import tempfile
storage_path = os.path.join(tempfile.gettempdir(), 'storage.data')
with open(STORAGE_PATH, 'w') as f:
f.write(data)
In which directory will the file be created? And is it possible to see it there or is it invisible?
You only created a string, nothing more.
The tempfile.gettempdir() function returns a string:
Return the name of the directory used for temporary files
It's just a name. Print it if you need to know the location on your system but don't want to step through the rules outlined in the documentation; there are 3 environment variables and another 3 standard locations to test to see where the current user has permission to create files.
Next, os.path.join() takes strings and outputs a string. Nothing is created on disk to build these strings. It'll depend on whether or not you actually do something with that string (like calling os.makedirs() or open()) on what will be created.
If you use a plain open() call on that string then a regular file is created in the tempfile.gettempdir() location that is visible to other processes.
If you need to create a temporary file securely, so in a manner that an attacker can't access the file or influence where it is created, use the tempfile.TemporaryFile() constructor or the tempfile.mkstemp() function; on UNIX systems that'll result in a file that is not listed in the directory any more (the file exists but can't be opened by other processes). It will be created in the tempfile.gettempdir() location however, so it'll use up disk space on the relevant disk partition that that directory lives on.
From this link...
The list is:
the directory named by the TMPDIR environment variable
the directory named by the TEMP environment variable
the directory named by the TMP environment variable
a platform-specific location:
on Windows, the directories C:\TEMP, C:\TMP, \TEMP, and \TMP, in that order.
on all other platforms, the directories /tmp, /var/tmp, and /usr/tmp, in that order.
as a last resort, the current working directory.
A sample from mac os :
/var/folders/4x/2rjwjpyn1pn9_0t1r5_m8blr0000gn/T/storage.data
When you go to the directory via finder the file should be there. As for Windows it might be hidden which I believe you can unlock.
There is currently no file being created. Check out the tempfile documentation to find more information about functions of the module such as the .gettempdir():
Python searches a standard list of directories to find one which the
calling user can create files in. The list is:
The directory named by the TMPDIR environment variable.
The directory named by the TEMP environment variable.
The directory named by the TMP environment variable.
A platform-specific location:
On Windows, the directories C:\TEMP, C:\TMP, \TEMP, and \TMP, in that order.
On all other platforms, the directories /tmp, /var/tmp, and /usr/tmp, in that order.
As a last resort, the current working directory.
Related
I am trying to open the file recentlyUpdated.yaml from my Python script. But when I try using:
open('recentlyUpdated.yaml')
I get an error that says:
IOError: [Errno 2] No such file or directory: 'recentlyUpdated.yaml'
Why? How can I fix the problem?
Ensure the file exists (and has the right file extension): use os.listdir() to see the list of files in the current working directory.
Ensure you're in the expected directory using os.getcwd().
(If you launch your code from an IDE, you may be in a different directory.)
You can then either:
Call os.chdir(dir) where dir is the directory containing the file. Then, open the file using just its name, e.g. open("file.txt").
Specify an absolute path to the file in your open call.
Use a raw string (r"") if your path uses backslashes, like
so: dir = r'C:\Python32'
If you don't use raw string, you have to escape every backslash: 'C:\\User\\Bob\\...'
Forward-slashes also work on Windows 'C:/Python32' and do not need to be escaped.
Let me clarify how Python finds files:
An absolute path is a path that starts with your computer's root directory, for example C:\Python\scripts if you're on Windows.
A relative path is a path that does not start with your computer's root directory, and is instead relative to something called the working directory. You can view Python's current working directory by calling os.getcwd().
If you try to do open('sortedLists.yaml'), Python will see that you are passing it a relative path, so it will search for the file inside the current working directory.
Calling os.chdir() will change the current working directory.
Example: Let's say file.txt is found in C:\Folder.
To open it, you can do:
os.chdir(r'C:\Folder')
open('file.txt') # relative path, looks inside the current working directory
or
open(r'C:\Folder\file.txt') # absolute path
Most likely, the problem is that you're using a relative file path to open the file, but the current working directory isn't set to what you think it is.
