I have this where it reads a file called source1.html, source2.html, source3.html, but when it cant find the next file (because it doesnt exist) it gives me a error. there can be an x amount of sourceX.html, so i need something to say if the next sourcex.html file can not be found, stop the loop.
Traceback (most recent call last): File "main.py", line 14, in
file = open(filename, "r") IOError: [Errno 2] No such file or
directory: 'source4.html
how can i stop the script looking for the next source file?
from bs4 import BeautifulSoup
import re
import os.path
n = 1
filename = "source" + str(n) + ".html"
savefile = open('OUTPUT.csv', 'w')
while os.path.isfile(filename):
strjpgs = "Extracted Layers: \n \n"
filename = "source" + str(n) + ".html"
n = n + 1
file = open(filename, "r")
soup = BeautifulSoup(file, "html.parser")
thedata = soup.find("div", class_="cplayer")
strdata = str(thedata)
DoRegEx = re.compile('/([^/]+)\.jpg')
jpgs = DoRegEx.findall(strdata)
strjpgs = strjpgs + "\n".join(jpgs) + "\n \n"
savefile.write(filename + '\n')
savefile.write(strjpgs)
print(filename)
print(strjpgs)
savefile.close()
print "done"
use a try / except and break
while os.path.isfile(filename):
try: # try to do this
# <your code>
except FileNotFoundError: # if this error occurs
break # exit the loop
The reason your code doesn't currently work is you're checking the previous file exists in your while loop. Not the next one. Hence you could also do
while True:
strjpgs = "Extracted Layers: \n \n"
filename = "source" + str(n) + ".html"
if not os.path.isfile(filename):
break
# <rest of your code>
you can try opening file, and break out of while loop once you catch an IOError exception.
from bs4 import BeautifulSoup
import re
import os.path
n = 1
filename = "source" + str(n) + ".html"
savefile = open('OUTPUT.csv', 'w')
while os.path.isfile(filename):
try:
strjpgs = "Extracted Layers: \n \n"
filename = "source" + str(n) + ".html"
n = n + 1
file = open(filename, "r")
except IOError:
print("file not found! breaking out of loop.")
break
soup = BeautifulSoup(file, "html.parser")
thedata = soup.find("div", class_="cplayer")
strdata = str(thedata)
DoRegEx = re.compile('/([^/]+)\.jpg')
jpgs = DoRegEx.findall(strdata)
strjpgs = strjpgs + "\n".join(jpgs) + "\n \n"
savefile.write(filename + '\n')
savefile.write(strjpgs)
print(filename)
print(strjpgs)
savefile.close()
print "done"
I'll suggest you to use os.path.exists() (which returns True/False) and os.path.isfile() both.
Use with statement to open file. It is Pythonic way to open files.
with statement is best preferred among the professional coders.
These are the contents of my current working directory.
H:\RishikeshAgrawani\Projects\Stk\ReadHtmlFiles>dir
Volume in drive H is New Volume
Volume Serial Number is C867-828E
Directory of H:\RishikeshAgrawani\Projects\Stk\ReadHtmlFiles
11/05/2018 16:12 <DIR> .
11/05/2018 16:12 <DIR> ..
11/05/2018 15:54 106 source1.html
11/05/2018 15:54 106 source2.html
11/05/2018 15:54 106 source3.html
11/05/2018 16:12 0 stopReadingIfNot.md
11/05/2018 16:11 521 stopReadingIfNot.py
5 File(s) 839 bytes
2 Dir(s) 196,260,925,440 bytes free
The below Python code shows how will you read files source1.html, source2.html, source.3.html and stop if there is no more files of the form sourceX.html (where X is 1, 2, 3, 4, ... etc.).
Sample code:
import os
n = 1;
html_file_name = 'source%d.html'
# It is necessary to check if sourceX.html is file or directory.
# If it is directory the check it if it exists or not.
# It it exists then perform operation (read/write etc.) on file.
while os.path.isfile(html_file_name % (n)) and os.path.exists(html_file_name % (n)):
print "Reading ", html_file_name % (n)
# The best way (Pythonic way) to open file
# You don't need to bother about closing the file
# It will be taken care by with statement
with open(html_file_name % (n), "r") as file:
# Make sure it works
print html_file_name % (n), " exists\n";
n += 1;
Output:
H:\RishikeshAgrawani\Projects\Stk\ReadHtmlFiles>python stopReadingIfNot.py
Reading source1.html
source1.html exists
Reading source2.html
source2.html exists
Reading source3.html
source3.html exists
So based on the above logic. you can modify your code. It will work.
