gz
What I want to do is use a loop to run a program with each file. I was thinking in something like this (just a sketch):
for i in {1:150}
gunzip file.$i.txt.gz
./my_program file.$i.txt output.$1.txt
gzip file.$1.txt
First of all, I don't know if something like this is gonna work, and second, I can't figure out how to keep the three digits numeration the file have ('001' instead of just '1').
Thanks a lot
The syntax for ranges in bash is
{1..150}
not {1:150}.
Moreover, if your bash is recent enough, you can add the leading zeroes:
{001..150}
The correct syntax of the for loop needs do and done.
for i in {001..150} ; do
# ...
done
It's unclear what $1 contains in your script.
To iterate over files I believe the simpler way is:
(assuming there are no files named 'file.*.txt' already in the directory and that your output file can have a different name)
for i in file.*.txt.gz; do
gunzip $i
./my_program $i $i-output.txt
gzip file.*.txt
done
Using find command:
# Path to the source directory
dir="./"
while read file
do
output="$(basename "$file")"
output="$(dirname "$file")/"${output/#file/output}
echo "$file ==> $output"
done < <(find "$dir" \
-regextype 'posix-egrep' \
-regex '.*file\.[0-9]{3}\.txt\.gz$')
The same via pipe:
find "$dir" \
-regextype 'posix-egrep' \
-regex '.*file\.[0-9]{3}\.txt\.gz$' | \
while read file
do
output="$(basename "$file")"
output="$(dirname "$file")/"${output/#file/output}
echo "$file ==> $output"
done
Sample output
/home/ruslan/tmp/file.001.txt.gz ==> /home/ruslan/tmp/output.001.txt.gz
/home/ruslan/tmp/file.002.txt.gz ==> /home/ruslan/tmp/output.002.txt.gz
(for $dir=/home/ruslan/tmp/).
Description
The scripts iterate the files in $dir directory. The $file variable is filled with the next line read from the find command.
The find command returns a list of paths corresponding to the regular expression '.*file\.[0-9]{3}\.txt\.gz$'.
The $output variable is built from two parts: basename (path without directories) and dirname (path to file's directory).
${output/#file/output} expression replaces file with output at the front end of $output variable (see Manipulating Strings)
Try-
for i in $(seq -w 1 150) #-w adds the leading zeroes
do
gunzip file."$i".txt.gz
./my_program file."$i".txt output."$1".txt
gzip file."$1".txt
done
The syntax for ranges is as choroba said, but when iterating over files you usually want to use a glob. If you know all the files have three digits in their names you can match on digits:
shopt -s nullglob
for i in file.0[0-9][0-9].txt.gz file.1[0-4][0-9] file.15[0].txt.gz; do
gunzip file.$i.txt.gz
./my_program file.$i.txt output.$i.txt
gzip file.$i.txt
done
This will only iterate through files that exist. If you use the range expression, you have to take extra care not to try to operate on files that don't exist.
for i in file.{000..150}.txt.gz; do
[[ -e "$i" ]] || continue
...otherstuff
done
Related
I have a directory with about 2000 files. How can I select a random sample of N files through using either a bash script or a list of piped commands?
Here's a script that uses GNU sort's random option:
ls |sort -R |tail -$N |while read file; do
# Something involving $file, or you can leave
# off the while to just get the filenames
done
You can use shuf (from the GNU coreutils package) for that. Just feed it a list of file names and ask it to return the first line from a random permutation:
ls dirname | shuf -n 1
# probably faster and more flexible:
find dirname -type f | shuf -n 1
# etc..
Adjust the -n, --head-count=COUNT value to return the number of wanted lines. For example to return 5 random filenames you would use:
find dirname -type f | shuf -n 5
Here are a few possibilities that don't parse the output of ls and that are 100% safe regarding files with spaces and funny symbols in their name. All of them will populate an array randf with a list of random files. This array is easily printed with printf '%s\n' "${randf[#]}" if needed.
