'Flatten complex directory structure in Python
I want to move files from a complex directory structure to just one place. For example i have this deep hierarchy:
foo/
foo2/
1.jpg
2.jpg
...
I want it to be:
1.jpg
2.jpg
...
My current solution:
def move(destination):
for_removal = os.path.join(destination, '\\')
is_in_parent = lambda x: x.find(for_removal) > -1
with directory(destination):
files_to_move = filter(is_in_parent,
glob_recursive(path='.'))
for file in files_to_move:
shutil.move(file, destination)
Definitions: directory
and glob_recursive
. Note, that my code only moves files to their common parent directory, not an arbitrary destination.
How can i move all files from a complex hierarchy to a single place succinctly and elegantly?
Solution 1:[1]
Run recursively through directory, move the files and launch move
for directories:
import shutil
import os
def move(destination, depth=None):
if not depth:
depth = []
for file_or_dir in os.listdir(os.path.join([destination] + depth, os.sep)):
if os.path.isfile(file_or_dir):
shutil.move(file_or_dir, destination)
else:
move(destination, os.path.join(depth + [file_or_dir], os.sep))
Solution 2:[2]
I don't like testing the name of the file about to be moved to see if we're already in the destination directory. Instead, this solution only scans the subdirectories of the destination
import os
import itertools
import shutil
def move(destination):
all_files = []
for root, _dirs, files in itertools.islice(os.walk(destination), 1, None):
for filename in files:
all_files.append(os.path.join(root, filename))
for filename in all_files:
shutil.move(filename, destination)
Explanation: os.walk walks recursively the destination in a "top down" manner. whole filenames are constructed with the os.path.join(root, filename) call. Now, to prevent scanning files at the top of the destination, we just need to ignore the first element of the iteration of os.walk. To do that I use islice(iterator, 1, None). One other more explicit way would be to do this:
def move(destination):
all_files = []
first_loop_pass = True
for root, _dirs, files in os.walk(destination):
if first_loop_pass:
first_loop_pass = False
continue
for filename in files:
all_files.append(os.path.join(root, filename))
for filename in all_files:
shutil.move(filename, destination)
Solution 3:[3]
this would do, it also renames files if they collide (I commented out the actual move and replaced with a copy):
import os
import sys
import string
import shutil
#Generate the file paths to traverse, or a single path if a file name was given
def getfiles(path):
if os.path.isdir(path):
for root, dirs, files in os.walk(path):
for name in files:
yield os.path.join(root, name)
else:
yield path
destination = "./newdir/"
fromdir = "./test/"
for f in getfiles(fromdir):
filename = string.split(f, '/')[-1]
if os.path.isfile(destination+filename):
filename = f.replace(fromdir,"",1).replace("/","_")
#os.rename(f, destination+filename)
shutil.copy(f, destination+filename)
Solution 4:[4]
import os.path, shutil
def move(src, dest):
not_in_dest = lambda x: os.path.samefile(x, dest)
files_to_move = filter(not_in_dest,
glob_recursive(path=src))
for f in files_to_move:
shutil.move(f, dest)
Source for glob_recursive
. Does not change name of file, if they collide.
samefile
is a safe way to compare paths. But it doesn't work on Windows, so check How to emulate os.path.samefile behaviour on Windows and Python 2.7?.
Solution 5:[5]
def splitPath(p):
a,b = os.path.split(p)
return (splitPath(a) if len(a) and len(b) else []) + [b]
def safeprint(s):
try:
print(s)
except UnicodeEncodeError:
if sys.version_info >= (3,):
print(s.encode('utf8').decode(sys.stdout.encoding))
else:
print(s.encode('utf8'))
def flatten(root, doit):
SEP = "¦"
REPL = "?"
folderCount = 0
fileCount = 0
if not doit:
print("Simulating:")
for path, dirs, files in os.walk(root, topdown=False):
if path != root:
for f in files:
sp = splitPath(path)
np = ""
for element in sp[1:]:
e2 = element.replace(SEP, REPL)
np += e2 + SEP
f2 = f.replace(SEP, REPL)
newName = np + f2
safeprint("Moved: "+ newName )
if doit:
shutil.move(os.path.join(path, f), os.path.join(root, f))
# Uncomment, if you want filenames to be based on folder hierarchy.
#shutil.move(os.path.join(path, f), os.path.join(root, newName))
fileCount += 1
safeprint("Removed: "+ path)
if doit:
os.rmdir(path)
folderCount += 1
if doit:
print("Done.")
else:
print("Simulation complete.")
print("Moved files:", fileCount)
print("Removed folders:", folderCount)
directory_path = r"C:\Users\jd\Documents\myFtpData"
flatten(directory_path, True)
Solution 6:[6]
Adding on to the answers, I believe my answer will satisfy all your needs, the other answers fail when there is a subdirectory and file with the same filename as the upper directory.
This was SOLVED here, Also look at my Github Repo for Structured File Copy and Flattened File Copy:
import os, fnmatch, shutil
PATTERN = '*.txt' # Regex Pattern to Match files
INPUT_FOLDER = "A" # os.getcwd()
INPUT_FOLDER = os.path.abspath(INPUT_FOLDER)
include_input_foldername = False
prepend = "_included" if include_input_foldername else ""
OUTPUT_FOLDER = f"Structured_Copy_{os.path.basename(INPUT_FOLDER)}{prepend}"
os.makedirs(OUTPUT_FOLDER, exist_ok=True)
def find(pattern, path):
"""Utility to find files wrt a regex search"""
result = []
for root, dirs, files in os.walk(path):
for name in files:
if fnmatch.fnmatch(name, pattern):
result.append(os.path.join(root, name))
return result
all_files = find(PATTERN, INPUT_FOLDER)
for each_path in all_files:
relative_path = os.path.relpath(each_path, os.path.dirname(INPUT_FOLDER)) if include_input_foldername else os.path.relpath(each_path, INPUT_FOLDER)
flattened_relative_fullpath = os.path.join(OUTPUT_FOLDER, relative_path)
os.makedirs(os.path.dirname(flattened_relative_fullpath), exist_ok=True)
shutil.copy(each_path, flattened_relative_fullpath)
print(f"Copied {each_path} to {flattened_relative_fullpath}")
print(f"Finished Copying {len(all_files)} Files from : {INPUT_FOLDER} to : {OUTPUT_FOLDER}")
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | S.A. |
Solution 2 | |
Solution 3 | DRC |
Solution 4 | Community |
Solution 5 | Janzaib M Baloch |
Solution 6 | Farhan Hai Khan |