Thanks.
This appears to be a sequence error. Let's look at a small fragment of your code, specifically lines dealing with filename:
filename = "source" + str(n) + ".html"
while os.path.isfile(filename):
filename = "source" + str(n) + ".html"
n = n + 1
file = open(filename, "r")
You're generating the next filename before you open the file (or really, checking the old filename then opening a new one). It's a little hard to see because you're really updating n while filename holds the previous number, but if we look at them in sequence it pops out:
n = 1
filename = "source1.html" # before loop
while os.path.isfile(filename):
filename = "source1.html" # first time inside loop
n = 2
open(filename)
while os.path.isfile(filename): # second time in loop - still source1
filename = "source2.html"
n = 3
open(filename) # We haven't checked if this file exists!
We can fix this a few ways. One is to move the entire updating, n before filename, to the end of the loop. Another is to let the loop mechanism update n, which is a sight easier (the real fix here is that we only use one filename value in each iteration of the loop):
for n in itertools.count(1):
filename = "source{}.html".format(n)
if not os.path.isfile(filename):
break
file = open(filename, "r")
#...
At the risk of looking rather obscure, we can also express the steps functionally (I'm using six here to avoid a difference between Python 2 and 3; Python 2's map wouldn't finish):
from six.moves import map
from itertools import count, takewhile
numbers = count(1)
filenames = map('source{}.html'.format, numbers)
existingfiles = takewhile(os.path.isfile, filenames)
for filename in existingfiles:
file = open(filename, "r")
#...
Other options include iterating over the numbers alone and using break when isfile returns False, or simply catching the exception when open fails (eliminating the need for isfile entirely).
My goal is to get to a txt file that is withing the second layer of zip files. The issue is that the txt file has the same name in all the .zip, so it overwrites the .txt and it only returns 1 .txt
from ftplib import *
import os, shutil, glob, zipfile, xlsxwriter
ftps = FTP_TLS()
ftps.connect(host='8.8.8.8', port=23)
ftps.login(user='xxxxxxx', passwd='xxxxxxx')
print ftps.getwelcome()
print 'Access was granted'
ftps.prot_p()
ftps.cwd('DirectoryINeed')
data = ftps.nlst() #Returns a list of .zip diles
data.sort() #Sorts the thing out
theFile = data[-2] #Its a .zip file #Stores the .zip i need to retrieve
fileSize = ftps.size(theFile) #gets the size of the file
print fileSize, 'bytes' #prints the size
def grabFile():
filename = 'the.zip'
localfile = open(filename, 'wb')
ftps.retrbinary('RETR ' + theFile, localfile.write)
ftps.quit()
localfile.close()
def unzipping():
zip_files = glob.glob('*.zip')
for zip_file in zip_files:
with zipfile.ZipFile(zip_file, 'r')as Z:
Z.extractall('anotherdirectory')
grabFile()
unzipping()
lastUnzip()
After this runs it grabs the .zip that I need and extracts the contents to a folder named anotherdirectory. Where it holds the second tier of .zips. This is where I get into trouble. When I try to extract the files from each zip. They all share the same name. I end up with a single .txt when I need one for each zip.
I think you're specifying the same output directory and filename each time. In the unzipping function,
change
Z.extractall('anotherdirectory')
to
Z.extractall(zip_file)
or
Z.extractall('anotherdirectory' + zip_file)
if the zip_file's are all the same, give each output folder a unique numbered name:
before unzipping function:
count = 1
then replace the other code with this:
Z.extractall('anotherdirectory/' + str(count))
count += 1
Thanks to jeremydeanlakey's response, I was able to get this part of my script. Here is how I did it:
folderUnzip = 'DirectoryYouNeed'
zip_files = glob.glob('*.zip')
count = 1
for zip_file in zip_files:
with zipfile.ZipFile(zip_file, 'r') as Z:
Z.extractall(folderUnzip + '/' + str(count))
count += 1
I need to find every instance of "translate" in a text file and replace a value 4 lines after finding the text:
"(many lines)
}
}
translateX xtran
{
keys
{
k 0 0.5678
}
}
(many lines)"
The value 0.5678 needs to be 0. It will always be 4 lines below the "translate" string
The file has up to about 10,000 lines.
example text file name: 01F.pz2.
