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.
Related
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
I'm using Glob.Glob to search a folder, and the sub-folders there in for all the invoices I have. To simplify that I'm going to add the program to the context menu, and have it take the path as the first part of,
import glob
for filename in glob.glob(path + "/**/*.pdf", recursive=True):
print(filename)
I'll have it keep the list and send those files to a Printer, in a later version, but for now just writing the name is a good enough test.
So my question is twofold:
Is there anything fundamentally wrong with the way I'm writing this?
Can anyone point me in the direction of how to actually capture folder path and provide it as path-variable?
You should have a look at this question: Python script on selected file. It shows how to set up a "Sent To" command in the context menu. This command calls a python script an provides the file name sent via sys.argv[1]. I assume that also works for a directory.
I do not have Python3.5 so that I can set the flag recursive=True, so I prefer to provide you a solution which you can run on any Python version (known up to day).
The solution consists in using calling os.walk() to run explore the directories and the set build-in type.
it is better to use set instead of list as with this later one you'll need more code to check if the directory you want to add is not listed already.
So basically you can keep two sets: one for the names of files you want to print and the other one for the directories and their sub folders.
So you can adapat this solution to your class/method:
import os
path = '.' # Any path you want
exten = '.pdf'
directories_list = set()
files_list = set()
# Loop over direcotries
for dirpath, dirnames, files in os.walk(path):
for name in files:
# Check if extension matches
if name.lower().endswith(exten):
files_list.add(name)
directories_list.add(dirpath)
You can then loop over directories_list and files_list to print them out.
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__)
I'm trying to get the list of files in a particular directory and count the number of files in the directory. I always get the following error:
WindowsError: [Error 3] The system cannot find the path specified: '/client_side/*.*'
My code is:
print len([name for name in os.listdir('/client_side/') if os.path.isfile(name)])
I followed the code example given here.
I am running the Python script on Pyscripter and the directory /client_side/ do exists. My python code is in the root folder and has a sub-folder called "client_side". Can someone help me out on this?
This error occurs when you use os.listdir on a path which does not refer to an existing path.
For example:
>>> os.listdir('Some directory does not exist')
Traceback (most recent call last):
File "<interactive input>", line 1, in <module>
WindowsError: [Error 3] : 'Some directory does not exist/*.*'
If you want to use os.listdir, you need to either guarantee the existence of the path that you would use, or use os.path.exists to check the existence first.
if os.path.exists('/client_side/'):
do something
else:
do something
Suppose your current working directory is c:\foobar, os.listdir('/client_side/') is equivalent to os.listdir('c:/client_side'), while os.listdir('client_side/') is equivalent to os.listdir('c:/foobar/client_side'). If your client_side directory is not in the root, such error will occur when using os.listdir.
For your 0 ouput problem, let us recall os.listdir(path)
Return a list containing the names of the entries in the directory given by path. The list is in arbitrary order. It does not include the special entries '.' and '..' even if they are present in the directory.
and os.path.isfile(path).
Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.
listdir returns neither the absolute paths nor relative paths, but a list of the name of your files, while isfile requires path. Therefore, all of those names would yield False.
To obtain the path, we can either use os.path.join , concat two strings directly.
print ([name for name in os.listdir(path)
if os.path.isfile(os.path.join(path, name))])
Or
print ([name for name in os.listdir('client_side/')
if os.path.isfile('client_side/' + name)])
I decided to change the code into:
def numOfFiles(path):
return len(next(os.walk(path))[2])
and use the following the call the code:
print numOfFiles("client_side")
Many thanks to everyone who told me how to pass the windows directory correctly in Python and to nrao91 in here for providing the function code.
EDIT: Thank you eryksun for correcting my code!
Two things:
os.listdir() does not do a glob pattern matching, use the glob module for that
probably you do not have a directory called '/client_side/*.*', but maybe one
without the . in the name
The syntax you used works fine, if the directory you look for exists, but there is no directory called '/client_side/.'.
In addition, be careful if using Python 2.x and os.listdir, as the results on windows are different when you use u'/client_side/' and just '/client_side'.
You can do just
os.listdir('client_side')
without slashes.
As I can see a WindowsError, Just wondering if this has something to do with the '/' in windows ! Ideally, on windows, you should have something like os.path.join('C:','client_side')
You want:
print len([name for name in os.listdir('./client_side/') if os.path.isfile(name)])
with a "." before "/client_side/".
The dot means the current path where you are working (i.e. from where you are calling your code), so "./client_side/" represents the path you want, which is specified relatively to your current directory.
If you write only "/client_side/", in unix, the program would look for a folder in the root of the system, instead of the folder that you want.
If you just want to see all the files in the directory where your script is located, you can use os.path.dirname(sys.argv[0]). This will give the path of the directory where your script is.
Then, with fnmatch function you can obtain the list of files in that directory with a name and/or extension specified in the filenamevariable.
import os,sys
from fnmatch import fnmatch
directory = os.path.dirname(sys.argv[0]) #this determines the directory
file_name= "*" #if you want the python files only, change "*" to "*.py"
for path, subdirs, files in os.walk(directory):
for name in files:
if fnmatch(name, file_name):
print (os.path.join(path, name))
I hope this helps.
Checking for existence is subject to a race. Better to handle the error (beg forgiveness instead of ask permission). Plus, in Python 3 you can suppress errors. Use suppress from contextlib:
with suppress(FileNotFoundError):
for name in os.listdir('foo'):
print(name)
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.