lately I started working with the Os module in python . And I finally arrived to this Os.path method . So here is my question . I ran this method in one of my kivy project just for testing and it actually didn't returned the correct output.The method consisted of finding if any directory exist and return a list of folders in the directory . otherwise print Invalid Path and return -1 . I passed in an existing directory and it returned -1 but the weird path is that when I run similar program out of my kivy project using the same path present in thesame folder as my python file it return the desired output .here is the image with the python file and the directory name image I have tested which returns invalid path.
and here is my code snippet
def get_imgs(self, img_path):
if not os.path.exists(img_path):
print("Invalid Path...")
return -1
else:
all_files = os.listdir(img_path)
imgs = []
for f in all_files:
if (
f.endswith(".png")
or f.endswith(".PNG")
or f.endswith(".jpg")
or f.endswith(".JPG")
or f.endswith(".jpeg")
or f.endswith(".JPEG")
):
imgs.append("/".join([img_path, f]))
return imgs
It's tough to tell without seeing the code with your function call. Whatever argument you're passing must not be a valid path. I use the os module regularly and have slowly learned a lot of useful methods. I always print out paths that I'm reading or where I'm writing before doing it in case anything unexpected happens, I can see that img_path variable, for example. Copy and paste the path in file explorer up to the directory and make sure that's all good.
Some other useful os.path methods you will find useful, based on your code:
os.join(<directory>, <file_name.ext>) is much more intuitive than imgs.append("/".join([img_path, f]))
os.getcwd() gets your working directory (which I print at the start of scripts in dev to quickly address issues before debugging). I typically use full paths to play it safe because Python pathing can cause differences/issues when running from cmd vs. PyCharm
os.path.basename(f) gives you the file, while os.path.dirname(f) gives you the directory.
It seems like a better approach to this is to use pathlib and glob. You can iterate over directories and use wild cards.
Look at these:
iterating over directories: How can I iterate over files in a given directory?
different file types: Python glob multiple filetypes
Then you don't even need to check whether os.path.exists(img_path) because this will read the files directly from your file system. There's also more wild cards in the glob library such as * for anything/any length, ? for any character, [0-9] for any number, found here: https://docs.python.org/3/library/glob.html
Related
Beginner at python. I'm trying to search users folders for illegal content saved in folders. I want to find all files that contain either one or a number of words from the below list and also the files also have an extension that's listed.
I can search the files using file.endswith but don't know how to add in the word condition.
I've looked through the site and how only come across how to search for a certain word and not a list of words.
Thank you in advance
import os
L = ['720p','aac','ac3','bdrip','brrip','demonoid','disc','hdtv','dvdrip',
'edition','sample','torrent','www','x264','xvid']
for root, dirs, files in os.walk("Y:\User Folders\"):
for file in files:
if file.endswith(('*.7z','.3gp','.alb','.ape','.avi','.cbr','.cbz','.cue','.divx','.epub','.flac',
'.flv','.idx','.iso','.m2ts','.m2v','.m3u','.m4a','.m4b','.m4p','.m4v','.md5',
'.mkv','.mobi','.mov','.mp3','.mp4','.mpeg','.mpg','.mta','.nfo','.ogg','.ogm',
'.pla','.rar','.rm','.rmvb','.sfap0','.sfk','.sfv','.sls','.smfmf','.srt,''.sub',
'.torrent','.vob','.wav','.wma','.wmv','.wpl','.zip')):
print(os.path.join(root, file))
Perhaps it might be better to do a reverse search, and display a warning about files that DON'T match the file types you want. For instance you could do this:
if file.endswith(".txt", ".py"):
print("File is ok!")
else:
print("File is not ok!")