I'd also like to cycle through the folder and repeat the process for every file with the pz2 extension (up to 40).
Any help would be appreciated!
Thanks.
I'm not quite sure about the logic for replacing 0.5678 in your file, therefore I use a function for that - change it to whatever you need, or explain more in details what you want. Last number in line? only floating-point number?
Try:
import os
dirname = "14432826"
lines_distance= 4
def replace_whatever(line):
# Put your logic for replacing here
return line.replace("0.5678", "0")
for filename in filter(lambda x:x.endswith(".pz2") and not x.startswith("m_"), os.listdir(dirname)):
print filename
with open(os.path.join(dirname, filename), "r") as f_in, open(os.path.join(dirname,"m_%s" % filename), "w") as f_out:
replace_tasks = []
for line in f_in:
# search marker in line
if line.strip().startswith("translate"):
print "Found marker in", line,
replace_tasks.append(lines_distance)
# replace if necessary
if len(replace_tasks)>0 and replace_tasks[0] == 0:
del replace_tasks[0]
print "line to change is", line,
line_to_write = replace_whatever(line)
else:
line_to_write = line
# Write to output
f_out.write(line_to_write)
# decrease counters
for i, task in enumerate(replace_tasks):
replace_tasks[i] -= 1
The comments within the code should help understanding. The main concept is the list replace_tasks that keeps record of when the next line to modify will come.
Remarks: Your code sample suggests that the data in your file are structured. It will definitely be saver to read this structure and work on it instead of search-and-replace approach on a plain text file.
Thorsten, I renamed my original files to have the .old extension and the following code works:
import os
target_dir = "."
# cycle through files
for path, dirs, files in os.walk(target_dir):
# file is the file counter
for file in files:
# get the filename and extension
filename, ext = os.path.splitext(file)
# see if the file is a pz2
if ext.endswith('.old') :
# rename the file to "old"
oldfilename = filename + ".old"
newfilename = filename + ".pz2"
old_filepath = os.path.join(path, oldfilename)
new_filepath = os.path.join(path, newfilename)
# open the old file for reading
oldpz2 = open (old_filepath,"r")
# open the new file for writing
newpz2 = open (new_filepath,"w")
# reset changeline
changeline = 0
currentline = 0
# cycle through old lines
for line in oldpz2 :
currentline = currentline + 1
if line.strip().startswith("translate"):
changeline = currentline + 4
if currentline == changeline :
print >>newpz2," k 0 0"
else :
print >>newpz2,line
I am trying to iterate through a number .rtf files and for each file: read the file, perform some operations, and then write new files into a sub-directory as plain text files with the same name as the original file, but with .txt extensions. The problem I am having is with the file naming.
If a file is named foo.rtf, I want the new file in the subdirectory to be foo.txt. here is my code:
import glob
import os
import numpy as np
dir_path = '/Users/me/Desktop/test/'
file_suffix = '*.rtf'
output_dir = os.mkdir('sub_dir')
for item in glob.iglob(dir_path + file_suffix):
with open(item, "r") as infile:
reader = infile.readlines()
matrix = []
for row in reader:
row = str(row)
row = row.split()
row = [int(value) for value in row]
matrix.append(row)
np_matrix = np.array(matrix)
inv_matrix = np.transpose(np_matrix)
new_file_name = item.replace('*.rtf', '*.txt') # i think this line is the problem?
os.chdir(output_dir)
with open(new_file_name, mode="w") as outfile:
outfile.write(inv_matrix)
When I run this code, I get a Type Error:
TypeError: coercing to Unicode: need string or buffer, NoneType found
How can I fix my code to write new files into a subdirectory and change the file extensions from .rtf to .txt? Thanks for the help.
