My book states:
Every program that runs on your computer has a current working directory, or cwd. Any filenames or paths that do not begin with the root folder are assumed to be under the current working directory
As I am on OSX, my root folder is /. When I type in os.getcwd() in my Python shell, I get /Users/apple/Documents. Why am I getting the Documents folder in my cwd? Is it saying that Python is using Documents folder? Isn't there any path heading to Python that begins with / (the root folder)? Also, does every program have a different cwd?
Every process has a current directory. When a process starts, it simply inherits the current directory from its parent process; and it's not, for example, set to the directory which contains the program you are running.
For a more detailed explanation, read on.
When disks became large enough that you did not want all your files in the same place, operating system vendors came up with a way to structure files in directories. So instead of saving everything in the same directory (or "folder" as beginners are now taught to call it) you could create new collections and other new collections inside of those (except in some early implementations directories could not contain other directories!)
Fundamentally, a directory is just a peculiar type of file, whose contents is a collection of other files, which can also include other directories.
On a primitive operating system, that was where the story ended. If you wanted to print a file called term_paper.txt which was in the directory spring_semester which in turn was in the directory 2021 which was in the directory studies in the directory mine, you would have to say
print mine/studies/2021/spring_semester/term_paper.txt
(except the command was probably something more arcane than print, and the directory separator might have been something crazy like square brackets and colons, or something;
lpr [mine:studies:2021:spring_semester]term_paper.txt
but this is unimportant for this exposition) and if you wanted to copy the file, you would have to spell out the whole enchilada twice:
copy mine/studies/2021/spring_semester/term_paper.txt mine/studies/2021/spring_semester/term_paper.backup
Then came the concept of a current working directory. What if you could say "from now on, until I say otherwise, all the files I am talking about will be in this particular directory". Thus was the cd command born (except on old systems like VMS it was called something clunkier, like SET DEFAULT).
cd mine/studies/2021/spring_semester
print term_paper.txt
copy term_paper.txt term_paper.backup
That's really all there is to it. When you cd (or, in Python, os.chdir()), you change your current working directory. It stays until you log out (or otherwise exit this process), or until you cd to a different working directory, or switch to a different process or window where you are running a separate command which has its own current working directory. Just like you can have your file browser (Explorer or Finder or Nautilus or whatever it's called) open with multiple windows in different directories, you can have multiple terminals open, and each one runs a shell which has its own independent current working directory.
So when you type pwd into a terminal (or cwd or whatever the command is called in your command language) the result will pretty much depend on what you happened to do in that window or process before, and probably depends on how you created that window or process. On many Unix-like systems, when you create a new terminal window with an associated shell process, it is originally opened in your home directory (/home/you on many Unix systems, /Users/you on a Mac, something more or less like C:\Users\you on recent Windows) though probably your terminal can be configured to open somewhere else (commonly Desktop or Documents inside your home directory on some ostensibly "modern" and "friendly" systems).
Many beginners have a vague and incomplete mental model of what happens when you run a program. Many will incessantly cd into whichever directory contains their script or program, and be genuinely scared and confused when you tell them that you don't have to. If frobozz is in /home/you/bin then you don't have to
cd /home/you/bin
./frobozz
because you can simply run it directly with
/home/you/bin/frobozz
and similarly if ls is in /bin you most definitely don't
cd /bin
./ls
just to get a directory listing.
Furthermore, like the ls (or on Windows, dir) example should readily convince you, any program you run will look in your current directory for files. Not the directory the program or script was saved in. Because if that were the case, ls could only produce a listing of the directory it's in (/bin) -- there is nothing special about the directory listing program, or the copy program, or the word processor program; they all, by design, look in the current working directory (though again, some GUI programs will start with e.g. your Documents directory as their current working directory, by design, at least if you don't tell them otherwise).
Many beginners write scripts which demand that the input and output files are in a particular directory inside a particular user's home directory, but this is just poor design; a well-written program will simply look in the current working directory for its input files unless instructed otherwise, and write output to the current directory (or perhaps create a new directory in the current directory for its output if it consists of multiple files).
Python, then, is no different from any other programs. If your current working directory is /Users/you/Documents when you run python then that directory is what os.getcwd() inside your Python script or interpreter will produce (unless you separately os.chdir() to a different directory during runtime; but again, this is probably unnecessary, and often a sign that a script was written by a beginner). And if your Python script accepts a file name parameter, it probably should simply get the operating system to open whatever the user passed in, which means relative file names are relative to the invoking user's current working directory.
python /home/you/bin/script.py file.txt
should simply open(sys.argv[1]) and fail with an error if file.txt does not exist in the current directory. Let's say that again; it doesn't look in /home/you/bin for file.txt -- unless of course that is also the current working directory of you, the invoking user, in which case of course you could simply write
python script.py file.txt
On a related note, many beginners needlessly try something like
with open(os.path.join(os.getcwd(), "input.txt")) as data:
...
which needlessly calls os.getcwd(). Why is it needless? If you have been following along, you know the answer already: the operating system will look for relative file names (like here, input.txt) in the current working directory anyway. So all you need is
with open("input.txt") as data:
...
