Hi I have a list of files ex. 0.png, 1.png ... 60.png, 61.png and I want to rename all the files to 10.png,11.png ... 70.png, 71.png however I do not know how I could do that.
In bash, you can use a parameter expansion to handle the rename, e.g.
for name in *.png; do
val="${name%.png}"
val=$((val+10))
mv "$name" "$val.png"
done
Explanation
val is created from the parameter expansion "${name%.png}" which simply trims ".png" from the right-hand side of the filename.
val=$((val+10)) adds 10 to the number.
mv "$name" "$val.png" moves the file from its original name to the new name with the value increased by 10.
If you want to eliminate the intermediate val variable, you can do it all in a single expression, e.g.
for name in *.png; do
mv "$name" "$((${name%.png} + 10)).png"
done
Look things over and let me know if you have further questions.
Assuming that the filenames are of the form number.ext, this function would do the trick.
#!/bin/bash
function rename_file() {
local file=$1
local fname=$(($(echo $file | cut -d. -f1) + 10))
local ext=$(echo $file | cut -d. -f2)
mv $file $fname.$ext
}
To rename a file, call rename_file file_name in your shell script.
Related
I have a list of files named with this format:
S2_7-CHX-2-5_Chr5.bed
S2_7-CHX-2-13_Chr27.bed
S2_7-CHX-2-0_Chr1.bed
I need to loop through each file to perform a task. Previously, I had named them without the step 2 indicator ("S2"), and this format had worked perfectly:
for FASTQ in *_clean.bam; do
SAMPLE=${FASTQ%_clean.bam}
echo $SAMPLE
echo $(samtools view -c ${SAMPLE}_clean.bam)
done
But now that I have the S2 preceding what I would like to set as the variable, this returns a list of empty "SAMPLE" variables. How can I rewrite the following code to specify only S2_*.bed?
for FASTQ in S2_*.bed; do
SAMPLE=${S2_FASTQ%.bed}
echo $SAMPLE
done
Edit: I'm trying to isolate the unique name from each file, for example "7-CHX-2-13_Chr27" so that I can refer to it later. I can't use the "S2" as part of this because I want to rename the file with "S3" for the next step, and so on.
Example of what I'm trying to use it for:
for FASTQ in S2_*.bed; do
SAMPLE=${S2_FASTQ%.bed}
echo $SAMPLE
#rename each mapping position with UCSC chromosome name using sed
while IFS=, read -r f1 f2; do
#rename each file
echo " sed "s/${f1}.1/chr${f2}/g" S2_${SAMPLE}_Chr${f2}.bed > S3_${SAMPLE}_Chr${f2}.bed" >> $SCRIPT
done < $INPUT
done
The name of the variable is still $FASTQ, the S2_ is not part of the variable name, but its value.
sample=${FASTQ%.bed}
# ~~~~~|~~~~
# | | |
# Variable | What to remove
# name |
# Remove
# from the right
If you want to remove the S2_ from the $sample, use left hand side removal:
sample=${sample#S2_}
The removals can't be combined, you have to proceed in two steps.
Note that I use lower case variable names. Upper case should be reserved for environment and internal shell variables.
There are n folders in the directory named after the date, for example:
20171002 20171003 20171005 ...20171101 20171102 20171103 ...20180101 20180102
tips: Dates are not continuous.
I want to compress every three folders in each month into one compression block.
For example:
tar jcvf mytar-20171002_1005.tar.bz2 20171002 20171003 20171005
How to write a shell to do this?
You need to do a for loop on your ls variable, then parse the directory name.
dir_list=$(ls)
prev_month=""
times=0
first_dir=""
last_dir=""
dir_list=()
for i in $dir_list; do
month=${i:0:6} #here month will be year plus month
if [ "$month" = "$prev_month" ]; then
i=$(($i+1))
if [ "$i" -eq "3" ]; then
#compress here
dir_list=()
first_dir=""
last_dir=""
else
last_dir=$i
dir_list+=($i)
fi
else
if [ "$first_dir" = "" ]; then
first_dir=$i
else
#compress here
first_dir="$i"
last_dir=""
dir_list=()
fi
fi
This code is not tested and may contain syntaxe error. '#compress here' need to be replace by a loop on the array to create a string to compress.