It's a common misconception that relative paths are relative to the location of the python script, but this is untrue. Relative file paths are always relative to the current working directory, and the current working directory doesn't have to be the location of your python script.
You have three options:
Use an absolute path to open the file:
file = open(r'C:\path\to\your\file.yaml')
Generate the path to the file relative to your python script:
from pathlib import Path
script_location = Path(__file__).absolute().parent
file_location = script_location / 'file.yaml'
file = file_location.open()
(See also: How do I get the path and name of the file that is currently executing?)
Change the current working directory before opening the file:
import os
os.chdir(r'C:\path\to\your\file')
file = open('file.yaml')
Other common mistakes that could cause a "file not found" error include:
Accidentally using escape sequences in a file path:
path = 'C:\Users\newton\file.yaml'
# Incorrect! The '\n' in 'Users\newton' is a line break character!
To avoid making this mistake, remember to use raw string literals for file paths:
path = r'C:\Users\newton\file.yaml'
# Correct!
(See also: Windows path in Python)
Forgetting that Windows doesn't display file extensions:
Since Windows doesn't display known file extensions, sometimes when you think your file is named file.yaml, it's actually named file.yaml.yaml. Double-check your file's extension.
The file may be existing but may have a different path. Try writing the absolute path for the file.
Try os.listdir() function to check that atleast python sees the file.
Try it like this:
file1 = open(r'Drive:\Dir\recentlyUpdated.yaml')
Possibly, you closed the 'file1'.
Just use 'w' flag, that create new file:
file1 = open('recentlyUpdated.yaml', 'w')
mode is an optional string that specifies the mode in which the file
is opened. It defaults to 'r' which means open for reading in text
mode. Other common values are 'w' for writing (truncating the file if
it already exists)...
(see also https://docs.python.org/3/library/functions.html?highlight=open#open)
If is VSCode see the workspace. If you are in other workspace this error can rise
Understanding absolute and relative paths
The term path means exactly what it sounds like. It shows the steps that need to be taken, into and out of folders, to find a file. Each step on the path is either a folder name, the special name . (which means the current folder), or the special name .. (which means to go back/out into the parent folder).
The terms absolute and relative also have their usual English meaning. A relative path shows where something is relative to some start point; an absolute path is a location starting from the top.
Paths that start with a path separator, or a drive letter followed by a path separator (like C:/foo) on Windows, are absolute. (On Windows there are also UNC paths, which are necessarily absolute. Most people will never have to worry about these.)
Paths that directly start with a file or folder name, or a drive letter followed directly by the file or folder name (like C:foo) on Windows, are relative.
Understanding the "current working directory"
Relative paths are "relative to" the so-called current working directory (hereafter abbreviated CWD). At the command line, Linux and Mac use a common CWD across all drives. (The entire file system has a common "root", and may include multiple physical storage devices.) Windows is a bit different: it remembers the most recent CWD for each drive, and has separate functionality to switch between drives, restoring those old CWD values.
Each process (this includes terminal/command windows) has its own CWD. When a program is started from the command line, it will get the CWD that the terminal/command process was using. When a program is started from a GUI (by double-clicking a script, or dragging something onto the script, or dragging the script onto a Python executable) or by using an IDE, the CWD might be any number of things depending on the details.
Importantly, the CWD is not necessarily where the script is located.
The script's CWD can be checked using os.getcwd, and modified using os.chdir. Each IDE has its own rules that control the initial CWD; check the documentation for details.
To set the CWD to the folder that contains the current script, determine that path and then set it:
os.chdir(os.path.dirname(os.path.abspath(__file__)))
Verifying the actual file name and path
There are many reasons why the path to a file might not match expectations. For example, sometimes people expect C:/foo.txt on Windows to mean "the file named foo.txt on the desktop". This is wrong. That file is actually - normally - at C:/Users/name/Desktop/foo.txt (replacing name with the current user's username). It could instead be elsewhere, if Windows is configured to put it elsewhere. To find the path to the desktop in a portable way, see How to get Desktop location?.
