Is there a way to get functionality similar to `mkdir -p` on the shell from within Python.  I am looking for a solution other than a system call.  I am sure the code is less than 20 lines, and I am wondering if someone has already written it?

mkdir -p functionality in Python

Imagine that you want to develop a non-trivial end-user desktop (not web) application in Python. What is the best way to structure the project&#39;s folder hierarchy?

Desirable features are ease of maintenance, IDE-friendliness, suitability for source control branching/merging, and easy generation of install packages.

In particular:

 1. Where do you put the source?
 1. Where do you put application startup scripts?
 1. Where do you put the IDE project cruft?
 1. Where do you put the unit/acceptance tests?
 1. Where do you put non-Python data such as config files?
 1. Where do you put non-Python sources such as C++ for pyd/so binary extension modules?


What is the best project structure for a Python application?

I want to describe directory &amp; file structures in some of my Jekyll blog posts, does Markdown provide a neat way of outputting such a thing?

For example, you can see **[at this link](http://jekyllrb.com/docs/structure/)** on the Jekyll website that the directory &amp; file structure is output on the page very neatly:

```
.
├── _config.yml
├── _drafts
│   ├── begin-with-the-crazy-ideas.textile
│   └── on-simplicity-in-technology.markdown
├── _includes
│   ├── footer.html
│   └── header.html
├── _layouts
│   ├── default.html
│   └── post.html
├── _posts
│   ├── 2007-10-29-why-every-programmer-should-play-nethack.textile
│   └── 2009-04-26-barcamp-boston-4-roundup.textile
├── _data
│   └── members.yml
├── _site
└── index.html
```

I believe the line block characters above are Unicode (as described in **[this answer here](https://softwareengineering.stackexchange.com/questions/118640/write-a-directory-structure-pseudo-code)**), but I am not sure how Markdown or different browsers will handle them. I was hoping that Markdown had included some way of doing this which outputs as the Unicode characters above perhaps.

Representing Directory &amp; File Structure in Markdown Syntax

I&#39;m really enjoying Rails (even though I&#39;m generally RESTless), and I enjoy Ruby being very OO. Still, the tendency to make huge ActiveRecord subclasses and huge controllers is quite natural (even if you do use a controller per resource). If you were to create deeper object worlds, where would you put the classes (and modules, I suppose)? I&#39;m asking about views (in the Helpers themselves?), controllers and models.

Lib is okay, and I&#39;ve found [some solutions to get it to reload in a dev environment][1], but I&#39;d like to know if there&#39;s a better way to do this stuff. I&#39;m really just concerned about classes growing too large. Also, what about Engines and how do they fit in?


  [1]: http://www.ruby-forum.com/topic/111410

OO Design in Rails: Where to put stuff

I know there is actually no single right way. However I&#39;ve found that it&#39;s hard to create a directory structure that works well and remain clean for every developer and administrator. There is some standard structure in most projects on github. But it does not show a way to organize another files and all projects on pc.

What is the most convenient way to organize all these directories on development machine? How do you name them, and how do you connect and deploy this to server?

 - projects (all projects that your are working on)
 - source files (the application itself)
 - working copy of repository (I use git)
 - virtual environment (I prefer to place this near the project)
 - static root (for compiled static files)
 - media root (for uploaded media files)
 - README
 - LICENSE
 - documents
 - sketches
 - examples (an example project that uses the application provided by this project)
 - database (in case sqlite is used)
 - anything else that you usually need for successful work on project

The problems that I want to solve:

 - Good names of directories so that their purpose is clear.
 - Keeping all project files (including virtualenv) in one place, so I can easily copy, move, archive, remove whole project or estimate disk space usage.
 - Creating multiple copies of some selected file sets such as entire application, repository or virtualenv, while keeping single copy of another files that I don&#39;t want to clone.
 - Deploying right set of files to the server simply by rsyncing selected one dir.

Best practice for Django project working directory structure

I know that we can use `os.walk()` to list all sub-directories or all files in a directory. However, I would like to list the full directory tree content:
```
- Subdirectory 1:
   - file11
   - file12
   - Sub-sub-directory 11:
         - file111
         - file112
- Subdirectory 2:
    - file21
    - sub-sub-directory 21
    - sub-sub-directory 22    
        - sub-sub-sub-directory 221
            - file 2211
```
How to best achieve this in Python?

List directory tree structure in python?