Assuming you don't have too many directories (I think the limit is several hundred), then you can use Bash's array manipulation.
So, you first load all your directory names into a Bash array:
dirs=( $(ls) )
(I'm going to assume files have no spaces in their names, otherwise it gets a bit dicey)
Then you can use Bash's array slice syntax to pop 3 elements at a time from the array:
while [ "${#dirs[#]}" -gt 0 ]; do
dirs_to_compress=( "${dirs[#]:0:3}" )
dirs=( "${dirs[#]:3}" )
# do something with dirs_to_compress
done
The rest should be pretty easy.
You can achieve this with xargs, a bash while loop, and awk:
ls | xargs -n3 | while read line; do
tar jcvf $(echo $line | awk '{print "mytar-"$1"_"substr($NF,5,4)".tar.bz2"}') $line
done
unset folders
declare -A folders
g=3
for folder in $(ls -d */); do
folders[${folder:0:6}]+="${folder%%/} "
done
for folder in "${!folders[#]}"; do
for((i=0; i < $(echo ${folders[$folder]} | tr ' ' '\n' | wc -l); i+=g)) do
group=(${folders[$folder]})
groupOfThree=(${group[#]:i:g})
tar jcvf mytar-${groupOfThree[0]}_${groupOfThree[-1]:4:4}.tar.bz2 ${groupOfThree[#]}
done
done
This script finds all folders in the current directory, seperates them in groups of months, makes groups of at most three folders and creates a .tar.bz2 for each of them with the name you used in the question.
I tested it with those folders:
20171101 20171102 20171103 20171002 20171003 20171005 20171007 20171009 20171011 20171013 20180101 20180102
And the created tars are:
mytar-20171002_1005.tar.bz2
mytar-20171007_1011.tar.bz2
mytar-20171013_1013.tar.bz2
mytar-20171101_1103.tar.bz2
mytar-20180101_0102.tar.bz2
Hope that helps :)
EDIT: If you are using bash version < 4.2 then replace the line:
tar jcvf mytar-${groupOfThree[0]}_${groupOfThree[-1]:4:4}.tar.bz2 ${groupOfThree[#]}
by:
tar jcvf mytar-${groupOfThree[0]}_${groupOfThree[`expr ${#groupOfThree[#]} - 1`]:4:4}.tar.bz2 ${groupOfThree[#]}
That's because bash version < 4.2 doesn't support negative indices for arrays.
I need to find the latest file - filename_YYYYMMDD in the directory DIR.
The below is not working as the position is shifting each time because of the spaces between(occurring mostly at file size field as it differs every time.)
please suggest if there is other way.
report =‘ls -ltr $DIR/filename_* 2>/dev/null | tail -1 | cut -d “ “ -f9’
You can use AWK to cut the last field . like below
report=`ls -ltr $DIR/filename_* 2>/dev/null | tail -1 | awk '{print $NF}'`
Cut may not be an option here
If I understand you want to loop though each file in the directory and file the largest 'YYYYMMDD' value and the filename associated with that value, you can use simple POSIX parameter expansion with substring removal to isolate the 'YYYYMMDD' and compare against a value initialized to zero updating the latest variable to hold the largest 'YYYYMMDD' as you loop over all files in the directory. You can store the name of the file each time you find a larger 'YYYYMMDD'.
For example, you could do something like:
#!/bin/sh
name=
latest=0
for i in *; do
test "${i##*_}" -gt "$latest" && { latest="${i##*_}"; name="$i"; }
done
printf "%s\n" "$name"
Example Directory
$ ls -1rt
filename_20120615
filename_20120612
filename_20120115
filename_20120112
filename_20110615
filename_20110612
filename_20110115
filename_20110112
filename_20100615
filename_20100612
filename_20100115
filename_20100112
Example Use/Output
$ name=; latest=0; \
> for i in *; do \
> test "${i##*_}" -gt "$latest" && { latest="${i##*_}"; name="$i"; }; \
> done; \
> printf "%s\n" "$name"
filename_20120615
Where the script selects filename_20120615 as the file with the greatest 'YYYYMMDD' of all files in the directory.
