Find a file 1 folder level down - python

I am trying to get full_path of places.sqlite file present in '%APPDATA%\Mozilla\Firefox\Profiles\<random_folder>\places.sqlite' using Python OS module. The issue as you can see that <random_folder> has a random name and there could be multiple folders inside the Profiles folder.
How do I navigate/find the path to the places.sqlite file?

You would ideally want to go through each folder to search for this file. In terminal 'locate file_name' command would do this for you. In python file you can use the following command:
import os
db_path = os.path.join(os.getenv('APPDATA'), r'Mozilla\Firefox\Profiles')
def find_file(file_name, path):
for root_folder, directory, file_names in os.walk(path):
if file_name in file_names:
return os.path.join(root_folder, file_name)
print(find_file('places.sqlite', db_path))

os.walk gives a list of all files in a path recusivly. Use it to search for 'places.sqlite' as follows.
path = ""
for root, dirs, files in os.walk("%APPDATA%\\Mozilla\\Firefox\\Profiles\\"):
if "places.sqlite" in files:
path = os.path.join(root, 'places.sqlite')
break

Use the os module to list out all directories in %APPDATA%\Mozilla\Firefox\Profiles\
loop over the directories until you find places.sqlite file (also using os module)

A glob might be simpler as in this case one expects the file to be there in level below the Profiles folder or not there at all.
import os
import pathlib
profiles = pathlib.Path(os.environ["APPDATA"]) / "Mozilla" / "Firefox" / "Profiles"
# rglob will recursively search as well
if places := list(profiles.rglob("places.sqlite")):
print(places[0]) # will print the sqllite file path
with places[0].open() as f:
# ....

Related

How to loop and optimise data extraction - Python [duplicate]

I need to iterate through all .asm files inside a given directory and do some actions on them.
How can this be done in a efficient way?
Python 3.6 version of the above answer, using os - assuming that you have the directory path as a str object in a variable called directory_in_str:
import os
directory = os.fsencode(directory_in_str)
for file in os.listdir(directory):
filename = os.fsdecode(file)
if filename.endswith(".asm") or filename.endswith(".py"):
# print(os.path.join(directory, filename))
continue
else:
continue
Or recursively, using pathlib:
from pathlib import Path
pathlist = Path(directory_in_str).glob('**/*.asm')
for path in pathlist:
# because path is object not string
path_in_str = str(path)
# print(path_in_str)
Use rglob to replace glob('**/*.asm') with rglob('*.asm')
This is like calling Path.glob() with '**/' added in front of the given relative pattern:
from pathlib import Path
pathlist = Path(directory_in_str).rglob('*.asm')
for path in pathlist:
# because path is object not string
path_in_str = str(path)
# print(path_in_str)
Original answer:
import os
for filename in os.listdir("/path/to/dir/"):
if filename.endswith(".asm") or filename.endswith(".py"):
# print(os.path.join(directory, filename))
continue
else:
continue
This will iterate over all descendant files, not just the immediate children of the directory:
import os
for subdir, dirs, files in os.walk(rootdir):
for file in files:
#print os.path.join(subdir, file)
filepath = subdir + os.sep + file
if filepath.endswith(".asm"):
print (filepath)
You can try using glob module:
import glob
for filepath in glob.iglob('my_dir/*.asm'):
print(filepath)
and since Python 3.5 you can search subdirectories as well:
glob.glob('**/*.txt', recursive=True) # => ['2.txt', 'sub/3.txt']
From the docs:
The glob module finds all the pathnames matching a specified pattern according to the rules used by the Unix shell, although results are returned in arbitrary order. No tilde expansion is done, but *, ?, and character ranges expressed with [] will be correctly matched.
Since Python 3.5, things are much easier with os.scandir() and 2-20x faster (source):
with os.scandir(path) as it:
for entry in it:
if entry.name.endswith(".asm") and entry.is_file():
print(entry.name, entry.path)
Using scandir() instead of listdir() can significantly increase the
performance of code that also needs file type or file attribute
information, because os.DirEntry objects expose this information if
the operating system provides it when scanning a directory. All
os.DirEntry methods may perform a system call, but is_dir() and
is_file() usually only require a system call for symbolic links;
os.DirEntry.stat() always requires a system call on Unix but only
requires one for symbolic links on Windows.
Python 3.4 and later offer pathlib in the standard library. You could do:
from pathlib import Path
asm_pths = [pth for pth in Path.cwd().iterdir()
if pth.suffix == '.asm']
Or if you don't like list comprehensions:
asm_paths = []
for pth in Path.cwd().iterdir():
if pth.suffix == '.asm':
asm_pths.append(pth)
Path objects can easily be converted to strings.
Here's how I iterate through files in Python:
import os
path = 'the/name/of/your/path'
folder = os.fsencode(path)
filenames = []
for file in os.listdir(folder):
filename = os.fsdecode(file)
if filename.endswith( ('.jpeg', '.png', '.gif') ): # whatever file types you're using...
filenames.append(filename)
filenames.sort() # now you have the filenames and can do something with them
NONE OF THESE TECHNIQUES GUARANTEE ANY ITERATION ORDERING
Yup, super unpredictable. Notice that I sort the filenames, which is important if the order of the files matters, i.e. for video frames or time dependent data collection. Be sure to put indices in your filenames though!
You can use glob for referring the directory and the list :
import glob
import os
#to get the current working directory name
cwd = os.getcwd()
#Load the images from images folder.
for f in glob.glob('images\*.jpg'):
dir_name = get_dir_name(f)
image_file_name = dir_name + '.jpg'
#To print the file name with path (path will be in string)
print (image_file_name)
To get the list of all directory in array you can use os :
os.listdir(directory)
I'm not quite happy with this implementation yet, I wanted to have a custom constructor that does DirectoryIndex._make(next(os.walk(input_path))) such that you can just pass the path you want a file listing for. Edits welcome!
import collections
import os
DirectoryIndex = collections.namedtuple('DirectoryIndex', ['root', 'dirs', 'files'])
for file_name in DirectoryIndex(*next(os.walk('.'))).files:
file_path = os.path.join(path, file_name)
I really like using the scandir directive that is built into the os library. Here is a working example:
import os
i = 0
with os.scandir('/usr/local/bin') as root_dir:
for path in root_dir:
if path.is_file():
i += 1
print(f"Full path is: {path} and just the name is: {path.name}")
print(f"{i} files scanned successfully.")
Get all the .asm files in a directory by doing this.
import os
path = "path_to_file"
file_type = '.asm'
for filename in os.listdir(path=path):
if filename.endswith(file_type):
print(filename)
print(f"{path}/{filename}")
# do something below
I don't understand why some answers are complicated. This is how I would do it with Python 2.7. Replace DIRECTORY_TO_LOOP with the directory you want to use.
import os
DIRECTORY_TO_LOOP = '/var/www/files/'
for root, dirs, files in os.walk(DIRECTORY_TO_LOOP, topdown=False):
for name in files:
print(os.path.join(root, name))