This one will possibly output the same file several times, and N needs to be known in advance. Here I chose N=42.
a=( * )
randf=( "${a[RANDOM%${#a[#]}]"{1..42}"}" )
This feature is not very well documented.
If N is not known in advance, but you really liked the previous possibility, you can use eval. But it's evil, and you must really make sure that N doesn't come directly from user input without being thoroughly checked!
N=42
a=( * )
eval randf=( \"\${a[RANDOM%\${#a[#]}]\"\{1..$N\}\"}\" )
I personally dislike eval and hence this answer!
The same using a more straightforward method (a loop):
N=42
a=( * )
randf=()
for((i=0;i<N;++i)); do
randf+=( "${a[RANDOM%${#a[#]}]}" )
done
If you don't want to possibly have several times the same file:
N=42
a=( * )
randf=()
for((i=0;i<N && ${#a[#]};++i)); do
((j=RANDOM%${#a[#]}))
randf+=( "${a[j]}" )
a=( "${a[#]:0:j}" "${a[#]:j+1}" )
done
Note. This is a late answer to an old post, but the accepted answer links to an external page that shows terrible bash practice, and the other answer is not much better as it also parses the output of ls. A comment to the accepted answer points to an excellent answer by Lhunath which obviously shows good practice, but doesn't exactly answer the OP.
ls | shuf -n 10 # ten random files
A simple solution for selecting 5 random files while avoiding to parse ls. It also works with files containing spaces, newlines and other special characters:
shuf -ezn 5 * | xargs -0 -n1 echo
Replace echo with the command you want to execute for your files.
This is an even later response to #gniourf_gniourf's late answer, which I just upvoted because it's by far the best answer, twice over. (Once for avoiding eval and once for safe filename handling.)
But it took me a few minutes to untangle the "not very well documented" feature(s) this answer uses. If your Bash skills are solid enough that you saw immediately how it works, then skip this comment. But I didn't, and having untangled it I think it's worth explaining.
Feature #1 is the shell's own file globbing. a=(*) creates an array, $a, whose members are the files in the current directory. Bash understands all the weirdnesses of filenames, so that list is guaranteed correct, guaranteed escaped, etc. No need to worry about properly parsing textual file names returned by ls.
Feature #2 is Bash parameter expansions for arrays, one nested within another. This starts with ${#ARRAY[#]}, which expands to the length of $ARRAY.
That expansion is then used to subscript the array. The standard way to find a random number between 1 and N is to take the value of random number modulo N. We want a random number between 0 and the length of our array. Here's the approach, broken into two lines for clarity's sake:
LENGTH=${#ARRAY[#]}
RANDOM=${a[RANDOM%$LENGTH]}
But this solution does it in a single line, removing the unnecessary variable assignment.
Feature #3 is Bash brace expansion, although I have to confess I don't entirely understand it. Brace expansion is used, for instance, to generate a list of 25 files named filename1.txt, filename2.txt, etc: echo "filename"{1..25}".txt".
The expression inside the subshell above, "${a[RANDOM%${#a[#]}]"{1..42}"}", uses that trick to produce 42 separate expansions. The brace expansion places a single digit in between the ] and the }, which at first I thought was subscripting the array, but if so it would be preceded by a colon. (It would also have returned 42 consecutive items from a random spot in the array, which is not at all the same thing as returning 42 random items from the array.) I think it's just making the shell run the expansion 42 times, thereby returning 42 random items from the array. (But if someone can explain it more fully, I'd love to hear it.)
The reason N has to be hardcoded (to 42) is that brace expansion happens before variable expansion.
Finally, here's Feature #4, if you want to do this recursively for a directory hierarchy:
shopt -s globstar
a=( ** )
This turns on a shell option that causes ** to match recursively. Now your $a array contains every file in the entire hierarchy.