It's also common to mis-count ..s in a relative path, or inappropriately repeat a folder name in a path. Take special care when constructing a path programmatically. Finally, keep in mind that .. will have no effect while already in a root directory (/ on Linux or Mac, or a drive root on Windows).
Take even more special care when constructing a path based on user input. If the input is not sanitized, bad things could happen (e.g. allowing the user to unzip a file into a folder where it will overwrite something important, or where the user ought not be allowed to write files).
Another common gotcha is that the special ~ shortcut for the current user's home directory does not work in an absolute path specified in a Python program. That part of the path must be explicitly converted to the actual path, using os.path.expanduser. See Why am I forced to os.path.expanduser in python? and os.makedirs doesn't understand "~" in my path.
Keep in mind that os.listdir will give only the file names, not paths. Trying to iterate over a directory listed this way will only work if that directory is the current working directory.
It's also important to make sure of the actual file name. Windows has an option to hide file name extensions in the GUI. If you see foo.txt in a window, it could be that the file's actual name is foo.txt.txt, or something else. You can disable this option in your settings. You can also verify the file name using the command line; dir will tell you the truth about what is in the folder. (The Linux/Mac equivalent is ls, of course; but the problem should not arise there in the first place.)
Backslashes in ordinary strings are escape sequences. This causes problems when trying to a backslash as the path separator on Windows. However, using backslashes for this is not necessary, and generally not advisable. See Windows path in Python.
When trying to create a new file using a file mode like w, the path to the new file still needs to exist - i.e., all the intervening folders. See for example Trying to use open(filename, 'w' ) gives IOError: [Errno 2] No such file or directory if directory doesn't exist. Also keep in mind that the new file name has to be valid. In particular, it will not work to try to insert a date in MM/DD/YYYY format into the file name, because the /s will be treated as path separators.
Check the path that has been mentioned, if it's absolute or relative.
If its something like-->/folder/subfolder/file -->Computer will search for folder in root directory.
If its something like--> ./folder/subfolder/file --> Computer will search for folder in current working directory.
If you are using IDE like VScode, make sure you have opened the IDE from the same directory where you have kept the file you want to access.
For example, if you want to access file.txt which is inside the Document, try opening the IDE from Document by right clicking in the directory and clicking "Open with "
I am trying to open the file recentlyUpdated.yaml from my Python script. But when I try using:
open('recentlyUpdated.yaml')
I get an error that says:
IOError: [Errno 2] No such file or directory: 'recentlyUpdated.yaml'
Why? How can I fix the problem?
Ensure the file exists (and has the right file extension): use os.listdir() to see the list of files in the current working directory.
Ensure you're in the expected directory using os.getcwd().
(If you launch your code from an IDE, you may be in a different directory.)
You can then either:
Call os.chdir(dir) where dir is the directory containing the file. Then, open the file using just its name, e.g. open("file.txt").
Specify an absolute path to the file in your open call.
Use a raw string (r"") if your path uses backslashes, like
so: dir = r'C:\Python32'
If you don't use raw string, you have to escape every backslash: 'C:\\User\\Bob\\...'
Forward-slashes also work on Windows 'C:/Python32' and do not need to be escaped.
Let me clarify how Python finds files:
An absolute path is a path that starts with your computer's root directory, for example C:\Python\scripts if you're on Windows.
A relative path is a path that does not start with your computer's root directory, and is instead relative to something called the working directory. You can view Python's current working directory by calling os.getcwd().
If you try to do open('sortedLists.yaml'), Python will see that you are passing it a relative path, so it will search for the file inside the current working directory.
Calling os.chdir() will change the current working directory.
Example: Let's say file.txt is found in C:\Folder.
To open it, you can do:
os.chdir(r'C:\Folder')
open('file.txt') # relative path, looks inside the current working directory
or
open(r'C:\Folder\file.txt') # absolute path
Most likely, the problem is that you're using a relative file path to open the file, but the current working directory isn't set to what you think it is.
It's a common misconception that relative paths are relative to the location of the python script, but this is untrue. Relative file paths are always relative to the current working directory, and the current working directory doesn't have to be the location of your python script.