os.listdir() shows nothing while it is not so

I am trying to sort my files for school automatically but when I try this:
import os, sys
path = "/Sorting for School/"
dirs = os.listdir(path)
#print all dirs
for file in dirs:
print file
Although there are two txt documents in dir but when I run this the output is:
[]
Thanks
The code is fine. You must be querying a wrong path.
/dir/ means a subdirectory named dir in the root directory (in UNIX) or in the root directory of the current drive (in Windows).
Your code is fine. The following should work too:
import os
path = "/Sorting for School/"
def handle_err(err):
print err
for root,dirs,files in os.walk(path,onerror=handle_err):
for name in files:
print(name)

All Files in Dir & Sub-Dir

I would like to find all the files in a directory and all sub-directories.
code used:
import os
import sys
path = "C:\\"
dirs = os.listdir(path)
filename = "C.txt"
FILE = open(filename, "w")
FILE.write(str(dirs))
FILE.close()
print dirs
The problem is - this code only lists files in directories, not sub-directories. What do I need to change in order to also list files in subdirectories?
To traverse a directory tree you want to use os.walk() for this.
Here's an example to get you started:
import os
searchdir = r'C:\root_dir' # traversal starts in this directory (the root)
for root, dirs, files in os.walk(searchdir):
for name in files:
(base, ext) = os.path.splitext(name) # split base and extension
print base, ext
which would give you access to the file names and the components.
You'll find the functions in the os and os.path module to be of great use for this sort of work.
This function will help you: os.path.walk() http://docs.python.org/library/os.path.html#os.path.walk

How can I iterate over files in a given directory?

