List directory tree structure in python?

A solution without your indentation:

for path, dirs, files in os.walk(given_path):
  print path
  for f in files:
    print f

os.walk already does the top-down, depth-first walk you are looking for.

Ignoring the dirs list prevents the overlapping you mention.

Similar to answers above, but for python3, arguably readable and arguably extensible:

from pathlib import Path

class DisplayablePath(object):
    display_filename_prefix_middle = '├──'
    display_filename_prefix_last = '└──'
    display_parent_prefix_middle = '    '
    display_parent_prefix_last = '│   '

    def __init__(self, path, parent_path, is_last):
        self.path = Path(str(path))
        self.parent = parent_path
        self.is_last = is_last
        if self.parent:
            self.depth = self.parent.depth + 1
        else:
            self.depth = 0

    @property
    def displayname(self):
        if self.path.is_dir():
            return self.path.name + '/'
        return self.path.name

    @classmethod
    def make_tree(cls, root, parent=None, is_last=False, criteria=None):
        root = Path(str(root))
        criteria = criteria or cls._default_criteria

        displayable_root = cls(root, parent, is_last)
        yield displayable_root

        children = sorted(list(path
                               for path in root.iterdir()
                               if criteria(path)),
                          key=lambda s: str(s).lower())
        count = 1
        for path in children:
            is_last = count == len(children)
            if path.is_dir():
                yield from cls.make_tree(path,
                                         parent=displayable_root,
                                         is_last=is_last,
                                         criteria=criteria)
            else:
                yield cls(path, displayable_root, is_last)
            count += 1

    @classmethod
    def _default_criteria(cls, path):
        return True

    @property
    def displayname(self):
        if self.path.is_dir():
            return self.path.name + '/'
        return self.path.name

    def displayable(self):
        if self.parent is None:
            return self.displayname

        _filename_prefix = (self.display_filename_prefix_last
                            if self.is_last
                            else self.display_filename_prefix_middle)

        parts = ['{!s} {!s}'.format(_filename_prefix,
                                    self.displayname)]

        parent = self.parent
        while parent and parent.parent is not None:
            parts.append(self.display_parent_prefix_middle
                         if parent.is_last
                         else self.display_parent_prefix_last)
            parent = parent.parent

        return ''.join(reversed(parts))

Example usage:

paths = DisplayablePath.make_tree(
    Path('doc'),
    criteria=is_not_hidden
)
for path in paths:
    print(path.displayable())

# With a criteria (skip hidden files)
def is_not_hidden(path):
    return not path.name.startswith(".")

paths = DisplayablePath.make_tree(Path('doc'), criteria=is_not_hidden)
for path in paths:
    print(path.displayable())

Example output:

doc/
├── _static/
│   ├── embedded/
│   │   ├── deep_file
│   │   └── very/
│   │       └── deep/
│   │           └── folder/
│   │               └── very_deep_file
│   └── less_deep_file
├── about.rst
├── conf.py
└── index.rst

Notes

• This uses recursion. It will raise a RecursionError on really deep folder trees
• The tree is lazily evaluated. It should behave well on really wide folder trees. Immediate children of a given folder are not lazily evaluated, though.

Edit:

• Added bonus! criteria callback for filtering paths.

Here's a function to do that with formatting:

import os

def list_files(startpath):
    for root, dirs, files in os.walk(startpath):
        level = root.replace(startpath, '').count(os.sep)
        indent = ' ' * 4 * (level)
        print('{}{}/'.format(indent, os.path.basename(root)))
        subindent = ' ' * 4 * (level + 1)
        for f in files:
            print('{}{}'.format(subindent, f))

List directory tree structure in Python?

We usually prefer to just use GNU tree, but we don't always have tree on every system, and sometimes Python 3 is available. A good answer here could be easily copy-pasted and not make GNU tree a requirement.

tree's output looks like this:

$ tree
.
├── package
│   ├── __init__.py
│   ├── __main__.py
│   ├── subpackage
│   │   ├── __init__.py
│   │   ├── __main__.py
│   │   └── module.py
│   └── subpackage2
│       ├── __init__.py
│       ├── __main__.py
│       └── module2.py
└── package2
    └── __init__.py

4 directories, 9 files

I created the above directory structure in my home directory under a directory I call pyscratch.

I also see other answers here that approach that sort of output, but I think we can do better, with simpler, more modern code and lazily evaluating approaches.