Instead of item.replace, check out some of the functions in the os.path module (http://docs.python.org/library/os.path.html). They're made for splitting up and recombining parts of filenames. For instance, os.path.splitext will split a filename into a file path and a file extension.
Let's say you have a file /tmp/foo.rtf and you want to move it to /tmp/foo.txt:
old_file = '/tmp/foo.rtf'
(file,ext) = os.path.splitext(old_file)
print 'File=%s Extension=%s' % (file,ext)
new_file = '%s%s' % (file,'.txt')
print 'New file = %s' % (new_file)
Or if you want the one line version:
old_file = '/tmp/foo.rtf'
new_file = '%s%s' % (os.path.splitext(old_file)[0],'.txt')
I've never used glob, but here's an alternative way without using a module:
You can easily strip the suffix using
name = name[:name.rfind('.')]
and then add the new suffix:
name = name + '.txt'
Why not using a function ?
def change_suffix(string, new_suffix):
i = string.rfind('.')
if i < 0:
raise ValueError, 'string does not have a suffix'
if not new_suffix[0] == '.':
new_suffix += '.'
return string[:i] + new_suffix
glob.iglob() yields pathnames, without the character '*'.
therefore your line should be:
new_file_name = item.replace('.rtf', '.txt')
consider working with clearer names (reserve 'filename' for a file name and use 'path' for a complete path to a file; use 'path_original' instead of 'item'), os.extsep ('.' in Windows) and os.path.splitext():
path_txt = os.extsep.join([os.path.splitext(path_original)[0], 'txt'])
now the best hint of all:
numpy can probably read your file directly:
data = np.genfromtxt(filename, unpack=True)
(see also here)
To better understand where your TypeError comes from, wrap your code in the following try/except block:
try:
(your code)
except:
import traceback
traceback.print_exc()
I am writing a Python program to find and remove duplicate files from a folder.
I have multiple copies of mp3 files, and some other files. I am using the sh1 algorithm.
How can I find these duplicate files and remove them?
Fastest algorithm - 100x performance increase compared to the accepted answer (really :))
The approaches in the other solutions are very cool, but they forget about an important property of duplicate files - they have the same file size. Calculating the expensive hash only on files with the same size will save tremendous amount of CPU; performance comparisons at the end, here's the explanation.
Iterating on the solid answers given by #nosklo and borrowing the idea of #Raffi to have a fast hash of just the beginning of each file, and calculating the full one only on collisions in the fast hash, here are the steps:
Buildup a hash table of the files, where the filesize is the key.
For files with the same size, create a hash table with the hash of their first 1024 bytes; non-colliding elements are unique
For files with the same hash on the first 1k bytes, calculate the hash on the full contents - files with matching ones are NOT unique.
The code:
#!/usr/bin/env python3
from collections import defaultdict
import hashlib
import os
import sys
def chunk_reader(fobj, chunk_size=1024):
"""Generator that reads a file in chunks of bytes"""
while True:
chunk = fobj.read(chunk_size)
if not chunk:
return
yield chunk
def get_hash(filename, first_chunk_only=False, hash=hashlib.sha1):
hashobj = hash()
file_object = open(filename, 'rb')
if first_chunk_only:
hashobj.update(file_object.read(1024))
else:
for chunk in chunk_reader(file_object):
hashobj.update(chunk)
hashed = hashobj.digest()
file_object.close()
return hashed
def check_for_duplicates(paths, hash=hashlib.sha1):
hashes_by_size = defaultdict(list) # dict of size_in_bytes: [full_path_to_file1, full_path_to_file2, ]
hashes_on_1k = defaultdict(list) # dict of (hash1k, size_in_bytes): [full_path_to_file1, full_path_to_file2, ]
hashes_full = {} # dict of full_file_hash: full_path_to_file_string
for path in paths:
for dirpath, dirnames, filenames in os.walk(path):