If you have Python installed (works with either Python 2 or Python 3):
To select one file (or line from an arbitrary command), use
ls -1 | python -c "import sys; import random; print(random.choice(sys.stdin.readlines()).rstrip())"
To select N files/lines, use (note N is at the end of the command, replace this by a number)
ls -1 | python -c "import sys; import random; print(''.join(random.sample(sys.stdin.readlines(), int(sys.argv[1]))).rstrip())" N
If you want to copy a sample of those files to another folder:
ls | shuf -n 100 | xargs -I % cp % ../samples/
make samples directory first obviously.
MacOS does not have the sort -R and shuf commands, so I needed a bash only solution that randomizes all files without duplicates and did not find that here. This solution is similar to gniourf_gniourf's solution #4, but hopefully adds better comments.
The script should be easy to modify to stop after N samples using a counter with if, or gniourf_gniourf's for loop with N. $RANDOM is limited to ~32000 files, but that should do for most cases.
#!/bin/bash
array=(*) # this is the array of files to shuffle
# echo ${array[#]}
for dummy in "${array[#]}"; do # do loop length(array) times; once for each file
length=${#array[#]}
randomi=$(( $RANDOM % $length )) # select a random index
filename=${array[$randomi]}
echo "Processing: '$filename'" # do something with the file
unset -v "array[$randomi]" # set the element at index $randomi to NULL
array=("${array[#]}") # remove NULL elements introduced by unset; copy array
done
If you have more files in your folder, you can use the below piped command I found in unix stackexchange.
find /some/dir/ -type f -print0 | xargs -0 shuf -e -n 8 -z | xargs -0 cp -vt /target/dir/
Here I wanted to copy the files, but if you want to move files or do something else, just change the last command where I have used cp.
This is the only script I can get to play nice with bash on MacOS. I combined and edited snippets from the following two links:
ls command: how can I get a recursive full-path listing, one line per file?
http://www.linuxquestions.org/questions/linux-general-1/is-there-a-bash-command-for-picking-a-random-file-678687/
#!/bin/bash
# Reads a given directory and picks a random file.
# The directory you want to use. You could use "$1" instead if you
# wanted to parametrize it.
DIR="/path/to/"
# DIR="$1"
# Internal Field Separator set to newline, so file names with
# spaces do not break our script.
IFS='
'
if [[ -d "${DIR}" ]]
then
# Runs ls on the given dir, and dumps the output into a matrix,
# it uses the new lines character as a field delimiter, as explained above.
# file_matrix=($(ls -LR "${DIR}"))
file_matrix=($(ls -R $DIR | awk '; /:$/&&f{s=$0;f=0}; /:$/&&!f{sub(/:$/,"");s=$0;f=1;next}; NF&&f{ print s"/"$0 }'))
num_files=${#file_matrix[*]}
# This is the command you want to run on a random file.
# Change "ls -l" by anything you want, it's just an example.
ls -l "${file_matrix[$((RANDOM%num_files))]}"
fi
exit 0
I use this: it uses temporary file but goes deeply in a directory until it find a regular file and return it.
# find for a quasi-random file in a directory tree:
# directory to start search from:
ROOT="/";
tmp=/tmp/mytempfile
TARGET="$ROOT"
FILE="";
n=
r=
while [ -e "$TARGET" ]; do
TARGET="$(readlink -f "${TARGET}/$FILE")" ;
if [ -d "$TARGET" ]; then
ls -1 "$TARGET" 2> /dev/null > $tmp || break;
n=$(cat $tmp | wc -l);
if [ $n != 0 ]; then
FILE=$(shuf -n 1 $tmp)
# or if you dont have/want to use shuf:
# r=$(($RANDOM % $n)) ;
# FILE=$(tail -n +$(( $r + 1 )) $tmp | head -n 1);
fi ;
else
if [ -f "$TARGET" ] ; then
rm -f $tmp
echo $TARGET
break;
else
# is not a regular file, restart:
TARGET="$ROOT"
FILE=""
fi
fi
done;
How about a Perl solution slightly doctored from Mr. Kang over here:
How can I shuffle the lines of a text file on the Unix command line or in a shell script?