You have three options:
Use an absolute path to open the file:
file = open(r'C:\path\to\your\file.yaml')
Generate the path to the file relative to your python script:
from pathlib import Path
script_location = Path(__file__).absolute().parent
file_location = script_location / 'file.yaml'
file = file_location.open()
(See also: How do I get the path and name of the file that is currently executing?)
Change the current working directory before opening the file:
import os
os.chdir(r'C:\path\to\your\file')
file = open('file.yaml')
Other common mistakes that could cause a "file not found" error include:
Accidentally using escape sequences in a file path:
path = 'C:\Users\newton\file.yaml'
# Incorrect! The '\n' in 'Users\newton' is a line break character!
To avoid making this mistake, remember to use raw string literals for file paths:
path = r'C:\Users\newton\file.yaml'
# Correct!
(See also: Windows path in Python)
Forgetting that Windows doesn't display file extensions:
Since Windows doesn't display known file extensions, sometimes when you think your file is named file.yaml, it's actually named file.yaml.yaml. Double-check your file's extension.
The file may be existing but may have a different path. Try writing the absolute path for the file.
Try os.listdir() function to check that atleast python sees the file.
Try it like this:
file1 = open(r'Drive:\Dir\recentlyUpdated.yaml')
Possibly, you closed the 'file1'.
Just use 'w' flag, that create new file:
file1 = open('recentlyUpdated.yaml', 'w')
mode is an optional string that specifies the mode in which the file
is opened. It defaults to 'r' which means open for reading in text
mode. Other common values are 'w' for writing (truncating the file if
it already exists)...
(see also https://docs.python.org/3/library/functions.html?highlight=open#open)
If is VSCode see the workspace. If you are in other workspace this error can rise
Understanding absolute and relative paths
The term path means exactly what it sounds like. It shows the steps that need to be taken, into and out of folders, to find a file. Each step on the path is either a folder name, the special name . (which means the current folder), or the special name .. (which means to go back/out into the parent folder).
The terms absolute and relative also have their usual English meaning. A relative path shows where something is relative to some start point; an absolute path is a location starting from the top.
Paths that start with a path separator, or a drive letter followed by a path separator (like C:/foo) on Windows, are absolute. (On Windows there are also UNC paths, which are necessarily absolute. Most people will never have to worry about these.)
Paths that directly start with a file or folder name, or a drive letter followed directly by the file or folder name (like C:foo) on Windows, are relative.
Understanding the "current working directory"
Relative paths are "relative to" the so-called current working directory (hereafter abbreviated CWD). At the command line, Linux and Mac use a common CWD across all drives. (The entire file system has a common "root", and may include multiple physical storage devices.) Windows is a bit different: it remembers the most recent CWD for each drive, and has separate functionality to switch between drives, restoring those old CWD values.
Each process (this includes terminal/command windows) has its own CWD. When a program is started from the command line, it will get the CWD that the terminal/command process was using. When a program is started from a GUI (by double-clicking a script, or dragging something onto the script, or dragging the script onto a Python executable) or by using an IDE, the CWD might be any number of things depending on the details.
Importantly, the CWD is not necessarily where the script is located.
The script's CWD can be checked using os.getcwd, and modified using os.chdir. Each IDE has its own rules that control the initial CWD; check the documentation for details.
To set the CWD to the folder that contains the current script, determine that path and then set it:
os.chdir(os.path.dirname(os.path.abspath(__file__)))
Verifying the actual file name and path
There are many reasons why the path to a file might not match expectations. For example, sometimes people expect C:/foo.txt on Windows to mean "the file named foo.txt on the desktop". This is wrong. That file is actually - normally - at C:/Users/name/Desktop/foo.txt (replacing name with the current user's username). It could instead be elsewhere, if Windows is configured to put it elsewhere. To find the path to the desktop in a portable way, see How to get Desktop location?.
It's also common to mis-count ..s in a relative path, or inappropriately repeat a folder name in a path. Take special care when constructing a path programmatically. Finally, keep in mind that .. will have no effect while already in a root directory (/ on Linux or Mac, or a drive root on Windows).
