I have some files in an array that I want to recursively search from many folders
An example of the filename array is ['A_010720_X.txt','B_120720_Y.txt']
Example of folder structure is as below which I can also provide as an array e.g ['A','B'] and ['2020-07-01','2020-07-12']. The "DL" remains the same for all.
C:\A\2020-07-01\DL
C:\B\2020-07-12\DL
etc
I have tried to use shutil but it doesn't seem to work effectively for my requirement as I can only pass in a full file name and not a wildcard. The code I have used with shutil which works but without wildcards and with absolute full file name and path e.g the code below will only give me A_010720_X.txt
I believe the way to go would be using glob or pathlib which i have not used before or cannot find some good examples similar to my use case
import shutil
filenames_i_want = ['A_010720_X.txt','B_120720_Y.txt']
RootDir1 = r'C:\A\2020-07-01\DL'
TargetFolder = r'C:\ELK\LOGS\ATH\DEST'
for root, dirs, files in os.walk((os.path.normpath(RootDir1)), topdown=False):
for name in files:
if name in filenames_i_want:
print ("Found")
SourceFolder = os.path.join(root,name)
shutil.copy2(SourceFolder, TargetFolder)
I think this should do what you need assuming they are all .txt files.
import glob
import shutil
filenames_i_want = ['A_010720_X.txt','B_120720_Y.txt']
TargetFolder = r'C:\ELK\LOGS\ATH\DEST'
all_files = []
for directory in ['A', 'B']:
files = glob.glob('C:\{}\*\DL\*.txt'.format(directory))
all_files.append(files)
for file in all_files:
if file in filenames_i_want:
shutil.copy2(file, TargetFolder)
I was looking for ways to loop over files in directory with python, and I found this question:
Loop through all CSV files in a folder
The point is that the files I have are binary files, with no file extension at the end.
What I want my program to do is to iterate through all the files that have no extension.
Anyway to apply this using wildcards? (Or any other way?)
You can use os.path.splitext to check if a file has an extension or not.
See this examples:
import os
os.path.splitext("foo.ext")
=> ('foo', '.ext')
os.path.splitext("foo")
=> ('foo', '')
So, you can do that:
import os
path = "path/to/files"
dirs = os.listdir(path)
for path in dirs:
if not os.path.splitext(path)[1]:
print(path)
But, beware of "hidden" files which name starts with a dot, ie.: ".bashrc".
You can also check for the existence of a dot in the filename:
for path in dirs:
if "." not in path:
print(path)
Sounds like what you are interested in is
[f for f in next(os.walk(folder))[2] if '.' not in f]
I would suggest using os.listdir(), and then check whether filename has an extension (check if there is a dot in a filename). Once You get all filenames without dots (that is, without extension), just be sure to check that the filename isn't actually directory name, and that's it.
You could use the glob module and filter out any files with extensions:
import glob
for filename in (filename for filename in glob.iglob('*') if '.' not in filename):
print(filename)
There is an mkv file in a folder named "export". What I want to do is to make a python script which fetches the file name from that export folder.
Let's say the folder is at "C:\Users\UserName\Desktop\New_folder\export".
How do I fetch the name?
I tried using this os.path.basename and os.path.splitext .. well.. didn't work out like I expected.
os.path implements some useful functions on pathnames. But it doesn't have access to the contents of the path. For that purpose, you can use os.listdir.
The following command will give you a list of the contents of the given path:
os.listdir("C:\Users\UserName\Desktop\New_folder\export")
Now, if you just want .mkv files you can use fnmatch(This module provides support for Unix shell-style wildcards) module to get your expected file names:
import fnmatch
import os
print([f for f in os.listdir("C:\Users\UserName\Desktop\New_folder\export") if fnmatch.fnmatch(f, '*.mkv')])
Also as #Padraic Cunningham mentioned as a more pythonic way for dealing with file names you can use glob module :
map(path.basename,glob.iglob(pth+"*.mkv"))
You can use glob:
from glob import glob
pth ="C:/Users/UserName/Desktop/New_folder/export/"
print(glob(pth+"*.mkv"))
path+"*.mkv" will match all the files ending with .mkv.
To just get the basenames you can use map or a list comp with iglob:
from glob import iglob
print(list(map(path.basename,iglob(pth+"*.mkv"))))
print([path.basename(f) for f in iglob(pth+"*.mkv")])
iglob returns an iterator so you don't build a list for no reason.