# get all files that have the same size - they are the collision candidates
for filename in filenames:
full_path = os.path.join(dirpath, filename)
try:
# if the target is a symlink (soft one), this will
# dereference it - change the value to the actual target file
full_path = os.path.realpath(full_path)
file_size = os.path.getsize(full_path)
hashes_by_size[file_size].append(full_path)
except (OSError,):
# not accessible (permissions, etc) - pass on
continue
# For all files with the same file size, get their hash on the 1st 1024 bytes only
for size_in_bytes, files in hashes_by_size.items():
if len(files) < 2:
continue # this file size is unique, no need to spend CPU cycles on it
for filename in files:
try:
small_hash = get_hash(filename, first_chunk_only=True)
# the key is the hash on the first 1024 bytes plus the size - to
# avoid collisions on equal hashes in the first part of the file
# credits to #Futal for the optimization
hashes_on_1k[(small_hash, size_in_bytes)].append(filename)
except (OSError,):
# the file access might've changed till the exec point got here
continue
# For all files with the hash on the 1st 1024 bytes, get their hash on the full file - collisions will be duplicates
for __, files_list in hashes_on_1k.items():
if len(files_list) < 2:
continue # this hash of fist 1k file bytes is unique, no need to spend cpy cycles on it
for filename in files_list:
try:
full_hash = get_hash(filename, first_chunk_only=False)
duplicate = hashes_full.get(full_hash)
if duplicate:
print("Duplicate found: {} and {}".format(filename, duplicate))
else:
hashes_full[full_hash] = filename
except (OSError,):
# the file access might've changed till the exec point got here
continue
if __name__ == "__main__":
if sys.argv[1:]:
check_for_duplicates(sys.argv[1:])
else:
print("Please pass the paths to check as parameters to the script")
And, here's the fun part - performance comparisons.
Baseline -
a directory with 1047 files, 32 mp4, 1015 - jpg, total size - 5445.998 MiB - i.e. my phone's camera auto upload directory :)
small (but fully functional) processor - 1600 BogoMIPS, 1.2 GHz 32L1 + 256L2 Kbs cache, /proc/cpuinfo:
Processor : Feroceon 88FR131 rev 1 (v5l)
BogoMIPS : 1599.07
(i.e. my low-end NAS :), running Python 2.7.11.
So, the output of #nosklo's very handy solution:
root#NAS:InstantUpload# time ~/scripts/checkDuplicates.py
Duplicate found: ./IMG_20151231_143053 (2).jpg and ./IMG_20151231_143053.jpg
Duplicate found: ./IMG_20151125_233019 (2).jpg and ./IMG_20151125_233019.jpg
Duplicate found: ./IMG_20160204_150311.jpg and ./IMG_20160204_150311 (2).jpg
Duplicate found: ./IMG_20160216_074620 (2).jpg and ./IMG_20160216_074620.jpg
real 5m44.198s
user 4m44.550s
sys 0m33.530s
And, here's the version with filter on size check, then small hashes, and finally full hash if collisions are found:
root#NAS:InstantUpload# time ~/scripts/checkDuplicatesSmallHash.py . "/i-data/51608399/photo/Todor phone"
Duplicate found: ./IMG_20160216_074620 (2).jpg and ./IMG_20160216_074620.jpg
Duplicate found: ./IMG_20160204_150311.jpg and ./IMG_20160204_150311 (2).jpg
Duplicate found: ./IMG_20151231_143053 (2).jpg and ./IMG_20151231_143053.jpg
Duplicate found: ./IMG_20151125_233019 (2).jpg and ./IMG_20151125_233019.jpg
real 0m1.398s
user 0m1.200s
sys 0m0.080s
Both versions were ran 3 times each, to get the avg of the time needed.
So v1 is (user+sys) 284s, the other - 2s; quite a diff, huh :)
With this increase, one could go to SHA512, or even fancier - the perf penalty will be mitigated by the less calculations needed.
Negatives:
More disk access than the other versions - every file is accessed once for size stats (that's cheap, but still is disk IO), and every duplicate is opened twice (for the small first 1k bytes hash, and for the full contents hash)
Will consume more memory due to storing the hash tables runtime
Recursive folders version:
This version uses the file size and a hash of the contents to find duplicates.