Tree in Python

To begin with, let's use an example that

• uses the Python 3 Path object
• uses the yield and yield from expressions (that create a generator function)
• uses recursion for elegant simplicity
• uses comments and some type annotations for extra clarity

from pathlib import Path

# prefix components:
space =  '    '
branch = '│   '
# pointers:
tee =    '├── '
last =   '└── '


def tree(dir_path: Path, prefix: str=''):
    """A recursive generator, given a directory Path object
    will yield a visual tree structure line by line
    with each line prefixed by the same characters
    """    
    contents = list(dir_path.iterdir())
    # contents each get pointers that are ├── with a final └── :
    pointers = [tee] * (len(contents) - 1) + [last]
    for pointer, path in zip(pointers, contents):
        yield prefix + pointer + path.name
        if path.is_dir(): # extend the prefix and recurse:
            extension = branch if pointer == tee else space 
            # i.e. space because last, └── , above so no more |
            yield from tree(path, prefix=prefix+extension)

and now:

for line in tree(Path.home() / 'pyscratch'):
    print(line)

prints:

├── package
│   ├── __init__.py
│   ├── __main__.py
│   ├── subpackage
│   │   ├── __init__.py
│   │   ├── __main__.py
│   │   └── module.py
│   └── subpackage2
│       ├── __init__.py
│       ├── __main__.py
│       └── module2.py
└── package2
    └── __init__.py

We do need to materialize each directory into a list because we need to know how long it is, but afterwards we throw the list away. For deep and broad recursion this should be lazy enough.

The above code, with the comments, should be sufficient to fully understand what we're doing here, but feel free to step through it with a debugger to better grock it if you need to.

More features

Now GNU tree gives us a couple of useful features that I'd like to have with this function:

• prints the subject directory name first (does so automatically, ours does not)
• prints the count of n directories, m files
• option to limit recursion, -L level
• option to limit to just directories, -d

Also, when there is a huge tree, it is useful to limit the iteration (e.g. with islice) to avoid locking up your interpreter with text, as at some point the output becomes too verbose to be useful. We can make this arbitrarily high by default - say 1000.

So let's remove the previous comments and fill out this functionality:

from pathlib import Path
from itertools import islice

space =  '    '
branch = '│   '
tee =    '├── '
last =   '└── '

def tree(dir_path: Path, level: int=-1, limit_to_directories: bool=False,
         length_limit: int=1000):
    """Given a directory Path object print a visual tree structure"""
    dir_path = Path(dir_path) # accept string coerceable to Path
    files = 0
    directories = 0
    def inner(dir_path: Path, prefix: str='', level=-1):
        nonlocal files, directories
        if not level: 
            return # 0, stop iterating
        if limit_to_directories:
            contents = [d for d in dir_path.iterdir() if d.is_dir()]
        else: 
            contents = list(dir_path.iterdir())
        pointers = [tee] * (len(contents) - 1) + [last]
        for pointer, path in zip(pointers, contents):
            if path.is_dir():
                yield prefix + pointer + path.name
                directories += 1
                extension = branch if pointer == tee else space 
                yield from inner(path, prefix=prefix+extension, level=level-1)
            elif not limit_to_directories:
                yield prefix + pointer + path.name
                files += 1
    print(dir_path.name)
    iterator = inner(dir_path, level=level)
    for line in islice(iterator, length_limit):
        print(line)
    if next(iterator, None):
        print(f'... length_limit, {length_limit}, reached, counted:')
    print(f'\n{directories} directories' + (f', {files} files' if files else ''))

And now we can get the same sort of output as tree:

tree(Path.home() / 'pyscratch')

prints:

pyscratch
├── package
│   ├── __init__.py
│   ├── __main__.py
│   ├── subpackage
│   │   ├── __init__.py
│   │   ├── __main__.py
│   │   └── module.py
│   └── subpackage2
│       ├── __init__.py
│       ├── __main__.py
│       └── module2.py
└── package2
    └── __init__.py

4 directories, 9 files

And we can restrict to levels:

tree(Path.home() / 'pyscratch', level=2)

prints:

pyscratch
├── package
│   ├── __init__.py
│   ├── __main__.py
│   ├── subpackage
│   └── subpackage2
└── package2
    └── __init__.py

4 directories, 3 files

And we can limit the output to directories:

tree(Path.home() / 'pyscratch', level=2, limit_to_directories=True)

prints:

pyscratch
├── package
│   ├── subpackage
│   └── subpackage2
└── package2

4 directories

Retrospective

In retrospect, we could have used path.glob for matching. We could also perhaps use path.rglob for recursive globbing, but that would require a rewrite. We could also use itertools.tee instead of materializing a list of directory contents, but that could have negative tradeoffs and would probably make the code even more complex.

Comments are welcome!