Take even more special care when constructing a path based on user input. If the input is not sanitized, bad things could happen (e.g. allowing the user to unzip a file into a folder where it will overwrite something important, or where the user ought not be allowed to write files).
Another common gotcha is that the special ~ shortcut for the current user's home directory does not work in an absolute path specified in a Python program. That part of the path must be explicitly converted to the actual path, using os.path.expanduser. See Why am I forced to os.path.expanduser in python? and os.makedirs doesn't understand "~" in my path.
Keep in mind that os.listdir will give only the file names, not paths. Trying to iterate over a directory listed this way will only work if that directory is the current working directory.
It's also important to make sure of the actual file name. Windows has an option to hide file name extensions in the GUI. If you see foo.txt in a window, it could be that the file's actual name is foo.txt.txt, or something else. You can disable this option in your settings. You can also verify the file name using the command line; dir will tell you the truth about what is in the folder. (The Linux/Mac equivalent is ls, of course; but the problem should not arise there in the first place.)
Backslashes in ordinary strings are escape sequences. This causes problems when trying to a backslash as the path separator on Windows. However, using backslashes for this is not necessary, and generally not advisable. See Windows path in Python.
When trying to create a new file using a file mode like w, the path to the new file still needs to exist - i.e., all the intervening folders. See for example Trying to use open(filename, 'w' ) gives IOError: [Errno 2] No such file or directory if directory doesn't exist. Also keep in mind that the new file name has to be valid. In particular, it will not work to try to insert a date in MM/DD/YYYY format into the file name, because the /s will be treated as path separators.
Check the path that has been mentioned, if it's absolute or relative.
If its something like-->/folder/subfolder/file -->Computer will search for folder in root directory.
If its something like--> ./folder/subfolder/file --> Computer will search for folder in current working directory.
If you are using IDE like VScode, make sure you have opened the IDE from the same directory where you have kept the file you want to access.
For example, if you want to access file.txt which is inside the Document, try opening the IDE from Document by right clicking in the directory and clicking "Open with "
I read a tutorial that you can compile all the libs files to .pyc, then pack all the .pyc as a zipped file. Then the python still works as magic and it becomes significantly smaller.
But when I zipped all the .pyc files as python36.zip and saved them under /lib/python3.6. Tried to start python, the python says it cannot find module codecs.
What did I do wrong? Can someone explain will this actually work?
To import modules from a .zip file, you need to add that file to sys.path - then it will act as a search directory. The zipimport module that does the job is a built-in one.
sys.path is constructed like this:
PYTHONPATH environment variable + value calculated from a compiled-in default (PYTHONPATH macro in pyconfig.h)
The calculated value includes a path to a .zip file that may or may not exist.
Then site is imported that does the following:
looks for and adds user-specific site-packages
looks for system-wide site-packages directories under sys.prefix and sys.exec_prefix.
when calculating both sys properties, the interpreter does a number of tests that includes looking for os.py file and lib-dynload directory where they should normally be
scans the site-packages directories for .pth files and appends their lines to sys.path, treating them as paths relative to the file's location.
So, you can move all the standard modules into the predefined .zip file. But you may need to leave an os.py or lib-dynload if sys.prefix and sys.exec_prefix become blank after that (the contents are irrelevant, the moved modules will be imported from the .zip because it's earlier on sys.path), or you will lose access to all 3rd-party modules.
Subdirectories that have their own entry in sys.path you need to handle separately so that their contents can still be found on sys.path.
(tested in Python 2.7-win32)
Though adding .pyc files to the archive is sufficient, pdb and stacktraces will be useless unless you place .pys there as well.
What is the best practice to get the path of the currently executed script?
path = os.path.dirname(os.path.abspath(__file__))
or
path = __path__[0]
When tried, they both output the same result.
But is one more rigorous than the other?
Is one more cross platform?
The doc says:
This variable can be modified; doing so affects future searches for
modules and subpackages contained in the package. While this feature
is not often needed, it can be used to extend the set of modules found
in a package.