You can pass it multiple paths, it will scan all paths recursively and report all duplicates found.
import sys
import os
import hashlib
def chunk_reader(fobj, chunk_size=1024):
"""Generator that reads a file in chunks of bytes"""
while True:
chunk = fobj.read(chunk_size)
if not chunk:
return
yield chunk
def check_for_duplicates(paths, hash=hashlib.sha1):
hashes = {}
for path in paths:
for dirpath, dirnames, filenames in os.walk(path):
for filename in filenames:
full_path = os.path.join(dirpath, filename)
hashobj = hash()
for chunk in chunk_reader(open(full_path, 'rb')):
hashobj.update(chunk)
file_id = (hashobj.digest(), os.path.getsize(full_path))
duplicate = hashes.get(file_id, None)
if duplicate:
print "Duplicate found: %s and %s" % (full_path, duplicate)
else:
hashes[file_id] = full_path
if sys.argv[1:]:
check_for_duplicates(sys.argv[1:])
else:
print "Please pass the paths to check as parameters to the script"
def remove_duplicates(dir):
unique = []
for filename in os.listdir(dir):
if os.path.isfile(filename):
filehash = md5.md5(file(filename).read()).hexdigest()
if filehash not in unique:
unique.append(filehash)
else:
os.remove(filename)
//edit:
For MP3 you may be also interested in this topic Detect duplicate MP3 files with different bitrates and/or different ID3 tags?
I wrote one in Python some time ago -- you're welcome to use it.
import sys
import os
import hashlib
check_path = (lambda filepath, hashes, p = sys.stdout.write:
(lambda hash = hashlib.sha1 (file (filepath).read ()).hexdigest ():
((hash in hashes) and (p ('DUPLICATE FILE\n'
' %s\n'
'of %s\n' % (filepath, hashes[hash])))
or hashes.setdefault (hash, filepath)))())
scan = (lambda dirpath, hashes = {}:
map (lambda (root, dirs, files):
map (lambda filename: check_path (os.path.join (root, filename), hashes), files), os.walk (dirpath)))
((len (sys.argv) > 1) and scan (sys.argv[1]))
Faster algorithm
In case many files of 'big size' should be analyzed (images, mp3, pdf documents), it would be interesting/faster to have the following comparison algorithm:
a first fast hash is performed on the first N bytes of the file (say 1KB). This hash would say if files are different without doubt, but will not say if two files are exactly the same (accuracy of the hash, limited data read from disk)
a second, slower, hash, which is more accurate and performed on the whole content of the file, if a collision occurs in the first stage
Here is an implementation of this algorithm:
import hashlib
def Checksum(current_file_name, check_type = 'sha512', first_block = False):
"""Computes the hash for the given file. If first_block is True,
only the first block of size size_block is hashed."""
size_block = 1024 * 1024 # The first N bytes (1KB)
d = {'sha1' : hashlib.sha1, 'md5': hashlib.md5, 'sha512': hashlib.sha512}
if(not d.has_key(check_type)):
raise Exception("Unknown checksum method")
file_size = os.stat(current_file_name)[stat.ST_SIZE]
with file(current_file_name, 'rb') as f:
key = d[check_type].__call__()
while True:
s = f.read(size_block)
key.update(s)
file_size -= size_block
if(len(s) < size_block or first_block):
break
return key.hexdigest().upper()
def find_duplicates(files):
"""Find duplicates among a set of files.
The implementation uses two types of hashes:
- A small and fast one one the first block of the file (first 1KB),
- and in case of collision a complete hash on the file. The complete hash
is not computed twice.
It flushes the files that seems to have the same content
(according to the hash method) at the end.
"""
print 'Analyzing', len(files), 'files'
# this dictionary will receive small hashes
d = {}
# this dictionary will receive full hashes. It is filled
# only in case of collision on the small hash (contains at least two
# elements)
duplicates = {}
for f in files:
# small hash to be fast
check = Checksum(f, first_block = True, check_type = 'sha1')
if(not d.has_key(check)):
# d[check] is a list of files that have the same small hash
d[check] = [(f, None)]
else:
l = d[check]
l.append((f, None))
for index, (ff, checkfull) in enumerate(l):
if(checkfull is None):
# computes the full hash in case of collision
checkfull = Checksum(ff, first_block = False)
l[index] = (ff, checkfull)
