I have a set of files that I would like to rename by using new filenames stored in a txt file. My original files:
7170M
7172M
7187P
7192N
7198P
I would like to add to each filename specific new names so that the above files become:
1956_26_7170M
1962_12_7172M
1989_32_7187P
1986_22_7192N
1943_13_7198P
I created a document new_names.txt containing new filenames and tried the following.
for f in *.txt; do mv "$f" "$new_name"; done < new_names.txt
But it changes only the first filename.
You can only store the new parts of the names in the file (provided you'll process the files in the sorted order):
1956_26
1962_12
1989_32
1986_22
1943_13
Then, iterate over the files, read one line for each of them:
for file in 7* ; do
read new
mv "$file" "$new"_"$file"
done < new_names.txt
Your main problem was than for ... in doesn't read from a file, it iterates over a list given after in. In your case, the list only had one member: new_names.txt. Also, You didn't populate $new_name anywhere.
Related
I have files in directories like:
./PBMCs/SRR1_1.fastq
./PBMCs/SRR1_2.fastq
./Monos/SRR2.fastq
./Monos/SRR3.fastq
I want to change the SRR# to a more informative name based on a file of key-value pairs:
SRR1 pbmc-1
SRR2 mono-1
SRR3 mono-2
And rename the files as:
./PBMCs/pbmc-1_1.fastq
./PBMCs/pbmc-1_2.fastq
./Monos/mono-1.fastq
./Monos/mono-2.fastq
All that I can think to do is loop through the list of original files and then loop through the lines of the name-change.txt file and replace the strings. However, I'm not sure how to implement this or if it's a good way to approach this.
Assuming all *.fastq are one subdirectory deep, this should work fine:
while read old new; do
for fastq in ./*/"$old"*.fastq; do
new_name=$new${fastq##*/"$old"}
echo mv "$fastq" "${fastq%/*}/$new_name"
done
done <name-change.txt
Remove echo if the output looks good.
My bash script retrieves a list of folders, and stores them under a projectlist parameter.
Later, I want to loop through each folder, using the project name as a variable in the file path. Each project folder contains a ProjectFile.txt file, from which I want to append the contents to a common .csv file.
projectlist=$(./some/project/filepath/projects)
echo ${projectlist[#]}
while read project;
do
while read line;
do
echo "$project" "$line" >> /data/allData.csv
done < ./some/project/filepath/projects/"$project"/ProjectFile.txt
done < $projectlist
What it currently ends up doing however, is printing each individual element in the projectlist, without writing anything to the .csv file. Any idea where the issue lies?
I'm writing a unix shell script that sorts data in ten subdirectories (labelled 1-10) of the home directory. In each subdirectory, the script needs to rename the files hehd.output and fort.hehd.time, as well as copy the file hehd.data to a .data file with a new name.
What I'd like it to do is rename each of these files in the following format:
AA.BB.CC
Where
AA = a variable in the hehd.data file within the subdirectory containing the file
BB = the name of the subdirectory containing the file (1-10)
CC = the original file name
Each subdirectory contains an hehd.data file, and each hehd.data file contains the string ij0=AA, where AA represents the variable I want to use to rename the files in the same subdirectory.
For example: When run, the script should search /home/4/hehd.data for the string ij0=2, then move /home/4/hehd.output to /home/4/2.4.hehd.output.
I'm currently using the grep command to have the script search for the string ij0=* and copy it to a new text file within the subdirectory. Next, the string ij0= is deleted from the text file, and then its contents are used to rename all target files in the same subdirectory. The last line of the shell script deletes the text file.
I'm looking for a better way to accomplish this, preferably such that all ten subdirectories can be sorted at once by the same script. My script seems incredibly inefficient, and doesn't do everything that I want it to by itself.
How can I improve this?
Any advice or suggestions would be appreciated; I'm trying to become a better computer user and that means learning better ways of doing things.