I assume you're basically asking how to list files in a given directory. What you want is:
import os
print os.listdir("""C:\Users\UserName\Desktop\New_folder\export""")
If there's multiple files and you want the one(s) that have a .mkv end you could do:
import os
files = os.listdir("""C:\Users\UserName\Desktop\New_folder\export""")
mkv_files = [_ for _ in files if _[-4:] == ".mkv"]
print mkv_files
If you are searching for recursive folder search, this method will help you to get filename using os.walk, also you can get those file's path and directory using this below code.
import os, fnmatch
for path, dirs, files in os.walk(os.path.abspath(r"C:/Users/UserName/Desktop/New_folder/export/")):
for filename in fnmatch.filter(files, "*.mkv"):
print(filename)
You can use glob
import glob
for file in glob.glob('C:\Users\UserName\Desktop\New_folder\export\*.mkv'):
print(str(file).split('\')[-1])
This will list out all the files having extention .mkv as
file.mkv, file2.mkv and so on.
From os.walk you can read file paths as a list
files = [ file_path for _, _, file_path in os.walk(DIRECTORY_PATH)]
for file_name in files[0]: #note that it has list of lists
print(file_name)
I am trying to write a simple function that returns a list of files in a directory and its subdirectories. I shamelessly stole the majority of this function from another SO poster. I am using Python 2.6.4.
def getFiles(Asite):
# returns a list of config files
from os import listdir
from os.path import isfile, join
mypath = '/etc/config/' + Asite
print mypath
files = [ f for f in listdir(mypath) if isfile(join(mypath,f)) ]
return files
The function simply returns an empty list, []. It appears that the mypath variable is not being interpolated by the listdir() and isfile() functions. Before anyone asks, yes, I have verified that there are in fact files located at mypath. Why is my files array empty?
Thanks all for your helpful comments. It turns out that the directory I was searching in only had sudirectories in it, but not files. So os.listdir() didn't work for searching subdirectories, as it only goes one level (thank you abhishekgarg). I ended up using the code below, which worked great (which I also found on SO, and modified a bit).
def getFiles(Asite):
# returns a list of config files
files = []
mypath = '/share/profile/base/sol-10-sparc-base/config/' + Asite
for dirname, dirnames, filenames in os.walk(mypath):
for filename in filenames:
pfile = os.path.join(dirname, filename)
files.append(pfile[len(mypath):])
return files
This question already has answers here:
How to use glob() to find files recursively?
(28 answers)
Closed 1 year ago.
I want to open a series of subfolders in a folder and find some text files and print some lines of the text files. I am using this:
configfiles = glob.glob('C:/Users/sam/Desktop/file1/*.txt')
But this cannot access the subfolders as well. Does anyone know how I can use the same command to access subfolders as well?
In Python 3.5 and newer use the new recursive **/ functionality:
configfiles = glob.glob('C:/Users/sam/Desktop/file1/**/*.txt', recursive=True)
When recursive is set, ** followed by a path separator matches 0 or more subdirectories.
In earlier Python versions, glob.glob() cannot list files in subdirectories recursively.
In that case I'd use os.walk() combined with fnmatch.filter() instead:
import os
import fnmatch
path = 'C:/Users/sam/Desktop/file1'
configfiles = [os.path.join(dirpath, f)
for dirpath, dirnames, files in os.walk(path)
for f in fnmatch.filter(files, '*.txt')]
This'll walk your directories recursively and return all absolute pathnames to matching .txt files. In this specific case the fnmatch.filter() may be overkill, you could also use a .endswith() test:
import os
path = 'C:/Users/sam/Desktop/file1'
configfiles = [os.path.join(dirpath, f)
for dirpath, dirnames, files in os.walk(path)
for f in files if f.endswith('.txt')]
There's a lot of confusion on this topic. Let me see if I can clarify it (Python 3.7):
glob.glob('*.txt') :matches all files ending in '.txt' in current directory
glob.glob('*/*.txt') :same as 1
glob.glob('**/*.txt') :matches all files ending in '.txt' in the immediate subdirectories only, but not in the current directory
glob.glob('*.txt',recursive=True) :same as 1
glob.glob('*/*.txt',recursive=True) :same as 3
glob.glob('**/*.txt',recursive=True):matches all files ending in '.txt' in the current directory and in all subdirectories
So it's best to always specify recursive=True.
To find files in immediate subdirectories:
configfiles = glob.glob(r'C:\Users\sam\Desktop\*\*.txt')
For a recursive version that traverse all subdirectories, you could use ** and pass recursive=True since Python 3.5:
configfiles = glob.glob(r'C:\Users\sam\Desktop\**\*.txt', recursive=True)
Both function calls return lists. You could use glob.iglob() to return paths one by one. Or use pathlib:
from pathlib import Path
path = Path(r'C:\Users\sam\Desktop')
txt_files_only_subdirs = path.glob('*/*.txt')
txt_files_all_recursively = path.rglob('*.txt') # including the current dir
Both methods return iterators (you can get paths one by one).
The glob2 package supports wild cards and is reasonably fast
code = '''
import glob2
glob2.glob("files/*/**")
'''
timeit.timeit(code, number=1)
On my laptop it takes approximately 2 seconds to match >60,000 file paths.