$ ls | perl -MList::Util=shuffle -e '#lines = shuffle(<>); print
#lines[0..4]'
I have a directory with more than 500 files, here's a sample of the files:
random-code_aa.log
random-code_aa_r-13.log
random-code_ab.log
random-code_ae.log
random-code_ag.log
random-code_ag_r-397.log
random-code_ah.log
random-code_ac.log
random-code_ac_r-41.log
random-code_ax.log
random-code_ax_r-273.log
random-code_az.log
what I would like to do, preferably using a bash loop, is look into the directory for the *_r-*.log files and if found then try to see if similar .log files exist but without whatever is preceding _r-*.log, if found then rename the .log files into their corresponding _r-*.log files but change the r into i.
Better demonstrate with an example from the files sample above:
if "random-code_aa_r-13.log" and "random-code_aa.log" exist then
rename "random-code_aa.log" to "random-code_aa_i-13.log"
I've tried with mv and rename but nothing worked.
This simple BASH script should take care of that:
for f in *_r-*.log; do
rf="${f/_r-*log/.log}"
[[ -f "$rf" ]] && mv "$rf" "${f/_r-/_i-}"
done
You can use sed:
for file in *_r-*.log ; do
barename=`echo $file | sed 's/_r-.*/.log/'`
newname=`echo $file | sed 's/_r-\(.*\)/_i-\1/'`
if [ -f $barename ] ; then
mv $barename $newname
fi
done
You can try to improve the regexes, as it is not safe for some file names. But it should work for file names that contain the minus sign only as the separator character.
You should be able to do that with a parameter substitution:
for f in *_r-*.log
do
stem="${f%_r-*.log}
num="${f%.log}"; num="${num##_r-}"
if test -e "${stem}_aa.log"
then mv "${stem}_aa.log" "${stem}_aa-${num}.log"
fi
done
This question already has answers here:
Rename filename to another name
(3 answers)
Closed 7 years ago.
Let´s say I have a bunch of files named something like this: bsdsa120226.nai bdeqa140223.nai and I want to rename them to 120226.nai 140223.nai. How can i achieve this using the script below?
#!/bin/bash
name1=`ls *nai*`
names=`ls *nai*| grep -Po '(?<=.{5}).+'`
for i in $name1
do
for y in $names
do
mv $i $y
done
done
Solution:
name1=`ls *nai*`
for i in $name1
do
y=$(echo "$i" | grep -Po '(?<=.{5}).+')
mv $i $y
done
This:
#!/bin/bash
shopt -s extglob nullglob
for file in *+([[:digit:]]).nai; do
echo mv -nv -- "$file" "${file##+([^[:digit:]])}"
done
Remove the echo if you're happy with the mv commands.
Note. This solution does not assume that there are 5 leading characters to delete. It will delete all the leading non-numeric characters.
Using only bash, you could do this:
for file in *nai* ; do
echo mv -- "$file" "${file:5}"
done
(Remove the echo when satisfied with the output.)
Avoid ls in scripts, except for displaying information. Use plain globbing instead.
See also How do I do string manipulations in bash? for more string manipulation techniques.
Your script can't work with that structure: if you have 5 files, it will call mv five times for the first file (once for each element in the second list), five times for the second, etc. You'd need to iterate over the two sets of names in lockstep. (It also doesn't deal with things like whitespace in filenames.)
You would be better off using rename (prename on some systems) since that allows you to use Perl regular expressions to do the renaming, along the lines of:
prename 's/^.{5}//' *.nai
The reason your script is not behaving is that, for every source file, you're attempting to rename it to every target file.
If you need to limit yourself to using that script, you need to work out the single target file for each source file, something like:
#!/bin/bash
for i in *.nai; do
y=$(echo "$i" | cut -c6-)
mv "$i" "$y"
done
If your system has rename tool, it's better to go with the simple rename command,
rename 's/^.{5}//' *.nai
It just remove the first 5 characters from the file name.