Since you are using only tools provided by the shell itself, it doesn't need to spawn subshells for each pipe or utility it calls.
Give it a test and let me know if that is what you intended, let me know if your intent was different, or if you have any further questions.
I have a lot of directories:
13R
613
AB1
ACT
AMB
ANI
Each directories contains a lots of file:
20140828.13R.file.csv.gz
20140829.13R.file.csv.gz
20140830.13R.file.csv.gz
20140831.13R.file.csv.gz
20140901.13R.file.csv.gz
20131114.613.file.csv.gz
20131115.613.file.csv.gz
20131116.613.file.csv.gz
20131117.613.file.csv.gz
20141114.ab1.file.csv.gz
20141115.ab1.file.csv.gz
20141116.ab1.file.csv.gz
20141117.ab1.file.csv.gz
etc..
The purpose if to have the first file from each directories
The result what I expect is:
13R|20140828
613|20131114
AB1|20141114
Which is the name of the directories pipe the date from the filename.
I guess I need a find and head command + awk but I can't make it, I need your help.
Here what I have test it
for f in $(ls -1);do ls -1 $f/ | head -1;done
But the folder name is missing.
When I mean the first file, is the first file returned in an alphabetical order within the folder.
Thanks.
You can do this with a Bash loop.
Given:
/tmp/test
/tmp/test/dir_1
/tmp/test/dir_1/file_1
/tmp/test/dir_1/file_2
/tmp/test/dir_1/file_3
/tmp/test/dir_2
/tmp/test/dir_2/file_1
/tmp/test/dir_2/file_2
/tmp/test/dir_2/file_3
/tmp/test/dir_3
/tmp/test/dir_3/file_1
/tmp/test/dir_3/file_2
/tmp/test/dir_3/file_3
/tmp/test/file_1
/tmp/test/file_2
/tmp/test/file_3
Just loop through the directories and form an array from a glob and grab the first one:
prefix="/tmp/test"
cd "$prefix"
for fn in dir_*; do
cd "$prefix"/"$fn"
arr=(*)
echo "$fn|${arr[0]}"
done
Prints:
dir_1|file_1
dir_2|file_1
dir_3|file_1
If your definition of 'first' is different that Bash's, just sort the array arr according to your definition before taking the first element.
You can also do this with find and awk:
$ find /tmp/test -mindepth 2 -print0 | awk -v RS="\0" '{s=$0; sub(/[^/]+$/,"",s); if (s in paths) next; paths[s]; print $0}'
/tmp/test/dir_1/file_1
/tmp/test/dir_2/file_1
/tmp/test/dir_3/file_1
And insert a sort (or use gawk) to sort as desired
sort has an unique option. Only the directory should be unique, so use the first field in sorting -k1,1. The solution works when the list of files is sorted already.
printf "%s\n" */* | sort -k1,1 -t/ -u | sed 's#\(.*\)/\([0-9]*\).*#\1|\2#'
You will need to change the sed command when the date field may be followed by another number.
This works for me:
for dir in $(find "$FOLDER" -type d); do
FILE=$(ls -1 -p $dir | grep -v / | head -n1)
if [ ! -z "$FILE" ]; then
echo "$dir/$FILE"
fi
done
I have the list of committed files on svn stored in a variable as the following:
REPOS="$1"
TXN="$2"
DETAILED_FILES=`svnlook changed -t $TXN $REPOS`
DETAILED_FILES looks like:
U data0.xml A data1.xml UU all_data.xml
How can I remove all change type prefixes? such as U |data0.xml
Also, is it possible to store these in an array?
And can I get the full path of these files by svnlook?
A more proper way would be:
repos=$1
txn=$2
files=()
while read -r _ f; do
files+=( "$f" )
done < <(svnlook -t "$txn" "$repos")
Mind the quotes! (you used quotes where they are useless—yet harmless—but omitted the mandatory ones!).
Yes, just do:
FILES=( $(echo $DETAILED_FILES | cut -c 3-) )
Now FILES is an array and you can access the array elements by iterating over them:
for i in "${FILES[#]}"; do echo "$i"; done
Explicitly, ${FILES[0]} will get you the first element, ${FILES[1]} second and so on.
I am not familiar with svnlook so can not answer your second question.