You can use Formic with Python 2.6
import formic
fileset = formic.FileSet(include="**/*.txt", directory="C:/Users/sam/Desktop/")
Disclosure - I am the author of this package.
Here is a adapted version that enables glob.glob like functionality without using glob2.
def find_files(directory, pattern='*'):
if not os.path.exists(directory):
raise ValueError("Directory not found {}".format(directory))
matches = []
for root, dirnames, filenames in os.walk(directory):
for filename in filenames:
full_path = os.path.join(root, filename)
if fnmatch.filter([full_path], pattern):
matches.append(os.path.join(root, filename))
return matches
So if you have the following dir structure
tests/files
├── a0
│ ├── a0.txt
│ ├── a0.yaml
│ └── b0
│ ├── b0.yaml
│ └── b00.yaml
└── a1
You can do something like this
files = utils.find_files('tests/files','**/b0/b*.yaml')
> ['tests/files/a0/b0/b0.yaml', 'tests/files/a0/b0/b00.yaml']
Pretty much fnmatch pattern match on the whole filename itself, rather than the filename only.
(The first options are of course mentioned in other answers, here the goal is to show that glob uses os.scandir internally, and provide a direct answer with this).
Using glob
As explained before, with Python 3.5+, it's easy:
import glob
for f in glob.glob('d:/temp/**/*', recursive=True):
print(f)
#d:\temp\New folder
#d:\temp\New Text Document - Copy.txt
#d:\temp\New folder\New Text Document - Copy.txt
#d:\temp\New folder\New Text Document.txt
Using pathlib
from pathlib import Path
for f in Path('d:/temp').glob('**/*'):
print(f)
Using os.scandir
os.scandir is what glob does internally. So here is how to do it directly, with a use of yield:
def listpath(path):
for f in os.scandir(path):
f2 = os.path.join(path, f)
if os.path.isdir(f):
yield f2
yield from listpath(f2)
else:
yield f2
for f in listpath('d:\\temp'):
print(f)
configfiles = glob.glob('C:/Users/sam/Desktop/**/*.txt")
Doesn't works for all cases, instead use glob2
configfiles = glob2.glob('C:/Users/sam/Desktop/**/*.txt")
If you can install glob2 package...
import glob2
filenames = glob2.glob("C:\\top_directory\\**\\*.ext") # Where ext is a specific file extension
folders = glob2.glob("C:\\top_directory\\**\\")
All filenames and folders:
all_ff = glob2.glob("C:\\top_directory\\**\\**")
If you're running Python 3.4+, you can use the pathlib module. The Path.glob() method supports the ** pattern, which means “this directory and all subdirectories, recursively”. It returns a generator yielding Path objects for all matching files.
from pathlib import Path
configfiles = Path("C:/Users/sam/Desktop/file1/").glob("**/*.txt")
You can use the function glob.glob() or glob.iglob() directly from glob module to retrieve paths recursively from inside the directories/files and subdirectories/subfiles.
Syntax:
glob.glob(pathname, *, recursive=False) # pathname = '/path/to/the/directory' or subdirectory
glob.iglob(pathname, *, recursive=False)
In your example, it is possible to write like this:
import glob
import os
configfiles = [f for f in glob.glob("C:/Users/sam/Desktop/*.txt")]
for f in configfiles:
print(f'Filename with path: {f}')
print(f'Only filename: {os.path.basename(f)}')
print(f'Filename without extensions: {os.path.splitext(os.path.basename(f))[0]}')
Output:
Filename with path: C:/Users/sam/Desktop/test_file.txt
Only filename: test_file.txt
Filename without extensions: test_file
Help:
Documentation for os.path.splitext and documentation for os.path.basename.
As pointed out by Martijn, glob can only do this through the **operator introduced in Python 3.5. Since the OP explicitly asked for the glob module, the following will return a lazy evaluation iterator that behaves similarly
import os, glob, itertools
configfiles = itertools.chain.from_iterable(glob.iglob(os.path.join(root,'*.txt'))
for root, dirs, files in os.walk('C:/Users/sam/Desktop/file1/'))
Note that you can only iterate once over configfiles in this approach though. If you require a real list of configfiles that can be used in multiple operations you would have to create this explicitly by using list(configfiles).
The command rglob will do an infinite recursion down the deepest sub-level of your directory structure. If you only want one level deep, then do not use it, however.
I realize the OP was talking about using glob.glob. I believe this answers the intent, however, which is to search all subfolders recursively.
The rglob function recently produced a 100x increase in speed for a data processing algorithm which was using the folder structure as a fixed assumption for the order of data reading. However, with rglob we were able to do a single scan once through all files at or below a specified parent directory, save their names to a list (over a million files), then use that list to determine which files we needed to open at any point in the future based on the file naming conventions only vs. which folder they were in.