Apart from me modifying it explicitly, can it be modified by the interpreter without me knowing?
If so, would that return a wrong path, or is the first element always pointing to the current directory of the script?
I need that to access a config file which would be stored inside a folder called config/ in my package.
But open("config/config.txt", 'rt') does not work, it returns a IOError: [Errno 2] No such file or directory
Thanks.
I am trying to open the file recentlyUpdated.yaml from my Python script. But when I try using:
open('recentlyUpdated.yaml')
I get an error that says:
IOError: [Errno 2] No such file or directory: 'recentlyUpdated.yaml'
Why? How can I fix the problem?
Ensure the file exists (and has the right file extension): use os.listdir() to see the list of files in the current working directory.
Ensure you're in the expected directory using os.getcwd().
(If you launch your code from an IDE, you may be in a different directory.)
You can then either:
Call os.chdir(dir) where dir is the directory containing the file. Then, open the file using just its name, e.g. open("file.txt").
Specify an absolute path to the file in your open call.
Use a raw string (r"") if your path uses backslashes, like
so: dir = r'C:\Python32'
If you don't use raw string, you have to escape every backslash: 'C:\\User\\Bob\\...'
Forward-slashes also work on Windows 'C:/Python32' and do not need to be escaped.
Let me clarify how Python finds files:
An absolute path is a path that starts with your computer's root directory, for example C:\Python\scripts if you're on Windows.
A relative path is a path that does not start with your computer's root directory, and is instead relative to something called the working directory. You can view Python's current working directory by calling os.getcwd().
If you try to do open('sortedLists.yaml'), Python will see that you are passing it a relative path, so it will search for the file inside the current working directory.
Calling os.chdir() will change the current working directory.
Example: Let's say file.txt is found in C:\Folder.
To open it, you can do:
os.chdir(r'C:\Folder')
open('file.txt') # relative path, looks inside the current working directory
or
open(r'C:\Folder\file.txt') # absolute path
Most likely, the problem is that you're using a relative file path to open the file, but the current working directory isn't set to what you think it is.
It's a common misconception that relative paths are relative to the location of the python script, but this is untrue. Relative file paths are always relative to the current working directory, and the current working directory doesn't have to be the location of your python script.
You have three options:
Use an absolute path to open the file:
file = open(r'C:\path\to\your\file.yaml')
Generate the path to the file relative to your python script:
from pathlib import Path
script_location = Path(__file__).absolute().parent
file_location = script_location / 'file.yaml'
file = file_location.open()
(See also: How do I get the path and name of the file that is currently executing?)
Change the current working directory before opening the file:
import os
os.chdir(r'C:\path\to\your\file')
file = open('file.yaml')
Other common mistakes that could cause a "file not found" error include:
Accidentally using escape sequences in a file path:
path = 'C:\Users\newton\file.yaml'
# Incorrect! The '\n' in 'Users\newton' is a line break character!
To avoid making this mistake, remember to use raw string literals for file paths:
path = r'C:\Users\newton\file.yaml'
# Correct!
(See also: Windows path in Python)
Forgetting that Windows doesn't display file extensions:
Since Windows doesn't display known file extensions, sometimes when you think your file is named file.yaml, it's actually named file.yaml.yaml. Double-check your file's extension.
The file may be existing but may have a different path. Try writing the absolute path for the file.
Try os.listdir() function to check that atleast python sees the file.
Try it like this:
file1 = open(r'Drive:\Dir\recentlyUpdated.yaml')
Possibly, you closed the 'file1'.
Just use 'w' flag, that create new file:
file1 = open('recentlyUpdated.yaml', 'w')
mode is an optional string that specifies the mode in which the file
is opened. It defaults to 'r' which means open for reading in text
mode. Other common values are 'w' for writing (truncating the file if
it already exists)...
(see also https://docs.python.org/3/library/functions.html?highlight=open#open)
If is VSCode see the workspace. If you are in other workspace this error can rise
Understanding absolute and relative paths
The term path means exactly what it sounds like. It shows the steps that need to be taken, into and out of folders, to find a file. Each step on the path is either a folder name, the special name . (which means the current folder), or the special name .. (which means to go back/out into the parent folder).