I need to iterate through all .asm files inside a given directory and do some actions on them.
How can this be done in a efficient way?
Python 3.6 version of the above answer, using os - assuming that you have the directory path as a str object in a variable called directory_in_str:
import os
directory = os.fsencode(directory_in_str)
for file in os.listdir(directory):
filename = os.fsdecode(file)
if filename.endswith(".asm") or filename.endswith(".py"):
# print(os.path.join(directory, filename))
continue
else:
continue
Or recursively, using pathlib:
from pathlib import Path
pathlist = Path(directory_in_str).glob('**/*.asm')
for path in pathlist:
# because path is object not string
path_in_str = str(path)
# print(path_in_str)
Use rglob to replace glob('**/*.asm') with rglob('*.asm')
This is like calling Path.glob() with '**/' added in front of the given relative pattern:
from pathlib import Path
pathlist = Path(directory_in_str).rglob('*.asm')
for path in pathlist:
# because path is object not string
path_in_str = str(path)
# print(path_in_str)
Original answer:
import os
for filename in os.listdir("/path/to/dir/"):
if filename.endswith(".asm") or filename.endswith(".py"):
# print(os.path.join(directory, filename))
continue
else:
continue
This will iterate over all descendant files, not just the immediate children of the directory:
import os
for subdir, dirs, files in os.walk(rootdir):
for file in files:
#print os.path.join(subdir, file)
filepath = subdir + os.sep + file
if filepath.endswith(".asm"):
print (filepath)
You can try using glob module:
import glob
for filepath in glob.iglob('my_dir/*.asm'):
print(filepath)
and since Python 3.5 you can search subdirectories as well:
glob.glob('**/*.txt', recursive=True) # => ['2.txt', 'sub/3.txt']
From the docs:
The glob module finds all the pathnames matching a specified pattern according to the rules used by the Unix shell, although results are returned in arbitrary order. No tilde expansion is done, but *, ?, and character ranges expressed with [] will be correctly matched.
Since Python 3.5, things are much easier with os.scandir() and 2-20x faster (source):
with os.scandir(path) as it:
for entry in it:
if entry.name.endswith(".asm") and entry.is_file():
print(entry.name, entry.path)
Using scandir() instead of listdir() can significantly increase the
performance of code that also needs file type or file attribute
information, because os.DirEntry objects expose this information if
the operating system provides it when scanning a directory. All
os.DirEntry methods may perform a system call, but is_dir() and
is_file() usually only require a system call for symbolic links;
os.DirEntry.stat() always requires a system call on Unix but only
requires one for symbolic links on Windows.
Python 3.4 and later offer pathlib in the standard library. You could do:
from pathlib import Path
asm_pths = [pth for pth in Path.cwd().iterdir()
if pth.suffix == '.asm']
Or if you don't like list comprehensions:
asm_paths = []
for pth in Path.cwd().iterdir():
if pth.suffix == '.asm':
asm_pths.append(pth)
Path objects can easily be converted to strings.
Here's how I iterate through files in Python:
import os
path = 'the/name/of/your/path'
folder = os.fsencode(path)
filenames = []
for file in os.listdir(folder):
filename = os.fsdecode(file)
if filename.endswith( ('.jpeg', '.png', '.gif') ): # whatever file types you're using...
filenames.append(filename)
filenames.sort() # now you have the filenames and can do something with them
NONE OF THESE TECHNIQUES GUARANTEE ANY ITERATION ORDERING
Yup, super unpredictable. Notice that I sort the filenames, which is important if the order of the files matters, i.e. for video frames or time dependent data collection. Be sure to put indices in your filenames though!
You can use glob for referring the directory and the list :
import glob
import os
#to get the current working directory name
cwd = os.getcwd()
#Load the images from images folder.
for f in glob.glob('images\*.jpg'):
dir_name = get_dir_name(f)
image_file_name = dir_name + '.jpg'
#To print the file name with path (path will be in string)
print (image_file_name)
To get the list of all directory in array you can use os :
os.listdir(directory)
I'm not quite happy with this implementation yet, I wanted to have a custom constructor that does DirectoryIndex._make(next(os.walk(input_path))) such that you can just pass the path you want a file listing for. Edits welcome!
import collections
import os
DirectoryIndex = collections.namedtuple('DirectoryIndex', ['root', 'dirs', 'files'])
for file_name in DirectoryIndex(*next(os.walk('.'))).files:
file_path = os.path.join(path, file_name)
I really like using the scandir directive that is built into the os library. Here is a working example:
import os
i = 0
with os.scandir('/usr/local/bin') as root_dir:
for path in root_dir:
if path.is_file():
i += 1
print(f"Full path is: {path} and just the name is: {path.name}")
print(f"{i} files scanned successfully.")
Get all the .asm files in a directory by doing this.
import os
path = "path_to_file"
file_type = '.asm'
for filename in os.listdir(path=path):
if filename.endswith(file_type):
print(filename)
print(f"{path}/{filename}")
# do something below
I don't understand why some answers are complicated. This is how I would do it with Python 2.7. Replace DIRECTORY_TO_LOOP with the directory you want to use.
import os
DIRECTORY_TO_LOOP = '/var/www/files/'
for root, dirs, files in os.walk(DIRECTORY_TO_LOOP, topdown=False):
for name in files:
print(os.path.join(root, name))