OR
for i in *.nai; do mv "$i" $(grep -oP '(?<=^.{5}).+' <<< "$i"); done
I've got a directory with a few thousand files in it, named things like:
filename.ext
filename (1).ext
filename (2).ext
otherfile.ext
otherfile (1).ext
etc.
Most of the files with bracketed numbers are duplicates of the original, but in some cases they're not.
How can I keep my original files, delete the duplicates, but not lose the files that are different?
I know that I could rm *\).ext, but that obviously doesn't make sure that files match the original.
I'm using OS X, so I have a md5 program that functions sort of like md5sum in Linux, though it puts the hash at the end of the line instead of the beginning. I was thinking I could use an awk script to take the output of md5 *.ext | awk 'some script', find duplicates by md5, and delete them, but the command line is too long (bash: /sbin/md5: Argument list too long).
And I don't know what to write in the script. I was thinking of storing things in an array with this:
awk '{a[$NF]++} a[$NF]>1{sub(/).*/,""); sub(/.*(/,""); system("rm " $0);}'
But that always seems to delete my original.
What am I doing wrong? How do I do it right?
Thanks.
Your awk script deletes original files because when you sort your files, . (period) sorts after (space). SO the first file that's seen is numbered, not the original, and subsequent checks (including the one against the original) compare files to the first numbered one.
Not only does rm *\).txt fail to match the original, it loses files that may not have an original in the first place.
I wouldn't do this quite this way. Rather than checking every numbered file and verifying whether it matches an original, you can go through your list of originals, then delete the numbered files that match them.
Instead:
$ for file in *[^\)].txt; do echo "-- Found: $file"; rm -v $(basename "$file" .txt)\ \(*\).txt; done
You can expand this to check MD5's along the way. But it's more code, so I'll break it into multiple lines, in a script:
#!/bin/bash
shopt -s nullglob # Show nothing if a fileglob matches no files
for file in *[^\)].ext; do
md5=$(md5 -q "$file") # The -q option gives you only the message digest
echo "-- Found: $file ($md5)"
for duplicate in $(basename "$file" .ext)\ \(*\).ext; do
if [[ "$md5" = "$(md5 -q "$duplicate")" ]]; then
rm -v "$duplicate"
fi
done
done
As an alternative, you can probably get away with doing this a little more simply, with less CPU overhead than calculating MD5 digests. Unix and Linux have a shell tool called cmp, which is like diff without the output. So:
#!/bin/bash
shopt -s nullglob
for file in *[^\)].ext; do
for duplicate in $(basename "$file" .ext)\ \(*\).ext; do
if cmp "$file" "$duplicate"; then
rm -v "$file"
fi
done
done
If you don't need to use AWK, you could maybe do something simpler in bash:
for file in *\([0-9]*\)*; do
[ -e "$(echo "$file" | sed -e 's/ ([0-9]\+)//')" ] && rm "$file"
done
Hope this helps a little =)
I have a directory with about 2000 files. How can I select a random sample of N files through using either a bash script or a list of piped commands?
Here's a script that uses GNU sort's random option:
ls |sort -R |tail -$N |while read file; do
# Something involving $file, or you can leave
# off the while to just get the filenames
done
You can use shuf (from the GNU coreutils package) for that. Just feed it a list of file names and ask it to return the first line from a random permutation:
ls dirname | shuf -n 1
# probably faster and more flexible:
find dirname -type f | shuf -n 1
# etc..
Adjust the -n, --head-count=COUNT value to return the number of wanted lines. For example to return 5 random filenames you would use:
find dirname -type f | shuf -n 5
Here are a few possibilities that don't parse the output of ls and that are 100% safe regarding files with spaces and funny symbols in their name. All of them will populate an array randf with a list of random files. This array is easily printed with printf '%s\n' "${randf[#]}" if needed.
This one will possibly output the same file several times, and N needs to be known in advance. Here I chose N=42.
a=( * )
randf=( "${a[RANDOM%${#a[#]}]"{1..42}"}" )
This feature is not very well documented.