So this doesn&#39;t seem like a terribly complicated question I have, but it&#39;s one I can&#39;t find the answer to.  I&#39;m confused about what the `-p` option does in Unix.  I used it for a lab assignment while creating a subdirectory and then another subdirectory within that one.  It looked like this:

    mkdir -p cmps012m/lab1

This is in a private directory with normal rights (`rlidwka`). Oh, and would someone mind giving a little explanation of what `rlidwka` means?  I&#39;m not a total noob to Unix, but I&#39;m not really familiar with what this means.  Hopefully that&#39;s not too vague of a question.

mkdir&#39;s &quot;-p&quot; option

Directory is something like:

    home/
        file1.html
        file2.html
    Another_Dir/
        file8.html
        Sub_Dir/
            file19.html

I am using the same PHP Zip class used in PHPMyAdmin [http://trac.seagullproject.org/browser/branches/0.6-bugfix/lib/other/Zip.php][1] . I&#39;m not sure how to zip a directory rather than just a file. Here&#39;s what I have so far:


    $aFiles = $this-&gt;da-&gt;getDirTree($target);
    /* $aFiles is something like, path =&gt; filetime
    Array
    (
        [home] =&gt; 
        [home/file1.html] =&gt; 1251280379
        [home/file2.html] =&gt; 1251280377
        etc...
    )
    
    */
    $zip = &amp; new Zip();
    foreach( $aFiles as $fileLocation =&gt; $time ){
        $file = $target . &quot;/&quot; . $fileLocation;
        if ( is_file($file) ){
            $buffer = file_get_contents($file);
            $zip-&gt;addFile($buffer, $fileLocation);
        }
    }
    THEN_SOME_PHP_CLASS::toDownloadData($zip); // this bit works ok


but when I try to unzip the corresponding downloaded zip file I get &quot;operation not permitted&quot;

This error only happens when I try to unzip on my mac, when I unzip through the command line the file unzips ok. Do I need to send a specific content type on download, currently &#39;application/zip&#39;

  [1]: http://trac.seagullproject.org/browser/branches/0.6-bugfix/lib/other/Zip.php

How to [recursively] Zip a directory in PHP?

Android Studio is giving me a Gradle build error that looks like this:

    Error:(3, 22) compileSdkVersion android-22 requires compiling with JDK 7

Now it gives me these clickable prompts:

    Download JDK 7
    Select a JDK from the File System
    Open build.gradle File

And I have already downloaded and installed JDK 7. The problem is when I go to select it in the &quot;File System&quot; i can only find a directory named 1.6.0 JDK. Furthermore, the installation of JDK 7 skipped the bullet point where I would&#39;ve selected the install directory, so I&#39;m really not sure where it is. My java control panel says I have &quot;Java 7 Update 79&quot; so I&#39;m pretty sure I&#39;m close, I just need to tell android studio where it is. I also ran the `java -version` command in the terminal and it says my version is &quot;1.7.0_79&quot;.

Any help would be appreciated!



How to specify the JDK version in Android Studio?

The use case is simple. I got the source files that were created using Eclipse. So, there is a deep directory structure, where any Java class could be referring to another Java class in the same, child, sibling or parent folder. 

How do I compile this whole thing from the terminal using javac ?

Java: How can I compile an entire directory structure of code ?

I am working on a C++ library. Ultimately, I would like to make it publicly available for multiple platforms (Linux and Windows at least), along with some examples and [Python][1] bindings. Work is progressing nicely, but at the moment the project is quite messy, built solely in and for [Visual C++][2] and not multi-platform at all.

Therefore, I feel a cleanup is in order. The first thing I&#39;d like to improve is the project&#39;s directory structure. I&#39;d like to create a structure that is suitable for the [Automake][3] tools to allow easy compilation on multiple platforms, but I&#39;ve never used these before. Since I&#39;ll still be doing (most of the) coding in Visual Studio, I&#39;ll need somewhere to keep my Visual Studio project and solution files as well.

I tried to google for terms like &quot;C++ library directory structure&quot;, but nothing useful seems to come up. I found some very basic guidelines, but no crystal clear solutions.

While looking at some open source libraries, I came up with the following:

    \mylib
        \mylib &lt;source files, read somewhere to avoid &#39;src&#39; directory&gt;
            \include? or just mix .cpp and .h
        \bin &lt;compiled examples, where to put the sources?&gt;
        \python &lt;Python bindings stuff&gt;
        \lib &lt;compiled library&gt;
        \projects &lt;VC++ project files, .sln goes in project root?&gt;
        \include? 
        README
        AUTHORS
        ...

I have no/little previous experience with multi-platform development/open source projects and am quite amazed that I cannot find any good guidelines on how to structure such a project.

How should one generally structure such a library project? What ca be recommended to read? Are there some good examples?