Using py.path.local from py package
The py package (install by $ pip install py) offers a very nice interface for working with files.
from py.path import local
def isbadname(path):
bad_extensions = [".pyc", "txt"]
bad_names = ["code", "xml"]
return (path.ext in bad_extensions) or (path.purebasename in bad_names)
for path in local(".").visit(isbadname):
print(path.strpath)
Explained:
Import
from py.path import local
py.path.local function creates "objectified" file names. To keep my code short, I import
it this way to use only local for objectifying file name strings.
Create objectified path to local directory:
local(".")
Created object is not a string, but an object, which has many interesting properties and methods.
Listing all files within some directory:
local(".").visit("*.txt")
returns a generator, providing all paths to files having extension ".txt"..
Alternative method to detect files to generate is providing a function, which gets argument path
(objectified file name) and returns True if the file is to be used, False otherwise.
The function isbadname serves exactly this purpose.
If you want to google for more information, use py path local (the name py is not giving good hits).
For more see https://py.readthedocs.io/en/latest/path.html
Note, that if you use pytest package, the py is installed with it (for good
reason - it makes tests related to file names much more readable and shorter).
I am using os.walk to run through a tree of directories check for some input files and then run a program if the proper inputs are there. I notice I am having a problem because of the away that os.walk is evaluating the root variable in the loop:
for root, dirs, files in os.walk('.'):# I use '.' because I want the walk to
# start where I run the script. And it
# will/can change
if "input.file" in files:
infile = os.path.join(root,"input.file")
subprocess.check_output("myprog input.file", Shell=True)
# if an input file is found, store the path to it
# and run the program
This is giving me an issue because the infile string looks like this
./path/to/input.file
When it needs to look like this for the program to be able to find it
/home/start/of/walk/path/to/input.file
I want to know if there is a better method/ a different way to use os.walk such that I can leave the starting directory arbitrary, but still be able to use the full path to any files that it finds for me. Thanks
The program I am using is written by me in c++ and I suppose I could modify it as well. But I am not asking about how to do that in this question just to clarify this question is about python's os.walk and related topics that is why there is no examples of my c++ code here.
Instead of using ., convert it to the absolute path by using os.path.abspath("."). That will convert your current path to an absolute path before you begin.
I'm doing a program in which Chimera needs to be opened, I'm opening it with:
def generate_files_bat(filename):
f = open(filename, 'w')
text = """echo off SET PATH=%PATH%;"C:\\Program Files (x86)\\Chimera 1.6.1\\bin" chimera colpeps.cmd"""
print >>f, text
f.close()
But I need to find Chimera apart from the computer the python program is running. Is there any way the path can be searched by the python program in any computer?
Generally speaking, I don't think it is such a good idea to search the path for a program. Imagine, for example that two different versions were installed on the machine. Are-you sure to find the right one? Maybe a configuraition file parsed with the standard module ConfigParser would be a better option?
Anyway, to go back to your question, in order to find a file or directory, you could try to use os.walk which recursively walks trough a directory tree.
Here is an example invoking os.walk from a generator, allowing you to collect either the first or all matching file names. Please note that the generator result is only based on file name. If you require more advanced filtering (say, to only keep executable files), you will probably use something like os.stat() to extend the test.
import os
def fileInPath(name, root):
for base, dirs, files in os.walk(root):
if name in files:
yield os.path.join(base, name)
print("Search for only one result:")
print(next(fileInPath("python", "/home/sylvain")))
print("Display all matching files:")
print([i for i in fileInPath("python", "/home/sylvain")])
There is which for Linux and where for Windows. They both give you the path to the executable, provided it lies in a directory that is 'searched' by the console (so it has to be in %PATH% in case of Windows)
There is a package called Unipath, that does elegant, clean path calculations.
Have look here for the AbstractPath constructor
Example:
from unipath import Path
prom_dir = Path(__file__)
So I'm sure this is a stupid question, but I've looked through Python's documentation and attempted a couple of Google codes and none of them has worked.
It seems like the following should work, but it returns "False" for
In my directory /foo/bar I have 3 items: 1 Folder "[Folder]", 1 file "test" (no extension), and 1 file "test.py".