If N is not known in advance, but you really liked the previous possibility, you can use eval. But it's evil, and you must really make sure that N doesn't come directly from user input without being thoroughly checked!
N=42
a=( * )
eval randf=( \"\${a[RANDOM%\${#a[#]}]\"\{1..$N\}\"}\" )
I personally dislike eval and hence this answer!
The same using a more straightforward method (a loop):
N=42
a=( * )
randf=()
for((i=0;i<N;++i)); do
randf+=( "${a[RANDOM%${#a[#]}]}" )
done
If you don't want to possibly have several times the same file:
N=42
a=( * )
randf=()
for((i=0;i<N && ${#a[#]};++i)); do
((j=RANDOM%${#a[#]}))
randf+=( "${a[j]}" )
a=( "${a[#]:0:j}" "${a[#]:j+1}" )
done
Note. This is a late answer to an old post, but the accepted answer links to an external page that shows terrible bash practice, and the other answer is not much better as it also parses the output of ls. A comment to the accepted answer points to an excellent answer by Lhunath which obviously shows good practice, but doesn't exactly answer the OP.
ls | shuf -n 10 # ten random files
A simple solution for selecting 5 random files while avoiding to parse ls. It also works with files containing spaces, newlines and other special characters:
shuf -ezn 5 * | xargs -0 -n1 echo
Replace echo with the command you want to execute for your files.
This is an even later response to #gniourf_gniourf's late answer, which I just upvoted because it's by far the best answer, twice over. (Once for avoiding eval and once for safe filename handling.)
But it took me a few minutes to untangle the "not very well documented" feature(s) this answer uses. If your Bash skills are solid enough that you saw immediately how it works, then skip this comment. But I didn't, and having untangled it I think it's worth explaining.
Feature #1 is the shell's own file globbing. a=(*) creates an array, $a, whose members are the files in the current directory. Bash understands all the weirdnesses of filenames, so that list is guaranteed correct, guaranteed escaped, etc. No need to worry about properly parsing textual file names returned by ls.
Feature #2 is Bash parameter expansions for arrays, one nested within another. This starts with ${#ARRAY[#]}, which expands to the length of $ARRAY.
That expansion is then used to subscript the array. The standard way to find a random number between 1 and N is to take the value of random number modulo N. We want a random number between 0 and the length of our array. Here's the approach, broken into two lines for clarity's sake:
LENGTH=${#ARRAY[#]}
RANDOM=${a[RANDOM%$LENGTH]}
But this solution does it in a single line, removing the unnecessary variable assignment.
Feature #3 is Bash brace expansion, although I have to confess I don't entirely understand it. Brace expansion is used, for instance, to generate a list of 25 files named filename1.txt, filename2.txt, etc: echo "filename"{1..25}".txt".
The expression inside the subshell above, "${a[RANDOM%${#a[#]}]"{1..42}"}", uses that trick to produce 42 separate expansions. The brace expansion places a single digit in between the ] and the }, which at first I thought was subscripting the array, but if so it would be preceded by a colon. (It would also have returned 42 consecutive items from a random spot in the array, which is not at all the same thing as returning 42 random items from the array.) I think it's just making the shell run the expansion 42 times, thereby returning 42 random items from the array. (But if someone can explain it more fully, I'd love to hear it.)
The reason N has to be hardcoded (to 42) is that brace expansion happens before variable expansion.
Finally, here's Feature #4, if you want to do this recursively for a directory hierarchy:
shopt -s globstar
a=( ** )
This turns on a shell option that causes ** to match recursively. Now your $a array contains every file in the entire hierarchy.
If you have Python installed (works with either Python 2 or Python 3):
To select one file (or line from an arbitrary command), use
ls -1 | python -c "import sys; import random; print(random.choice(sys.stdin.readlines()).rstrip())"
To select N files/lines, use (note N is at the end of the command, replace this by a number)
ls -1 | python -c "import sys; import random; print(''.join(random.sample(sys.stdin.readlines(), int(sys.argv[1]))).rstrip())" N
If you want to copy a sample of those files to another folder:
ls | shuf -n 100 | xargs -I % cp % ../samples/
make samples directory first obviously.