  [1]: http://en.wikipedia.org/wiki/Python_%28programming_language%29
  [2]: http://en.wikipedia.org/wiki/Visual_C++#32-bit_versions
  [3]: https://en.wikipedia.org/wiki/Automake


Directory structure for a C++ library

If I have a custom shell script or programs, that I created myself or downloaded from the web, and I want to be able to execute this from the CLI, is there the standard location to put this in Linux/Unix directory structure?

    /usr/bin ?
    /usr/local/bin ?
    /usr/lib ?
    /usr/sbin ?
    /bin ?
    /sbin ?
    /var ?

I usually put it under my ~/bin folder and put it in PATH, but it doesn&#39;t seem clean. And everytime I downloaded a new program, I have to put it in the PATH again.

Unix standard directory to put custom executables or scripts?

I&#39;ve been struggling with imports in my package for the last hour.

I&#39;ve got a directory structure like so:

    main_package
     |
     | __init__.py
     | folder_1
     |  | __init__.py
     |  | folder_2
     |  |  | __init__.py
     |  |  | script_a.py
     |  |  | script_b.py
     |
     | folder_3
     |  | __init__.py
     |  | script_c.py

I want to access code in `script_b.py` as well as code from `script_c.py` from `script_a.py`. How can I do this?

If I put a simple `import script_b` inside `script_a.py`, when I run

    from main_package.folder_1.folder_2 import script_b

I am met with an

    ImportError: no module named &quot;script_b&quot;

For accessing `script_c.py`, I have no clue. I wasn&#39;t able to find any information about accessing files two levels up, but I know I can import files one level up with

    from .. import some_module

How can I access both these files from `script_a.py`?

Python relative-import script two levels up

I&#39;m developing a LAMP online store, which will allow admins to upload multiple images for each item.

My concern is - right off the bat there will be 20000 items meaning roughly 60000 images.

Questions:

1. What is the maximum number of files and/or directories on Linux?

2. What is the usual way of handling this situation (best practice)?


My idea was to make a directory for each item, based on its unique ID, but then I&#39;d still have 20000 directories in a main *uploads* directory, and it will grow indefinitely as old items won&#39;t be removed.

Thanks for any help.


Maximum number of files/directories on Linux?

I am trying to do something to all the files under a given path. I don&#39;t want to collect all the file names beforehand then do something with them, so I tried this:
    
    import os
    import stat
    
    def explore(p):
      s = &#39;&#39;
      list = os.listdir(p)
      for a in list:
        path = p + &#39;/&#39; + a
        stat_info = os.lstat(path )
        if stat.S_ISDIR(stat_info.st_mode):
         explore(path)
        else:
          yield path

    if __name__ == &quot;__main__&quot;:
      for x in explore(&#39;.&#39;):
        print &#39;--&gt;&#39;, x

But this code skips over directories when it hits them, instead of yielding their contents. What am I doing wrong?
 

Yield in a recursive function

I&#39;ve entered a command `cd //` instead of `cd /` by mistake and instead of getting an error as I&#39;d expect the shell (Bash) showed a prompt as if I&#39;m in the `//` directory:-

    &lt;username&gt;@&lt;hostname&gt;://$
    &lt;username&gt;@&lt;hostname&gt;://$ pwd
    //

Listing the content of the directory shows that it is the same as `/`.

So what exactly `cd //` do and what&#39;s the difference from `cd /`?


What does double slash // in `cd //` mean in Linux?

(*Probably a noob question, but I didn&#39;t find a solution after googling for 20 minutes.*)

I created a new *pure Python* project with PyCharm which yielded the following folder structure

    myproject
    └── venv
        ├── bin
        │&#160;&#160; ├── activate
        │&#160;&#160; ├── activate.csh
        │&#160;&#160; ├── activate.fish
        │&#160;&#160; ├── easy_install
        │&#160;&#160; ├── easy_install-3.5
        │&#160;&#160; ├── pip
        │&#160;&#160; ├── pip3
        │&#160;&#160; ├── pip3.5
        │&#160;&#160; ├── python
        │&#160;&#160; ├── python3
        │&#160;&#160; └── python3.5
        ├── include
        ├── lib
        │&#160;&#160; └── python3.5
        ├── lib64 -&gt; lib
        └── pyvenv.cfg

Where do I put `myproject.py` or the `myproject` folder now?

- Inside or outside of `venv`?
- In the `venv/bin`folder?
- Just inside `venv`, i.e. `myproject/venv/myproject.py`?

All Directory Structure Solutions on Gang of Coders

Total of 17 Directory Structure Solutions