I have created a package and am now creating my tests within the package. For one test my inputs are a set of files and my outputs will be a different set a files created within the test.
I am saving the input files in the test directory of my package and would like to save the output files there too. Since others may run this test, I do not want to specify the input/output file location using my own path eg /home/myname/.julia/v4.0/MyPackage/test/MyInputFile.txt
How do I specify that the input location is within the package's test folder?
So basically how do I tell Julia to look in the packages's folder under the test directory and not have to worry about specifying the entire path including user name etc?
For example currently I have to say
readtable(/home/myname/.julia/v4.0/MyPackage/test/MyInputFile.txt, separator = '\t', header = false)
But I'd like to just be able to say
readtable(/MyPackage/test/MyInputFile.txt, separator = '\t', header = false)
so that no matter who the user of the package is and where they may store the package, they can still run the test?
I know that LOAD_PATH gives the path Julia looks for packages but I can't find any information on where it looks when importing files.
joinpath(Pkg.dir("MyPackage"), "test") is what you need.
As #GnimucK mentioned in a comment, a better solution is
dirname(#__FILE__)
Why is this better? A package could be installed and used from somewhere else (not the standard package directory). Pkg.dir is "stupid" and does not know better. This is rare, of course, and in most cases it won't matter.
Related
I'm using python to download an application from a content distribution network. The application downloads as self extract file cabinet. When executed it creates a directory using a version naming format. For example app-version2/../app.exe, Thus I cannot rely on the folder name as it may change in the future. I'm trying to find the best way to work with the content inside the folder without depending on the actual folder name.
My idea was to rename the folder using os.listdir() and then os.rename(app-version2, myapp) This would work but is not automated. What would be the best automated method to find a folder name that contains version numbers and change that to something more static?
Assuming you want to find the path of the directory which begins with app, you can accomplish this using pathlib.Path:
from pathlib import Path
app_path = next(Path().glob('app*'))
This will give you the path to the first file or directory in your current directory whose name begins with "app".
TLDR
I need to get paths to system directories like "Screenshots":
On an English system. I can just use this one:
C:/Users/User/Pictures/Screenshots
How do I get the path to "Screenshots" directory on a non-English system?
C:/Users/User/Pictures/[NAME]
Description
I have a file manager app, it displays system directories and loads them on click.
The app can run system commands via Powershell and use Node.js (preferred)
Problem
The problem is, it only works if the system has English system language.
Currently, to resolve the "Screenshots" directory path, the app simply joins the User directory with the word "Screenshots"
const pictures = electronRemote.app.getPath('pictures')
const screenshots = PATH.join(pictures, 'Screenshots')
link to the line in code
Expectedly, the C:/Users/User/Screenshots path only exists on English systems.
One way to solve this is to use short names, at least on Windows, I know that system directories have short names like SCREEN~1 and WALLPA~1 for Screenshots and Wallpapers directories, but if I use these names the paths will look like this:
C:/Users/User/SCREEN~1 instead of C:/Users/User/Screenshots throughout the app.
And even if I were to run these paths through a function to convert it to readable name, how would I know which word to replace it with? I need to get the name in the system's language.
Are these translations stored somewhere on the system? Can I just retrieve the translated directory name and use that in the code above?
Question
How do I make it to get / resolve the actual path of system directories like Screenshots and Wallpapers, independently of system locale?
If you know how to do it, could you please suggest the solution for all platforms (Win, Mac, Linux)?
Should I just use the short names like SCREEN~1 and then automatically replace all the occurrences in UI and also filter all paths through a function that replaces this short name with the actual path throughout the whole app? Seems like a lot of work, this approach
In the piece of code I have, there are many instances where I have the following line
'/home/myname/directory'
For example, I have the following lines of code
filepath = os.listdir('/home/myname/directory')
for content in filepath
# do something
In the next part of the project, I have to share the code with some one else. I know this person runs openSUSE. If I want code to create that specific directory with the same path to the directory as mine, what do I need to include?
I know its going to involve the OS module but i am not sure which functions and methods to use.
Your code can check the existence of the directory and create it if not found:
if not os.path.exists("/home/myname/directory"):
os.makedirs("/home/myname/directory")
# do something
I am using the latest version of pyRevit, v45.
I'm writing some info in temporary files with
myTempFile = script.get_instance_data_file("id")
This creates a file named pyRevit_2018_xxxx_id.tmp in which I store useful info. If I'm not mistaken, the "xxxx" part is changing every time I reload Revit. Now, I need to get access to this information from another pyRevit script.
How can I retrieve the name of the temp file I need to read? In other words, how do I access "myTempFile" from within the second script, which has no idea of the name of "myTempFile"?
I guess I can share somehow that variable between my script, but what's the proper way to do this? I know this must be a very basic programming question, but I'm indeed not a programmer ;)
Thanks a lot,
Arnaud.
Ok, I realise now that my variables in the 1st script cease to exist after its execution.
So for now I wrote the file name in another file, of which I know the name.. That works.
But if there's a cleaner way to do this, I'd be glad to learn ;)
Arnaud
pyrevit.script module provides 4 different methods for creating temporary files based on their use case:
get_instance_data_file:
for data files marked with Revit instance pid. This means that scripts running on another instance will not see this temp file.
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_instance_data_file
get_universal_data_file:
for temp files accessible to all Revit instances and versions
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_universal_data_file
get_data_file:
Base method to get a standard temp file for current revit version
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_data_file
get_document_data_file:
temp file marked with active document (so scripts working on another document will not see this)
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_document_data_file
Each method uses a pattern to create the temp file name. So as long as the call to the method is the same of different scripts, the method generates the same file name.
Example:
Script 1:
from pyrevit import script
tfile = script.get_data_file('mydata')
Script 2:
from pyrevit import script
tempfile = script.get_data_file('mydata')
In this example tempfile = tfile since the file id is the same.
There is documentation on each so make sure you take a look at those and pick the flavor that serves your purpose.
I'm processing a data set and running into a problem - although I xlswrite all the relevant output variables to a big Excel file that is timestamped, I don't save the code that actually generated that result. So if I try to recreate a certain set of results, I can't do it without relying on memory (which is obviously not a good plan). I'd like to know if there's a command(s) that will help me save the m-files used to generate the output Excel file, as well as the Excel file itself, in a folder I can name and timestamp so I don't have to do this manually.
In my perfect world I would run the master code file that calls 4 or 5 other function m-files, then all those m-files would be saved along with the Excel output to a folder names results_YYYYMMDDTIME. Does this functionality exist? I can't seem to find it.
There's no such functionality built in.
You could build a dependency tree of your main function by using depfun with mfilename.
depfun(mfilename()) will return a list of all functions/m-files that are called by the currently executing m-file.
This will include all files that come as MATLAB builtins, you might want to remove those (and only record the MATLAB version in your excel sheet).
As pseudocode:
% get all files:
dependencies = depfun(mfilename());
for all dependencies:
if not a matlab-builtin:
copyfile(dependency, your_folder)
As a "long term" solution you might want to check if using a version control system like subversion, mercurial (or one of many others) would be applicable in your case.
In larger projects this is preferred way to record the version of source code used to produce a certain result.