I have a folder of data folders with the following structure:
sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/data1.gz
sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/data2.gz
sampleName2-randomNumbers/subfolder1/subfolder2/subfolder3/data1.gz
I want to modify all the data.gz within each sample folder by appending the sample name but not the random numbers to get:
sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/sampleName1_data1.gz
sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/sampleName1_data2.gz
sampleName2-randomNumbers/subfolder1/subfolder2/subfolder3/sampleName2_data1.gz
It seems like this should be a simple mv for loop but I haven't been able to figure out how to pull part of a folder name using basename.
for i in */Data/Intensities/BaseCalls/*.gz; do mv $i "fastq""/"${i%%-*}"."`basename $i`; done
I couldn't figure out how to make the files stay in their original folder but for my purposes it works to have all the files go to a new folder ("fastq")
I suppose the "sampleName" part doesn't include dashes. In that case, use the standard pattern removal expansion: %%. That is, suppose your full path (relative to directory root) is stored in $path, just do ${path%%-*} to extract the "sampleName" part. Search for %% in the Bash Reference Manual for more details. As a simple example:
> path=sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/data1.gz
> echo ${path%%-*}
sampleName1
Otherwise, you could also use more advanced substring extraction based on regex. See BashFAQ/100 or Manipulating Strings from the TLDP Advanced Bash Scripting Guide.
Update. Here's the full command to perform the job described, and it is entirely native to the shell:
for file in */Data/Intensities/BaseCalls/*.gz; do
mv "$file" "${file%/*}/${file%%-*}_${file##*/}"
done
Related
This has to be a duplicate but I have read and tried at least a dozen of Q&As here on SO, and I cannot get any of them working for my case.
Really hope this won't result in downvotes because of it.
So I'm on Windows (10) and have a Bash terminal that I want to use for my task. The MINGW64 one I downloaded when I started working with Git.
I would prefer the solution with this program, but will be perfectly happy with one in Command Prompt Terminal or even PowerShell.
I created a TemplateApp which is in C:\Apps\TemplateApp folder which has multiple folders and subfolders named TemplateApp or TemplateApp.something as well as a lot of files that have TemplateApp as a part of their name.
Could be:
TemplateApp.ext
TemplateApp.something.ext
something.TemplateApp.something.ext
Then I copied the uppermost folder to C:\Apps\TemplateApp - Copy and in turn renamed it to C:\Apps\ProductionApplication.
Now for the love of whomever, I cannot make any of the scripts I found on SO to work for my case, ie. to rename all the above mentioned files and folders by replacing TemplateApp with ProductionApplication.
Here is a bash function I wrote that I think does very much like what you are wanting to do.
function func_CreateSourceAndDestination() {
#
for (( i = 0 ; i < ${#files_syncSource[#]} ; i++ )) ; do
files_syncDestination[${i}]="${files_syncSource[${i}]#${directory_MusicLibraryRoot_source}}"
file_destinationPath="$( dirname -- "${directory_PMPRoot_destination}${files_syncDestination[${i}]}" )"
if [ ! -d "${file_destinationPath}" ] ; then
mkdir -p "${file_destinationPath}"
fi
rsync -rltDvPmz "${files_syncSource[${i}]}" "${directory_PMPRoot_destination}${files_syncDestination[${i}]}"
done
}
In my case I'm feeding into rsync for a source and a destination. I'm pulling all the file paths from an array that has been split into path segments. I have to make certain character substitutions for FAT and NTFS file systems. I do this recursively.
files_syncDestination[${i}]="${files_syncDestination[${i}]//\:/__}"
That's the magic. I load a new array with the character substituted. You could do the same with a loaded variable including your phrases for change.
files_syncDestination[${i}]="${files_syncDestination[${i}]//${targetPhrase}/${subPhrase}}"
After that change in the function, you could use rsync or cp or mv as you prefer to go from your source array to your destination array.
(The double-slash in the substitution makes the substitution global.)
I would like to copy a series of similar files from the current directory to the target directory, the files under the current directory are:
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0001_ux.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0001_uz.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0002_ux.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0002_uz.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0003_ux.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0003_uz.hst
Where sim is from sim0001 to sim0500 and f is from f0001 to f0009. I only need f0002, f0005 and f0008. I write the following code:
target_dir="projects/data"
for i in {0001..0500}; do
for s in f000{2,5,8}; do
files="[*]$i[*]$s[*]"
cp $files target_dir
done
done
I am very new to Shell, and wondering how to write the $files="[*]$i[*]$s[*]"$, so that it could match only the f0002, f0005 and f0008. The reason why I also use for i in {0001..0500}; do is that the files are too large and I would like to make sure I could access some completed ones (for example, including all sim0001) in the beginning.
Edit: changed for s in f0002 f0005 f0008; do to f000{2,5,8}.
What you need is globbing and a bit different quoting:
cp *"$i"*"$s"* "$target_dir"
Not storing this in a variable is intentional - it's faster and it's safe. If you end up with such a large list of files that you start running into system limits you'll have to look into xargs.
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
In shell, what is a good way to duplicating files in an existing directory so that the result gives the same file but with a different extension? So taking something like:
path/view/blah.html.erb
And adding:
path/view/blah.mobile.erb
So that in the path/view directory, there would be:
path/view/blah.html.erb
path/view/blah.mobile.erb
I'd ideally like to perform this at a directory level and not create the file if it already has both extensions but that isn't necessary.
You can do:
cd /path/view/
for f in *.html.erb; do
cp "$f" "${f/.html./.mobile.}"
done
PS: This replaces first instance of .html. with .mobile., syntax is bash specific (let me know if you're not using BASH).
I have to use bash scripting to copy files from one folder to another. If the destination folder has a file with the same name but older timestamp, it should not copy. Only newer files should be copied. I could have used cp -u, but I was asked not to use it. Essentially I have to use the test command testing for "ot". Please let me know how could this be done. I believe two for loops one to read the files in the source and one for the destination directories can be used and the the time stamp compared. The problem is that both for loops produce the absolute path names along with the file name. So not sure how to compare them
Thanks
You can profit from the parameter substitution:
for file in "$folder1"/* ; do
filename=${file##*/} # Remove everything to the last slash.
Or, you can change the directory:
cd "$folder1"
for file in * ; do
## you have to use full or relative path to $folder2 here