MacOS does not have the sort -R and shuf commands, so I needed a bash only solution that randomizes all files without duplicates and did not find that here. This solution is similar to gniourf_gniourf's solution #4, but hopefully adds better comments.
The script should be easy to modify to stop after N samples using a counter with if, or gniourf_gniourf's for loop with N. $RANDOM is limited to ~32000 files, but that should do for most cases.
#!/bin/bash
array=(*) # this is the array of files to shuffle
# echo ${array[#]}
for dummy in "${array[#]}"; do # do loop length(array) times; once for each file
length=${#array[#]}
randomi=$(( $RANDOM % $length )) # select a random index
filename=${array[$randomi]}
echo "Processing: '$filename'" # do something with the file
unset -v "array[$randomi]" # set the element at index $randomi to NULL
array=("${array[#]}") # remove NULL elements introduced by unset; copy array
done
If you have more files in your folder, you can use the below piped command I found in unix stackexchange.
find /some/dir/ -type f -print0 | xargs -0 shuf -e -n 8 -z | xargs -0 cp -vt /target/dir/
Here I wanted to copy the files, but if you want to move files or do something else, just change the last command where I have used cp.
This is the only script I can get to play nice with bash on MacOS. I combined and edited snippets from the following two links:
ls command: how can I get a recursive full-path listing, one line per file?
http://www.linuxquestions.org/questions/linux-general-1/is-there-a-bash-command-for-picking-a-random-file-678687/
#!/bin/bash
# Reads a given directory and picks a random file.
# The directory you want to use. You could use "$1" instead if you
# wanted to parametrize it.
DIR="/path/to/"
# DIR="$1"
# Internal Field Separator set to newline, so file names with
# spaces do not break our script.
IFS='
'
if [[ -d "${DIR}" ]]
then
# Runs ls on the given dir, and dumps the output into a matrix,
# it uses the new lines character as a field delimiter, as explained above.
# file_matrix=($(ls -LR "${DIR}"))
file_matrix=($(ls -R $DIR | awk '; /:$/&&f{s=$0;f=0}; /:$/&&!f{sub(/:$/,"");s=$0;f=1;next}; NF&&f{ print s"/"$0 }'))
num_files=${#file_matrix[*]}
# This is the command you want to run on a random file.
# Change "ls -l" by anything you want, it's just an example.
ls -l "${file_matrix[$((RANDOM%num_files))]}"
fi
exit 0
I use this: it uses temporary file but goes deeply in a directory until it find a regular file and return it.
# find for a quasi-random file in a directory tree:
# directory to start search from:
ROOT="/";
tmp=/tmp/mytempfile
TARGET="$ROOT"
FILE="";
n=
r=
while [ -e "$TARGET" ]; do
TARGET="$(readlink -f "${TARGET}/$FILE")" ;
if [ -d "$TARGET" ]; then
ls -1 "$TARGET" 2> /dev/null > $tmp || break;
n=$(cat $tmp | wc -l);
if [ $n != 0 ]; then
FILE=$(shuf -n 1 $tmp)
# or if you dont have/want to use shuf:
# r=$(($RANDOM % $n)) ;
# FILE=$(tail -n +$(( $r + 1 )) $tmp | head -n 1);
fi ;
else
if [ -f "$TARGET" ] ; then
rm -f $tmp
echo $TARGET
break;
else
# is not a regular file, restart:
TARGET="$ROOT"
FILE=""
fi
fi
done;
How about a Perl solution slightly doctored from Mr. Kang over here:
How can I shuffle the lines of a text file on the Unix command line or in a shell script?
$ ls | perl -MList::Util=shuffle -e '#lines = shuffle(<>); print
#lines[0..4]'