One final remark. On Unix-like systems, all files are ultimately inside the root directory / which contains a number of other directories (and usually regular users are not allowed to write anything there, and system administrators with the privilege to do it typically don't want to). Every relative file name can be turned into an absolute file name by tracing the path from the root directory to the current directory. So if the file we want to access is in /home/you/Documents/file.txt it means that home is in the root directory, and contains you, which contains Documents, which contains file.txt. If your current working directory were /home you could refer to the same file by the relative path you/Documents/file.txt; and if your current directory was /home/you, the relative path to it would be Documents/file.txt (and if your current directory was /home/you/Music you could say ../Documents/file.txt but let's not take this example any further now).
Windows has a slightly different arrangement, with a number of drives with single-letter identifiers, each with its own root directory; so the root of the C: drive is C:\ and the root of the D: drive is D:\ etc. (and the directory separator is a backslash instead of a slash, although you can use a slash instead pretty much everywhere, which is often a good idea for preserving your sanity).
Your python interpreter location is based off of how you launched it, as well as subsequent actions taken after launching it like use of the os module to navigate your file system. Merely starting the interpreter will place you in the directory of your python installation (not the same on different operating systems). On the other hand, if you start by editing or running a file within a specific directory, your location will be the folder of the file you were editing. If you need to run the interpreter in a certain directory and you are using idle for example, it is easiest to start by creating a python file there one way or another and when you edit it you can start a shell with Run > Python Shell which will already be in that directory. If you are using the command line interpreter, navigate to the folder where you want to run your interpreter before running the python/python3/py command. If you need to navigate manually, you can of course use the following which has already been mentioned:
import os
os.chdir('full_path_to_your_directory')
This has nothing to do with osx in particular, it's more of a concept shared by all unix-based systems, and I believe Windows as well. os.getcwd() is the equivalent of the bash pwd command - it simply returns the full path of the current location in which you are in. In other words:
alex#suse:~> cd /
alex#suse:/> python
Python 2.7.12 (default, Jul 01 2016, 15:34:22) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.getcwd()
'/'
It depends from where you started the python shell/script.
Python is usually (except if you are working with virtual environments) accessible from any of your directory. You can check the variables in your path and Python should be available. So the directory you get when you ask Python is the one in which you started Python. Change directory in your shell before starting Python and you will see you will it.
os.getcwd() has nothing to do with OSX in particular. It simply returns the directory/location of the source-file. If my source-file is on my desktop it would return C:\Users\Dave\Desktop\ or let say the source-file is saved on an external storage device it could return something like G:\Programs\. It is the same for both unix-based and Windows systems.
I am trying to set up a correct PATH, but I'm wondering what it should contain. If I have
/usr/bin/ls
/usr/local/bin/ls
and I want to prefer the one in /usr/local/bin, which of the following should I use?
PATH=/usr/local/bin/ls:/usr/bin/ls
PATH=/usr/local/bin:/usr/bin
or something else entirely?
This is not per se a suitable question for Stack Overflow.
I expect this to be closed as General Computing or Too Broad;
but the answer is frequently needed by beginners, so I hope this won't be deleted.
PATH works only with directories, not with single files
From the POSIX standard (emphasis mine)
PATH
This variable shall represent the sequence of path prefixes that certain functions and utilities apply in searching for an executable file known only by a filename. The prefixes shall be separated by a colon ( ':' ). [...] The list shall be searched from beginning to end, applying the filename to each prefix, until an executable file with the specified name and appropriate execution permissions is found.
When you type in ls into your shell and your PATH is set to /usr/local/bin/ls:/usr/bin/ls then your shell will …
… look for an executable with the path /usr/local/bin/ls/ls (note the double ls at the end).
As that path does not exist on your system your shell will proceed to look for an executable with the path /usr/bin/ls/ls (double ls again). That path also doesn't exist.
The shell couldn't find an executable using all paths in PATH so your shell will print something like bash: ls: command not found.
So, what did we learn? Paths listed by PATH have to be directories. You cannot list single files. Therefore the correct answer in your case is PATH=/usr/local/bin:/usr/bin.
Where things get interesting
Imagine the following situation. You have two versions of program c1 and two versions of program c2. The different versions are stored in the directories /a/ and /b/.
/a/c1
/a/c2
/b/c1
/b/c2
How can we set PATH to prefer /a/c1 over /b/c1/ but at the same time /b/c2 over /a/c2?