I'm look to have a script that can distinguish folders from files for a bunch of functions, but I can't figure out anything that works.
#!/usr/bin/python
import os, re
for f in os.listdir('/foo/bar'):
print f, os.path.isdir(f)
Currently returns false for everything.
This is because listdir() returns the names of the files in /foo/bar. When you later do os.path.isdir() on one of these, the OS interprets it relative to the current working directory which is probably the directory your script is in, not /foo/bar, and it probably does not contain a directory of the specified name. A path that doesn't exist is not a directory and so isdir() returns False..
Use the complete pathname. Best way is to use os.path.join, e.g., os.path.isdir(os.path.join('/foo/bar', f)).
You might want to use os.walk instead: http://docs.python.org/library/os.html#os.walk
When it returns the contents of the directory, it returns files and directories in separate lists, negating the need for checking.
So you could do:
import os
root, dirs, files = next(os.walk('/foo/bar'))
print 'directories:', dirs
print 'files:', files
I suppose that os.path.isdir(os.path.join('/foo/bar', f)) should work.
I have a file, for example "something.exe" and I want to find path to this file
How can I do this in python?
Perhaps os.path.abspath() would do it:
import os
print os.path.abspath("something.exe")
If your something.exe is not in the current directory, you can pass any relative path and abspath() will resolve it.
use os.path.abspath to get a normalized absolutized version of the pathname
use os.walk to get it's location
import os
exe = 'something.exe'
#if the exe just in current dir
print os.path.abspath(exe)
# output
# D:\python\note\something.exe
#if we need find it first
for root, dirs, files in os.walk(r'D:\python'):
for name in files:
if name == exe:
print os.path.abspath(os.path.join(root, name))
# output
# D:\python\note\something.exe
if you absolutely do not know where it is, the only way is to find it starting from root c:\
import os
for r,d,f in os.walk("c:\\"):
for files in f:
if files == "something.exe":
print os.path.join(r,files)
else, if you know that there are only few places you store you exe, like your system32, then start finding it from there. you can also make use of os.environ["PATH"] if you always put your .exe in one of those directories in your PATH variable.
for p in os.environ["PATH"].split(";"):
for r,d,f in os.walk(p):
for files in f:
if files == "something.exe":
print os.path.join(r,files)
Just to mention, another option to achieve this task could be the subprocess module, to help us execute a command in terminal, like this:
import subprocess
command = "find"
directory = "/Possible/path/"
flag = "-iname"
file = "something.foo"
args = [command, directory, flag, file]
process = subprocess.run(args, stdout=subprocess.PIPE)
path = process.stdout.decode().strip("\n")
print(path)
With this we emulate passing the following command to the Terminal:
find /Posible/path -iname "something.foo".
After that, given that the attribute stdout is binary string, we need to decode it, and remove the trailing "\n" character.
I tested it with the %timeit magic in spyder, and the performance is 0.3 seconds slower than the os.walk() option.
I noted that you are in Windows, so you may search for a command that behaves similar to find in Unix.
Finally, if you have several files with the same name in different directories, the resulting string will contain all of them. In consequence, you need to deal with that appropriately, maybe using regular expressions.
This is really old thread, but might be useful to someone who stumbles across this. In python 3, there is a module called "glob" which takes "egrep" style search strings and returns system appropriate pathing (i.e. Unix\Linux and Windows).
https://docs.python.org/3/library/glob.html
Example usage would be:
results = glob.glob('./**/FILE_NAME')
Then you get a list of matches in the result variable.
Uh... This question is a bit unclear.
What do you mean "have"? Do you have the name of the file? Have you opened it? Is it a file object? Is it a file descriptor? What???
If it's a name, what do you mean with "find"? Do you want to search for the file in a bunch of directories? Or do you know which directory it's in?
If it is a file object, then you must have opened it, reasonably, and then you know the path already, although you can get the filename from fileob.name too.