This question already has an answer here:
How to refer to the file currently being loaded in Emacs Lisp?
(1 answer)
Closed 3 years ago.
When I edit a file, say current-file-path.el
I can have the following code:
(message (format "Here is the path of the current file %s" (buffer-file-name)))
when execting the statement in the buffer of the file when the file is open in a buffer, I got the correct message
of the path of the file:
Here is the path of the current file /home/yubrshen/tmp/current-file-path.el
However, if I just load the file, then the message becomes:
Here is the path of the current file nil
What would be the proper way to find out the path of the file where my code is?
Actually, I'm interested in knowing the directory of the file of my program so that I can load the other files at the same directory through the program.
You seem to be looking for symbol-file.
Of course, not every symbol is defined in Lisp code loaded from a named file; some are defined in the C source code for Emacs, and some are defined interactively by yourself.
There is also no guarantee that your data files will be packaged in the same location as your source code, so what you are describing should probably be implemented with a package variable instead.
(defvar foo-directory (file-name-directory load-file-name)
"*Directory for data files belonging to package \`foo'.")
This should probably use defcustom actually, but I'd have to guess too many things about your code to create a meaningful example.
Related
I'm trying to use Atom to run a Lua script. However, when I try to load files via the require() command, it always says it's unable to locate them. The files are all in the same folder. For example, to load utils.lua I have tried
require 'utils'
require 'utils.lua'
require 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua'
require 'D:\\Users\\Mike\\Dropbox\\Lua Modeling\\utils.lua'
require 'D:/Users/Mike/Dropbox/Lua Modeling/utils.lua'
I get errors like
Lua: D:\Users\Mike\Dropbox\Lua Modeling\main.lua:12: module 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua' not found:
no field package.preload['D:\Users\Mike\Dropbox\Lua Modeling\utils.lua']
no file '.\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
no file 'D:\Program Files (x86)\Lua\5.1\lua\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
no file 'D:\Program Files (x86)\Lua\5.1\lua\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua\init.lua'
no file 'D:\Program Files (x86)\Lua\5.1\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
The messages says on the first line that 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua' was not found, even though that is the full path of the file. What am I doing wrong?
Thanks.
The short answer
You should be able to load utils.lua by using the following code:
require("utils")
And by starting your program from the directory that utils.lua is in:
cd "D:\Users\Mike\Dropbox\Lua Modeling"
lua main.lua
The long answer
To understand what is going wrong here, it is helpful to know a little bit about how require works. The first thing that require does is to search for the module in the module path. From Programming in Lua chapter 8.1:
The path used by require is a little different from typical paths. Most programs use paths as a list of directories wherein to search for a given file. However, ANSI C (the abstract platform where Lua runs) does not have the concept of directories. Therefore, the path used by require is a list of patterns, each of them specifying an alternative way to transform a virtual file name (the argument to require) into a real file name. More specifically, each component in the path is a file name containing optional interrogation marks. For each component, require replaces each ? by the virtual file name and checks whether there is a file with that name; if not, it goes to the next component. The components in a path are separated by semicolons (a character seldom used for file names in most operating systems). For instance, if the path is
?;?.lua;c:\windows\?;/usr/local/lua/?/?.lua
then the call require"lili" will try to open the following files:
lili
lili.lua
c:\windows\lili
/usr/local/lua/lili/lili.lua
Judging from your error message, your Lua path seems to be the following:
.\?.lua;D:\Program Files (x86)\Lua\5.1\lua\?.lua;D:\Program Files (x86)\Lua\5.1\lua\?\init.lua;D:\Program Files (x86)\Lua\5.1\?.lua
To make that easier to read, here are each the patterns separated by line breaks:
.\?.lua
D:\Program Files (x86)\Lua\5.1\lua\?.lua
D:\Program Files (x86)\Lua\5.1\lua\?\init.lua
D:\Program Files (x86)\Lua\5.1\?.lua
From this list you can see that when calling require
Lua fills in the .lua extension for you
Lua fills in the rest of the file path for you
In other words, you should just specify the module name, like this:
require("utils")
Now, Lua also needs to know where the utils.lua file is. The easiest way is to run your program from the D:\Users\Mike\Dropbox\Lua Modeling folder. This means that when you run require("utils"), Lua will expand the first pattern .\?.lua into .\utils.lua, and when it checks that path it will find the utils.lua file in the current directory.
In other words, running your program like this should work:
cd "D:\Users\Mike\Dropbox\Lua Modeling"
lua main.lua
An alternative
If you can't (or don't want to) change your working directory to run the program, you can use the LUA_PATH environment variable to add new patterns to the path that require uses to search for modules.
set LUA_PATH=D:\Users\Mike\Dropbox\Lua Modeling\?.lua;%LUA_PATH%;
lua "D:\Users\Mike\Dropbox\Lua Modeling\main.lua"
There is a slight trick to this. If the LUA_PATH environment variable already exists, then this will add your project's folder to the start of it. If LUA_PATH doesn't exist, this will add ;; to the end, which Lua fills in with the default path.
This question already has an answer here:
All files has FILE_ATTRIBUTE_ARCHIVE attribute
(1 answer)
Closed 1 year ago.