# for each new full hash computed, check if their is
# a collision in the duplicate dictionary.
if(not duplicates.has_key(checkfull)):
duplicates[checkfull] = [ff]
else:
duplicates[checkfull].append(ff)
# prints the detected duplicates
if(len(duplicates) != 0):
print
print "The following files have the same sha512 hash"
for h, lf in duplicates.items():
if(len(lf)==1):
continue
print 'Hash value', h
for f in lf:
print '\t', f.encode('unicode_escape') if \
type(f) is types.UnicodeType else f
return duplicates
The find_duplicates function takes a list of files. This way, it is also possible to compare two directories (for instance, to better synchronize their content.) An example of function creating a list of files, with specified extension, and avoiding entering in some directories, is below:
def getFiles(_path, extensions = ['.png'],
subdirs = False, avoid_directories = None):
"""Returns the list of files in the path :'_path',
of extension in 'extensions'. 'subdir' indicates if
the search should also be performed in the subdirectories.
If extensions = [] or None, all files are returned.
avoid_directories: if set, do not parse subdirectories that
match any element of avoid_directories."""
l = []
extensions = [p.lower() for p in extensions] if not extensions is None \
else None
for root, dirs, files in os.walk(_path, topdown=True):
for name in files:
if(extensions is None or len(extensions) == 0 or \
os.path.splitext(name)[1].lower() in extensions):
l.append(os.path.join(root, name))
if(not subdirs):
while(len(dirs) > 0):
dirs.pop()
elif(not avoid_directories is None):
for d in avoid_directories:
if(d in dirs): dirs.remove(d)
return l
This method is convenient for not parsing .svn paths for instance, which surely will trigger colliding files in find_duplicates.
Feedbacks are welcome.
#IanLee1521 has a nice solution here. It is very efficient because it checks the duplicate based on the file size first.
#! /usr/bin/env python
# Originally taken from:
# http://www.pythoncentral.io/finding-duplicate-files-with-python/
# Original Auther: Andres Torres
# Adapted to only compute the md5sum of files with the same size
import argparse
import os
import sys
import hashlib
def find_duplicates(folders):
"""
Takes in an iterable of folders and prints & returns the duplicate files
"""
dup_size = {}
for i in folders:
# Iterate the folders given
if os.path.exists(i):
# Find the duplicated files and append them to dup_size
join_dicts(dup_size, find_duplicate_size(i))
else:
print('%s is not a valid path, please verify' % i)
return {}
print('Comparing files with the same size...')
dups = {}
for dup_list in dup_size.values():
if len(dup_list) > 1:
join_dicts(dups, find_duplicate_hash(dup_list))
print_results(dups)
return dups
def find_duplicate_size(parent_dir):
# Dups in format {hash:[names]}
dups = {}
for dirName, subdirs, fileList in os.walk(parent_dir):
print('Scanning %s...' % dirName)
for filename in fileList:
# Get the path to the file
path = os.path.join(dirName, filename)
# Check to make sure the path is valid.
if not os.path.exists(path):
continue
# Calculate sizes
file_size = os.path.getsize(path)
# Add or append the file path
if file_size in dups:
dups[file_size].append(path)
else:
dups[file_size] = [path]
return dups
def find_duplicate_hash(file_list):
print('Comparing: ')
for filename in file_list:
print(' {}'.format(filename))
dups = {}
for path in file_list:
file_hash = hashfile(path)
if file_hash in dups:
dups[file_hash].append(path)
else:
dups[file_hash] = [path]
return dups
# Joins two dictionaries
def join_dicts(dict1, dict2):
for key in dict2.keys():
if key in dict1:
dict1[key] = dict1[key] + dict2[key]
else:
dict1[key] = dict2[key]
def hashfile(path, blocksize=65536):
afile = open(path, 'rb')
hasher = hashlib.md5()
buf = afile.read(blocksize)
while len(buf) > 0:
hasher.update(buf)
buf = afile.read(blocksize)
afile.close()
return hasher.hexdigest()
def print_results(dict1):
results = list(filter(lambda x: len(x) > 1, dict1.values()))
if len(results) > 0:
print('Duplicates Found:')
print(
'The following files are identical. The name could differ, but the'
' content is identical'
)
print('___________________')
for result in results:
for subresult in result:
print('\t\t%s' % subresult)
print('___________________')
else:
print('No duplicate files found.')