Try this:
fromdir=/home
for i in {1..10};do
AA=$(sed 's/ij0=\([0-9]*\)/\1/' "$fromdir/$i/hehd.data")
BB="$i"
for f in "$fromdir/$i/"*;do
CC="${f##*/}"
if [[ "$CC" = "hehd.data" ]]; then
echo cp "$f" "$fromdir/$i/$AA.$BB.$CC"
else
echo mv "$f" "$fromdir/$i/$AA.$BB.$CC"
fi
done
done
It loops over directories using Bash sequence {1..10].
In each directory, with the sed command the ij0 value is assigned to AA variable, the directory name is assigned to BB.
In the file loop, if the file is hehd.data it's copied, else it's renamed with the new name.
You can remove the echo before cp and mv commands if the output meets your needs.
Im trying to move multiple files from one directory to another directory.
File name is with sequence and will be varying.
Example:
/global/userhome/usrsats/---------directory which has file names as below:
fl_cl_filename1
fl_cl_filename2
fl_cl_filename3
...
...
Now when moved to another directory, i need to get only the file name and delimit the fl_cl part.
Please help
Assuming you're using bash, I would do this with the remove the matching prefix pattern facility like this (with DEST_DIR set to the destination directory):
cd /global/userhome/usrsats
for f in *; do mv $f ${DEST_DIR}/${f#fl_cl_}; done
I'm not very good in shell scripting and would like to ask you some question about looping of files big dataset: in my example I have alot of files with the common .pdb extension in the work dir. I need to loop all of them and i) to print name (w.o pdb extension) of each looped file and make some operation after this. E.g I need to make new dir for EACH file outside of the workdir with the name of each file and copy this file to that dir. Below you can see example of my code which are not worked- it's didn't show me the name of the file and didn't create folder for each of them. Please correct it and show me where I was wrong
#!/bin/bash
# set the work dir
receptors=./Receptors
for pdb in $receptors
do
filename=$(basename "$pdb")
echo "Processing of $filename file"
cd ..
mkdir ./docking_$filename
done
Many thanks for help,
Gleb
If all your files are contained within the .Repectors folder, you can loop each of them like so:
#!/bin/bash
for pdb in ./Receptors/*.pdb ; do
filename=$(basename "$pdb")
filenamenoextention=${filename/.pdb/}
mkdir "../docking_${filenamenoextention}"
done
Btw:
filenamenoextention=${filename/.pdb/}
Does a search replace in the variable $pdb. The syntax is ${myvariable/FOO/BAR}, and replaces all "FOO" substrings in $myvariable with "BAR". In your case it replaces ".pdb" with nothing, effectively removing it.
Alternatively, and safer (in case $filename contains multiple ".pdb"-substrings) is to remove the last four characters, like so: filenamenoextention=${filename:0:-4}
The syntax here is ${myvariable:s:e} where s and e correspond to numbers for the start and end index (not inclusive). It also let's you use negative numbers, which are offsets from the end. In other words: ${filename:0:-4} says: extract the substring from $filename starting from index 0, until you reach fourth-to-the-last character.
A few problems you have had with your script:
for pdb in ./Receptors loops only "./Receptors", and not each of the files within the folder.
When you change to parent directory (cd ..), you do so for the current shell session. This means that you keep going to the parent directory each time. Instead, you can specify the parent directory in the mkdir call. E.g mkdir ../thedir
You're looping over a one-item list, I think what you wanted to get is the list of the content of ./Receptors:
...
for pdb in $receptors/*
...
to list only file with .pdb extension use $receptors/*.pdb
So instead of just giving the path in for loop, give this:
for pdb in $receptors/*.pdb
To remove the extension :
set the variable ext to the extension you want to remove and using shell expansion operator "%" remove the extension from your filename eg:
ext=.pdb
filename=${filename%${ext}}
You can create the new directory without changing your current directory:
So to create a directory outside your current directory use the following command
mkdir ../docking_$filename
And to copy the file in the new directory use cp command
After correction
Your script should look like:
receptors=./Receptors
ext=.pdb
for pdb in $receptors/*.pdb
do
filename=$(basename "$pdb")
filename=${filename%${ext}}
echo "Processing of $filename file"
mkdir ../docking_$filename
cp $pdb ../docking_$filename
done