Find all CSV files in a directory using Python

How can I find all files in directory with the extension .csv in python?
import os
import glob
path = 'c:\\'
extension = 'csv'
os.chdir(path)
result = glob.glob('*.{}'.format(extension))
print(result)
from os import listdir
def find_csv_filenames( path_to_dir, suffix=".csv" ):
filenames = listdir(path_to_dir)
return [ filename for filename in filenames if filename.endswith( suffix ) ]
The function find_csv_filenames() returns a list of filenames as strings, that reside in the directory path_to_dir with the given suffix (by default, ".csv").
Addendum
How to print the filenames:
filenames = find_csv_filenames("my/directory")
for name in filenames:
print name
By using the combination of filters and lambda, you can easily filter out csv files in given folder.
import os
all_files = os.listdir("/path-to-dir")
csv_files = list(filter(lambda f: f.endswith('.csv'), all_files))
# lambda returns True if filename (within `all_files`) ends with .csv or else False
# and filter function uses the returned boolean value to filter .csv files from list files.
use Python OS module to find csv file in a directory.
the simple example is here :
import os
# This is the path where you want to search
path = r'd:'
# this is the extension you want to detect
extension = '.csv'
for root, dirs_list, files_list in os.walk(path):
for file_name in files_list:
if os.path.splitext(file_name)[-1] == extension:
file_name_path = os.path.join(root, file_name)
print file_name
print file_name_path # This is the full path of the filter file
I had to get csv files that were in subdirectories, therefore, using the response from tchlpr I modified it to work best for my use case:
import os
import glob
os.chdir( '/path/to/main/dir' )
result = glob.glob( '*/**.csv' )
print( result )
import os
path = 'C:/Users/Shashank/Desktop/'
os.chdir(path)
for p,n,f in os.walk(os.getcwd()):
for a in f:
a = str(a)
if a.endswith('.csv'):
print(a)
print(p)
This will help to identify path also of these csv files
While solution given by thclpr works it scans only immediate files in the directory and not files in the sub directories if any. Although this is not the requirement but just in case someone wishes to scan sub directories too below is the code that uses os.walk
import os
from glob import glob
PATH = "/home/someuser/projects/someproject"
EXT = "*.csv"
all_csv_files = [file
for path, subdir, files in os.walk(PATH)
for file in glob(os.path.join(path, EXT))]
print(all_csv_files)
Copied from this blog.
Use the python glob module to easily list out the files we need.
import glob
path_csv=glob.glob("../data/subfolrder/*.csv")
You could just use glob with recursive = true, the pattern ** will match any files and zero or more directories, subdirectories and symbolic links to directories.
import glob, os
os.chdir("C:\\Users\\username\\Desktop\\MAIN_DIRECTORY")
for file in glob.glob("*/.csv", recursive = true):
print(file)
This solution uses the python function filter. This function creates a list of elements for which a function returns true. In this case, the anonymous function used is partial matching '.csv' on every element of the directory files list obtained with os.listdir('the path i want to look in')
import os
filepath= 'filepath_to_my_CSVs' # for example: './my_data/'
list(filter(lambda x: '.csv' in x, os.listdir('filepath_to_my_CSVs')))
Many (linked) answers change working directory with os.chdir(). But you don't have to.
Recursively print all CSV files in /home/project/ directory:
pathname = "/home/project/**/*.csv"
for file in glob.iglob(pathname, recursive=True):
print(file)
Requires python 3.5+. From docs [1]:
pathname can be either absolute (like /usr/src/Python-1.5/Makefile) or relative (like ../../Tools/*/*.gif)
pathname can contain shell-style wildcards.
Whether or not the results are sorted depends on the file system.
If recursive is true, the pattern ** will match any files and zero or more directories, subdirectories and symbolic links to directories
[1] https://docs.python.org/3/library/glob.html#glob.glob
You could just use glob with recursive = True, the pattern ** will match any files and zero or more directories, subdirectories and symbolic links to directories.
import glob, os
os.chdir("C:\\Users\\username\\Desktop\\MAIN_DIRECTORY")
for file in glob.glob("*/*.csv", recursive = True):
print(file)
Please use this tested working code. This function will return a list of all the CSV files with absolute CSV file paths in your specified path.
import os
from glob import glob
def get_csv_files(dir_path, ext):
os.chdir(dir_path)
return list(map(lambda x: os.path.join(dir_path, x), glob(f'*.{ext}')))
print(get_csv_files("E:\\input\\dir\\path", "csv"))

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