def main():
parser = argparse.ArgumentParser(description='Find duplicate files')
parser.add_argument(
'folders', metavar='dir', type=str, nargs='+',
help='A directory to parse for duplicates',
)
args = parser.parse_args()
find_duplicates(args.folders)
if __name__ == '__main__':
sys.exit(main())
import hashlib
import os
import sys
from sets import Set
def read_chunk(fobj, chunk_size = 2048):
""" Files can be huge so read them in chunks of bytes. """
while True:
chunk = fobj.read(chunk_size)
if not chunk:
return
yield chunk
def remove_duplicates(dir, hashfun = hashlib.sha512):
unique = Set()
for filename in os.listdir(dir):
filepath = os.path.join(dir, filename)
if os.path.isfile(filepath):
hashobj = hashfun()
for chunk in read_chunk(open(filepath,'rb')):
hashobj.update(chunk)
# the size of the hashobj is constant
# print "hashfun: ", hashfun.__sizeof__()
hashfile = hashobj.hexdigest()
if hashfile not in unique:
unique.add(hashfile)
else:
os.remove(filepath)
try:
hashfun = hashlib.sha256
remove_duplicates(sys.argv[1], hashfun)
except IndexError:
print """Please pass a path to a directory with
duplicate files as a parameter to the script."""
Python has a standard library called filecmp to compare files and directories.
It checks for file size. It checks content in 8k chunks.
It works on binary files.
It does not hash.
python docs for filecmp
In order to be safe (removing them automatically can be dangerous if something goes wrong!), here is what I use, based on #zalew's answer.
Pleas also note that the md5 sum code is slightly different from #zalew's because his code generated too many wrong duplicate files (that's why I said removing them automatically is dangerous!).
import hashlib, os
unique = dict()
for filename in os.listdir('.'):
if os.path.isfile(filename):
filehash = hashlib.md5(open(filename, 'rb').read()).hexdigest()
if filehash not in unique:
unique[filehash] = filename
else:
print filename + ' is a duplicate of ' + unique[filehash]
I have found a 100% working code for removing duplicate files recursively inside a folder. Just replace the folder name in the clean method with your folder name.
import time
import os
import shutil
from hashlib import sha256
class Duplython:
def __init__(self):
self.home_dir = os.getcwd()
self.File_hashes = []
self.Cleaned_dirs = []
self.Total_bytes_saved = 0
self.block_size = 65536
self.count_cleaned = 0
def welcome(self) -> None:
print('******************************************************************')
print('**************** DUPLYTHON ****************************')
print('********************************************************************\n\n')
print('---------------- WELCOME ----------------------------')
time.sleep(3)
print('\nCleaning .................')
return None
def generate_hash(self, Filename: str) -> str:
Filehash = sha256()
try:
with open(Filename, 'rb') as File:
fileblock = File.read(self.block_size)
while len(fileblock) > 0:
Filehash.update(fileblock)
fileblock = File.read(self.block_size)
Filehash = Filehash.hexdigest()
return Filehash
except:
return False
def clean(self) -> None:
all_dirs = [path[0] for path in os.walk('E:\\songs')]
for path in all_dirs:
os.chdir(path)
All_Files = [file for file in os.listdir() if os.path.isfile(file)]
for file in All_Files:
filehash = self.generate_hash(file)
if not filehash in self.File_hashes:
if filehash:
self.File_hashes.append(filehash)
# print(file)
else:
byte_saved = os.path.getsize(file)
self.count_cleaned += 1
self.Total_bytes_saved += byte_saved
os.remove(file)
filename = file.split('/')[-1]
print(filename, '.. cleaned ')
os.chdir(self.home_dir)
def cleaning_summary(self) -> None:
mb_saved = self.Total_bytes_saved / 1048576
mb_saved = round(mb_saved, 2)
print('\n\n--------------FINISHED CLEANING ------------')
print('File cleaned : ', self.count_cleaned)
print('Total Space saved : ', mb_saved, 'MB')
print('-----------------------------------------------')
def main(self) -> None:
self.welcome()
self.clean()
self.cleaning_summary()
#
# if __name__ == '__main__':
# App = Duplython()
# App.main()
def dedupe_bing_images():
App = Duplython()
App.main()
return True
dedupe_bing_images()