I have tried to append a record on the next line in the file using the following code(please note that the file has been created already). But, it does not insert any records at all. The file remains empty.
with open(utmppath+'/'+tmpfile, "a") as myfile:
myfile.write(record+'\n')
myfile.close()
Any suggestion would be great. Thanks
Check additionally if you set your path correctly:
import os
path = utmppath+'/'+tmpfile
assert os.path.isfile(path), path
The assertion checks if the file exists and raises an AssertionError if you used a wrong path. Additionally the used path is included in the error message thanks to the variable
after the comma.
Additionally I recommend you to join files with the help of os.path.join and os.path.abspath. os.path.join concatenates path strings correctly for you and os.path.abspath creates an absolute path.
path = os.path.join(utmppath, tmpfile)
Let's say the wished file is in the same directory like your script and called your_output.txt - you can use this:
path = os.path.abspath(os.path.join(os.path.dirname(__file__), 'your_output.txt'))
By the way, __file__ gives you the name of your script file.
Related
how could do print the path directory of my file saved with python.
For ad example, i start with idle3 (Fedora 24), i save the IDLE code on: /home/jetson/Desktop/Python/Programs and in the code is written:
f = open("Hello.txt","w")
f.write("This is a test","\n")
f.close()
My question is: after f.close i would print the directory grabbed with
import sys.os
print("The file is saved in ",direcotrysaved)
It's possible?
Convert the relative path to absolute path:
path = os.path.abspath(filename)
Then take just the directory from the absolute path:
directory = os.path.dirname(path)
That will work regardless of what filename is. It could by just a filename, a relative path or absolute path.
See os.path documentation for other useful functions.
If no path is given then the file is opened in the current working directory. This is returned by os.getcwd().
You can use os.getcwd() for getting the current working directory:
import os
os.getcwd()
Otherwise, you could also specify the full path for the file when opening it, e.g.
f.open('/path/to/the/folder/Hello.txt', 'w')
Using f.name provides the file name if you didn't supply it as a separate variable.
Using os.path.abspath() provides the full file name including it's directory path i.e. in this case "/home/rolf/def"
Using os.path.dirname() strips the file name from the path, as that you already know.
>>> import os
>>> f=open('xxxx.txt','w')
>>> f.close()
>>> print(f.name,'Stored in', os.path.dirname(os.path.abspath(f.name)))
xxxx.txt Stored in /home/rolf/def
I'm doing a Python course on Udacity. And there is a class called Rename Troubles, which presents the following incomplete code:
import os
def rename_files():
file_list = os.listdir("C:\Users\Nick\Desktop\temp")
print (file_list)
for file_name in file_list:
os.rename(file_name, file_name.translate(None, "0123456789"))
rename_files()
As the instructor explains it, this will return an error because Python is not attempting to rename files in the right folder. He then proceeds to check the "current working directory", and then goes on to specify to Python which directory to rename files in.
This makes no sense to me. We are using the for loop to specifically tell Python that we want to rename the contents of file_list, which we have just pointed to the directory we need, in rename_files(). So why does it not attempt to rename in that folder? Why do we still need to figure out cwd and then change it?? The code looks entirely logical without any of that.
Look closely at what os.listdir() gives you. It returns only a list of names, not full paths.
You'll then proceed to os.rename one of those names, which will be interpreted as a relative path, relative to whatever your current working directory is.
Instead of messing with the current working directory, you can os.path.join() the path that you're searching to the front of both arguments to os.rename().
I think your code needs some formatting help.
The basic issue is that os.listdir() returns names relative to the directory specified (in this case an absolute path). But your script can be running from any directory. Manipulate the file names passed to os.rename() to account for this.
Look into relative and absolute paths, listdir returns names relative to the path (in this case absolute path) provided to listdir. os.rename is then given this relative name and unless the app's current working directory (usually the directory you launched the app from) is the same as provided to listdir this will fail.
There are a couple of alternative ways of handling this, changing the current working directory:
os.chdir("C:\Users\Nick\Desktop\temp")
for file_name in os.listdir(os.getcwd()):
os.rename(file_name, file_name.translate(None, "0123456789"))
Or use absolute paths:
directory = "C:\Users\Nick\Desktop\temp"
for file_name in os.listdir(directory):
old_file_path = os.path.join(directory, file_name)
new_file_path = os.path.join(directory, file_name.translate(None, "0123456789"))
os.rename(old_file_path, new_file_path)
You can get a file list from ANY existing directory - i.e.
os.listdir("C:\Users\Nick\Desktop\temp")
or
os.listdir("C:\Users\Nick\Desktop")
or
os.listdir("C:\Users\Nick")
etc.
The instance of the Python interpreter that you're using to run your code is being executed in a directory that is independent of any directory for which you're trying to get information. So, in order to rename the correct file, you need to specify the full path to that file (or the relative path from wherever you're running your Python interpreter).
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)
If I have the following file:
file = '/Users/david542/Desktop/work.txt'
I can use os.path.basename(file) to get the file name.
What command would I use to get the directory of the file (i.e., to get "/Users/david542/Desktop") ?
os.path.dirname(file) returns the directory of the passed file name. Alternatively, you can use os.path.split(file) which will give you a tuple containing the directory name and the file name in one call.
>>> os.path.dirname(os.path.realpath('/Users/david542/Desktop/work.txt'))
os.path.dirname(file) will yield directory name.
import os
print(os.path.dirname("c:/windows/try.txt"))
I think you're searching for os.path.dirname. Otherwise you could use os.path.split which returns the path and the filename in a tuple.
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.