I want to call the NCO command ncrcat via the Python NCO bindings. From what I can tell there is only the ability to specify a single input file argument, and my ncrcat command requires two.
The command I'm trying to replicate in my Python code:
$ ncrcat -O -L 4 input_1.nc input_2.nc output
Maybe I include an iterable in the place of the single input, like so:
from nco import Nco
nco = Nco()
nco.ncrcat(input=[input_1,input_2],
output=output_netcdf,
options=['-O', '-4', '-L 4', '-h'])
Thanks in advance for any suggestions.
Yep thats the right syntax for specifying more than one input file
...Henry
Related
I have two kinds of variables in two different nc files. The dimension and other things are same, I just have to add one more variable in the existing nc file, How can I do this (using CDO or R or any other)
I used the command line (cdo selvar,varname in.nc out.nc) but it doesn't help. This command does work but deletes the existing variables. Any suggestions on how can I add new variables without deleting the variable inside the nc file?
Many thanks.
cdo solution
From your comment of clarification, I think the cdo command you need is cat
cdo cat aaa.nc bbb.nc output.nc
This will concatenate the fields in bbb.nc to the ones in aaa.nc and put the result in output.nc
nco solution
As an alternative you can also use ncrcat:
ncrcat aaa.nc bbb.nc output.nc
The NCO solution is
ncks -A -v yyy bbb.nc aaa.nc
as documented here. (Adrian's suggested NCO command would concatenate the files in time, not append one variable to the other file)
I am writing a code in R and I want to input values through the command line.
I am using get opt and I intend to read the values as such:
ScriptName -n File1 File2 -w 2...
I need an option which can read more than one value, in this case more than one file name, but I discovered that getopt doesn't allow this.
Any idea or any other package that will allow this?
I will use ls output for pipe input, so I need to escape the file name. when I use GNU ls, It works well. what's the equivalent in BSD ls? I hoping the output is like this.
$ gls --quoting-style escape t*1
text\ 1 text1
Why are/were you trying to use ls in a pipeline? You should probably be using find (maybe with -print0 and xargs -0, or -exec).
I suppose you could use ls -1f and then run the output through vis (or some similar filter) with some appropriate options to add the necessary quoting or escaping of your choice, but without knowing what you are feeding filenames into, and what (if any) other options you would want to use with ls, it's impossible to give much better guidance.
From the freebsd man page on ls there is no such option, however, you can try -m which will give you a comma separated streamed output:
-m Stream output format; list files across the page, separated by
commas.
I tried it on osx and it gave me:
$ ls -m
Hello World, Hello World.txt, foo.txt
That is a lot easier to parse from a script.
if have any xml file as below:
<soap env="abc" id="xyz">
<emp>acdf</emp>
<Workinstance name="ab" id="ab1">
<x>1</x>
<y>2</y>
</Workinstance>
<projectinstance name="cd" id="cd1">
<u>1</u>
<v>2</v>
</projectinstance>
</soap>
I want to extract the id field in workinstance using unix script
I tried grep but, it is retrieving the whole xml file.
Can someone help me how to get it?
You might want to consider something like XMLStarlet, which implements the XPath/XQuery specifications.
Parsing XML with regular expressions is essentially impossible even under the best of conditions, so the sooner you give up on trying to do this with grep, the better off you're likely to be.
XmlStarlet seems the tool I was looking for!
To do extract your tag, try to do the following:
cat your_file.xml | xmlstarlet sel -t -v 'soap/Workinstance/#id'
The "soap/Workinstance/#id" is an XPath expression that will get the id attribute inside Workinstance tag. By using "-v" flag, you ask xmlstarlet to print the extracted text to the standard output.
If you have Ruby
$ ruby -ne 'print $_.gsub(/.*id=\"|\".*$/,"" ) if /<Workinstance/' file
ab1
Im looking at a dictionary file (".dic") and its associated "aff" file. What I'm trying to do is combine the rules in the "aff" file with the words in the "dic" file to create a global list of all words contained within the dictionary file.
The documentation behind these files is difficult to find. Does anyone know of a resource that I can learn from?
Is there any code out there that will already do this (am I duplicating an effort that I don't need to)?
thanks!
According to Pillowcase, here it's an example of usage:
# Download dictionary
wget -O ./dic/es_ES.aff "https://raw.githubusercontent.com/sbosio/rla-es/master/source-code/hispalabras-0.1/hispalabras/es_ES.aff"
wget -O ./dic/es_ES.dic "https://raw.githubusercontent.com/sbosio/rla-es/master/source-code/hispalabras-0.1/hispalabras/es_ES.dic"
# Compile program
wget -O ./dic/unmunch.cxx "https://raw.githubusercontent.com/hunspell/hunspell/master/src/tools/unmunch.cxx"
wget -O ./dic/unmunch.h "https://raw.githubusercontent.com/hunspell/hunspell/master/src/tools/unmunch.h"
g++ -o ./dic/unmunch ./dic/unmunch.cxx
# Generate dictionary
./dic/unmunch ./dic/es_ES.dic ./dic/es_ES.aff 2> /dev/null > ./dic/es_ES.txt.bk
sort ./dic/es_ES.txt.bk > ./dic/es_ES.txt # Opcional
rm ./dic/es_ES.txt.bk # Opcional
You need a utility called munch.exe to apply the aff rules to the dic file.
These could be Hunspell dictionary files. Unfortunately, the command to create a "global" or unmunched wordlist only fully support simple .aff and .dic files.
From the documentation.
unmunch: list all recognized words of a MySpell dictionary
Syntax:
unmunch dic_file affix_file
Try it and see what happens. For generating all wordforms for one word only, look here.