Sadly, there is no way to achieve this directly as we can only specify directories in PATH. We have to move/rename some files or create symlinks and use the symlinks inside the paths. One possible solution:
mkdir /c
ln -s /a/c1 /c/c1
ln -s /b/c2 /c/c2
export PATH=/c:/a:/b
The trailing :/a:/b is not really necessary here. I included them under the assumption that /a and /b contain more executables than just c1 and c2.
Indeed, as you can easily find out through experimentation, the variable PATH should already contain a list of directories which are consulted in that order. In fact, you should already find that you have /usr/local/bin and /usr/bin in the default PATH, usually in this order (though perhaps with other directories between them, and certainly with more directories around them).
To inspect your current PATH value, try
echo "$PATH"
or for a slightly more human-readable rendering
echo "${PATH//:/$'\n'}" # bash only
or
echo "$PATH" | tr ':' '\012' # any POSIX system
If you managed to set your PATH to an invalid value (which would cause simple commands like ls and cat to no longer be found, and produce command not found errors) you can try
PATH=/usr/local/bin:/usr/bin:/usr
to hopefully at least restore the essential functionality so that you can use cp or a simple system editor to get back to the original, safe, system default PATH.
I'm not great in the terminal, and I can't figure out why its returning this. It's probably really obvious so apologies for asking, but the executable file I'm referencing is definitely in that file path, and after researching I can't seem to find an answer:
/home/user/protoc-3.5.1-linux-x86_64/bin/protoc object_detection/protos /*.proto --python_out=.
object_detection/protos/*.proto: No such file or directory
(I can't cd into it as I need to do this in a particular directory)
Thanks
It seems bash is looking for a specific file that has the name "[star]" instead of using this as a wildcard.
I think you might need to use a pipe to get the desired result.
From your command line, it looks like protoc is the executable, located at /home/user/protoc-3.5.1-linux-x86_64/bin/protoc. And that you're giving it two arguments separated by a space: object_detection/protos and /*.proto. If you have spaces in the file path, you'll need to escape them or double-quote them:
protoc object_detection/protos\ /*.proto or
protoc "object_detection/protos /*.proto"
The odd thing is that the error message indicates differently:
object_detection/protos/*.proto: No such file or directory
Or possibly the protoc executable needs the absolute (complete) path for file arguments. If from your current working directory the command ls object_detection/protos/*.proto shows results for you, then you can try running your command like this to use absolute file paths:
/home/user/protoc-3.5.1-linux-x86_64/bin/protoc $PWD/object_detection/protos/*.proto
$PWD is an environment variable that contains your working directory path.
I have build a set of libraries and many of my Fortran programs will use them. This creates a problem in that if I ever need to change the location of the libraries then I will need to individually update the path directories in each make file.
How is this usually overcome? I have planned instead to have each make file read a path from a single master path file in the home or root directory (this files location will never change). Within this file is the path for each Library and if any path changes only this file needs to updated.
So I wrote a bash script file, called Master_Library_Paths:
export Library1_Name = {Library1_Name_Path}
echo $Library1_Name
export Library2_Name = {Library2_Name_Path}
echo $Library2_Name
export Library3_Name = {Library3_Name_Path}
echo $Library3_Name
And placed it in my home directory. Then in the make files, I have a line:
$(shell . {Path for Master_Library_Paths} ) \
And load the libraries:
-I$(Library1_Name)
-I$(Library2_Name)
-I$(Library3_Name)
This works great if I run ./Master_Library_Paths in the terminal session first and then go to the directory to compile the program, however that is quite time consuming, How can I fix it so that these arguments Library1_Name, Library2_Name ect are known throughout the system?
New system wide LD_LIBRARY_PATH´s can be added in /etc/ld.so.conf , /etc/ld.so.conf.d/
Or may be in /etc/profile.d/
-
I am using the 2003 textbook - http://www.amazon.com/Unix-Shell-Programming-3rd-Edition/dp/0672324903
My OS is linux L-ubuntu 13 which is not based on POSIX (I think)
It says that I can store who | wc -l in a file called nu and then execute nu. But, before that I need to make this file executable by using chmod +x file(s). This does not work. How do I make the nu "command" work ? I know I can do it by naming nu as nu.sh and then doing bash nu.sh, but I want to try this way also.
To execute a file that is not in the PATH, you must give a properly qualified directory name. While giving the name of the file in the current directory is sufficient as an argument to a program, in order to execute a shell script or other executable file, you must give at least a relative path. For example, if the file is in your home directory, which is also the working directory, any of the following are acceptable:
./nu
~/nu
/home/username/nu
However, simply nu will only attempt to search the PATH, which probably includes places such as /bin, /usr/bin, and so on.