When I run my code, the file attribute is 32 for all of my files.
According to this Microsoft docs page:
FILE_ATTRIBUTE_ARCHIVE, 32 (0x20), A file or directory that is an archive file or directory. Applications typically use this attribute to mark files for backup or removal .
But those are normal .jpg files. I would have expected something like this:
FILE_ATTRIBUTE_NORMAL, 128 (0x80), A file that does not have other attributes set. This attribute is valid only when used alone.
Is this just my setup, or is this the expected value for normal files?
There's nothing wrong with it. All files/folders in Windows have 4 basic attributes: Read-only, System, Hidden, Archive. The Archive attribute is pretty much useless these days because it's only used for backup tools to recognize whether a file has been backed up or not in the CP/M and DOS era and has nothing to do with the file type. Any files can have it enabled
It's also explained in the MSDN doc you linked above:
A file or directory that is an archive file or directory. Applications typically use this attribute to mark files for backup or removal .
I have a Julia program that inputs a csv and transforms the data via a bunch of functions, and outputs a csv file. I want to turn this into a binary so that I can run on different machines without having the source code on different machines.
I am looking at PackageCompiler.jl, but I can't find any understandable documentation for creating a binary app. I am trying:
using PackageCompiler
#time create_app("JuliaPrograms", "test"; precompile_execution_file="script.jl")
The file that contains all my code is script.jl and it lives in the dir JuliaPrograms, and I want the compiled binary to be named test.
When I run julia script.jl it performs as I want. I want to be able to run ./test with the same result.
However, I get this error:
ERROR: could not find project at "/Users/userx/JuliaPrograms/"
What am I doing wrong? Do I need some special project directory?
Per the docs here: https://julialang.github.io/PackageCompiler.jl/dev/apps.html#Creating-an-app-1 you need to make sure you define:
function julia_main()::Cint
# do something based on ARGS?
return 0 # if things finished successfully
end
a function called julia_main as the entry point to the app. You can find an example app here: https://github.com/JuliaLang/PackageCompiler.jl/tree/master/examples/MyApp
You may also want to check the location of the code itself. Is it being saved at "/Users/userx/JuliaPrograms/"? You can switch your directory in the Julia Reply by typing ; which will enter you into shell mode and then you can cd into the directory where your code is.
I'm new in Bioinformatics and Biopython, so I have some difficulties with it.
I was reading the Biopython (SeqIO) documentation, but when I try to execute some SeqIO.parse() commands I get FileNotFoundError.
For example, I want to get "example.fasta" file (which I don't have it on my PC). I try to do it with this command:
for record in SeqIO.parse("example.fasta", "fasta"):
print(record.id)
But, all I get is FileNotFoundError: [Errno 2] No such file or directory
Can someone help me with this?
My understanding is that FileNotFoundError occurs when the code tries to open a file on your computer and does not find it.
This can happen either because you simply do not have this file, or you gave the name with a typo, or the path to the file is not correct (This is an important notion: the path to the file should be absolute, or relative to the current working directory (usually the one from which you executed the python script)).
As suggested in the comments to your question, you seem to be expecting SeqIO.parse to get the file for you. This is not the case. The first argument you give to this function (in the example "example.fasta") is the path to an existing file that you want to "parse", that is, interpret its information content and make this content available to the rest of your program in a convenient form.
So in order to get this example working, you first need to get a fasta file. If you do not already have one, you can download some manually from genbank, or find one in the biopython installation (if you installed it from source and know where the source code is located), for instance in Tests/Quality/example.fasta.
I have created a project in /Projects/test that have the following files:
/Projects/test/first.rb
/Projects/test/second.rb
In first.rb, I do:
load 'second.rb'
And it gets loaded correctly. However, if I open the console and I type $:, I don't see the current directory "." in the load path. How does Ruby know where to load that 'second.rb' from?
See the documentation of Kernel#load clearly :
Loads and executes the Ruby program in the file filename. If the filename does not resolve to an absolute path, the file is searched for in the library directories listed in $:. If the optional wrap parameter is true, the loaded script will be executed under an anonymous module, protecting the calling program’s global namespace. In no circumstance will any local variables in the loaded file be propagated to the loading environment.
In case load 'second.rb' - second.rb has been internally resolved to the absolute path /Projects/test/second.rb,as your requiring file in the directory is same as required file directory. Nothing has been searched to the directories listed in$: for your case.
Just remember another way always
- The load method looks first in the current directory for files
Contrary to the currently accepted answer, the argument 'second.rb' does not resolve to an absolute path. If that were what was meant, you would also be able to require 'second.rb', since require has exactly the same wording about absolute paths.
I think what's happening here is just that the phrasing in the documentation for load is not clear at all about what the actual steps are. When it says "Loads and executes the Ruby program in the file filename," it means that literally — it treats the argument as a file name and attempts to load it as a Ruby program. If isn't an absolute path†, then Ruby goes through $LOAD_PATH and looks for it in those places. If that doesn't turn anything up, then it just goes ahead and tries to open it just as you passed it in. That's the logic that MRI actually follows.
† The actual check that Ruby does is essentially "Does the path start with '/', '~' or './'?".