I try to open annotation file from mimic2 db for patient a40017 that called a40017.alM.
I have this link for the data: http://www.physionet.org/pn5/mimic2db/a40017/
and I don't find the exact command in Cygwin that export the file to csv or text.
I try to use this command:
rdann -r mimic2db/a40017/a40017 -f 0 -t 216647.728 -a alM -v >annotations.txt
but I got an empty file
Is anyone know how can I do that?
Thanks,
Gal
Anwering myself. Rdann is a mingw32/64 program
https://physionet.org/physiotools/binaries/windows/
If you are in the same directory of the program and it is not in the PATH you need to run:
./rdann -r mimic2db/a40017/a40017 -f 0 -t 216647.728 -a alM -v
or
/<fullpath>/rdann ...
Related
I am trying "cp -r flange/ $FOAM_RUN/ "in openfoan10 ubuntu terminal enter image description here but its showing can not create directory so please help to solve it
You're getting that error because the destination directory $FOAM_RUN does not exist. You need to create it first:
mkdir -p $FOAM_RUN
Then, you can try your command:
cp -r flange/ $FOAM_RUN/
I am trying to send a file in cygwin commandline with rsync and openssh.
So the command I'm using is rsync -P -e "ssh -p 2222" deborahtrez#209.6.204.90:/home/dell/cygdrive/c/Users/dell/Videos/Movavi/installation_tutorial.mp4
Then it asks for my password, which I enter. I press ENTER and I always get returned No such file or directory
The file path to my video "installation_tutorial.mp4" is in C:\Users\dell\Videos\Movavi so the full path becomes C:\Users\dell\Videos\Movavi\installation_tutorial.mp4. So this is what I have tried so far.
Please, what am I doing wrong? Is it my file path? If so, what is the correct file path that i should be using? Help!
The path is wrong. The best way to convert from Windows to Posix is to use cygpath
$ cygpath -u 'C:\Users\dell\Videos\Movavi\installation_tutorial.mp4'
/cygdrive/c/Users/dell/Videos/Movavi/installation_tutorial.mp4
I am using wget to try and download two .zip files (SWVF_1_44.zip and SWVF_44_88.zip) from this site: http://www2.sos.state.oh.us/pls/voter/f?p=111:1:0::NO:RP:P1_TYPE:STATE
when I run:
wget -r -l1 -H -t1 -nd -N -np -A.zip -erobots=off "http://www2.sos.state.oh.us/pls/voter/f?p=111:1:0::NO:RP:P1_TYPE:STATE/SWVF_1_44.zip"
I get a downloaded zip file that has a screwed up name (f#p=111%3A1%3A0%3A%3ANO%3ARP%3AP1_TYPE%3ASTATE%2FSWVF_1_44) and it cannot be opened.
Any thoughts on where my code is wrong?
There's nothing "wrong" with your code. Wget is simply assuming you want to save the file in the same name that appears in the url. Use the -O option to specify an output file:
wget blahblahblah -O useablefilename.zip
With this command, I'm able to ZIP all files from the folders:
wzzip.exe -a -p -r C:\DestinationPath\DataFiles_20130903.zip C:\SourcePath\*.*
But, my folder has .dat,.bat,.txt,.xls files.I want to ZIP only .dat and .bat file.How to do this?
Thanks.
use this command (for the particular scenario in the question):
wzzip.exe -a -p -r C:\DestinationPath\DataFiles_20130903.zip C:\SourcePath\*.dat C:\SourcePath\*.bat
for more command line options for winZip refer to the following links:
winZip command line Reference 1
winZip command line Reference 2
To provide multiple file names you can also use #filename where the filename is a file which contains the list of files which you want to include in the zip file.
If you are making the command configurable then you can ask the user/ other program which is calling your command to select the file extensions and then write these selected extensions into the "filename" file using java code or any other language you prefer.
For example if the user selects bat and dat , then write "C:\SourcePath\*.bat" and "C:\SourcePath\*.dat" into the file(assume filename is fileExtensions.txt) and call the command
wzzip.exe -a -p -r "C:\DestinationPath\DataFiles_20130903.zip" #"C:\SourcePath\fileExtensions.txt"
You can use the D7zip
An excellent zipador file and folders D7zip.exe
link to download
https://drive.google.com/file/d/0B4bu9X3c-WZqdlVlZFV4Wl9QWDA/edit?usp=sharing
How to use
compressing files
D7Zip.exe -z "c:\fileout.zip" -f "C:\filein.txt"
compressing files and putting password
D7Zip.exe -z "c:\fileout.zip" -f "C:\filein.txt" -s "123"
compressing folders
D7Zip.exe -z "c:\folderout.zip" -f "C:\folderin\"
unzipping files
D7Zip.exe -u "c:\fileout.zip" -f "c:\folderout\"
unzipping files that have password
D7Zip.exe -u "c:\fileout.zip" -f "c:\folderout\" -s "123"
decompressing files by extension
D7Zip.exe -u "c:\fileout.zip" -f "c:\folderout\*.txt"
decompressing files without asking for confirmation to replace
D7Zip.exe -u "c:\fileout.zip" -f "c:\folderout\" -r
help
D7Zip.exe -?
D7Zip.exe by Delmar Grande.
If the command line given above is right then give this a go: but check the paths.
#echo off
pushd "C:\SourcePath"
"c:\program files\winzip\wzzip.exe" -a -p -r "C:\DestinationPath\DataFiles_20130903.zip" *.dat *.bat
popd
It is an odd behaviour seen only on Solaris that when I try to copy a symbolic link with the "cp -R -P" command to some other folder with a different name, it copies the entire directory/file it's pointing to.
For example:
link -> dir
cp -R -P link folder/new_link
I believe the "-d" argument is what you need.
As per the cp man page:
-d same as --no-dereference --preserve=link
Example:
cp -d -R -P link folder/new_link
I was using "cp -d" and that worked for me.
The cp man page seems to say that you want to use an '-H' to preserve symlinks within the source directory.
You might consider copying via tar, like tar -cf - srcdir|(cd somedir;tar -xf -)
Try using cpio (with the -p (pass) option) or the old tar in a pipe trick.