The terms absolute and relative also have their usual English meaning. A relative path shows where something is relative to some start point; an absolute path is a location starting from the top.
Paths that start with a path separator, or a drive letter followed by a path separator (like C:/foo) on Windows, are absolute. (On Windows there are also UNC paths, which are necessarily absolute. Most people will never have to worry about these.)
Paths that directly start with a file or folder name, or a drive letter followed directly by the file or folder name (like C:foo) on Windows, are relative.
Understanding the "current working directory"
Relative paths are "relative to" the so-called current working directory (hereafter abbreviated CWD). At the command line, Linux and Mac use a common CWD across all drives. (The entire file system has a common "root", and may include multiple physical storage devices.) Windows is a bit different: it remembers the most recent CWD for each drive, and has separate functionality to switch between drives, restoring those old CWD values.
Each process (this includes terminal/command windows) has its own CWD. When a program is started from the command line, it will get the CWD that the terminal/command process was using. When a program is started from a GUI (by double-clicking a script, or dragging something onto the script, or dragging the script onto a Python executable) or by using an IDE, the CWD might be any number of things depending on the details.
Importantly, the CWD is not necessarily where the script is located.
The script's CWD can be checked using os.getcwd, and modified using os.chdir. Each IDE has its own rules that control the initial CWD; check the documentation for details.
To set the CWD to the folder that contains the current script, determine that path and then set it:
os.chdir(os.path.dirname(os.path.abspath(__file__)))
Verifying the actual file name and path
There are many reasons why the path to a file might not match expectations. For example, sometimes people expect C:/foo.txt on Windows to mean "the file named foo.txt on the desktop". This is wrong. That file is actually - normally - at C:/Users/name/Desktop/foo.txt (replacing name with the current user's username). It could instead be elsewhere, if Windows is configured to put it elsewhere. To find the path to the desktop in a portable way, see How to get Desktop location?.
It's also common to mis-count ..s in a relative path, or inappropriately repeat a folder name in a path. Take special care when constructing a path programmatically. Finally, keep in mind that .. will have no effect while already in a root directory (/ on Linux or Mac, or a drive root on Windows).
Take even more special care when constructing a path based on user input. If the input is not sanitized, bad things could happen (e.g. allowing the user to unzip a file into a folder where it will overwrite something important, or where the user ought not be allowed to write files).
Another common gotcha is that the special ~ shortcut for the current user's home directory does not work in an absolute path specified in a Python program. That part of the path must be explicitly converted to the actual path, using os.path.expanduser. See Why am I forced to os.path.expanduser in python? and os.makedirs doesn't understand "~" in my path.
Keep in mind that os.listdir will give only the file names, not paths. Trying to iterate over a directory listed this way will only work if that directory is the current working directory.
It's also important to make sure of the actual file name. Windows has an option to hide file name extensions in the GUI. If you see foo.txt in a window, it could be that the file's actual name is foo.txt.txt, or something else. You can disable this option in your settings. You can also verify the file name using the command line; dir will tell you the truth about what is in the folder. (The Linux/Mac equivalent is ls, of course; but the problem should not arise there in the first place.)
Backslashes in ordinary strings are escape sequences. This causes problems when trying to a backslash as the path separator on Windows. However, using backslashes for this is not necessary, and generally not advisable. See Windows path in Python.
When trying to create a new file using a file mode like w, the path to the new file still needs to exist - i.e., all the intervening folders. See for example Trying to use open(filename, 'w' ) gives IOError: [Errno 2] No such file or directory if directory doesn't exist. Also keep in mind that the new file name has to be valid. In particular, it will not work to try to insert a date in MM/DD/YYYY format into the file name, because the /s will be treated as path separators.
Check the path that has been mentioned, if it's absolute or relative.
If its something like-->/folder/subfolder/file -->Computer will search for folder in root directory.
If its something like--> ./folder/subfolder/file --> Computer will search for folder in current working directory.
If you are using IDE like VScode, make sure you have opened the IDE from the same directory where you have kept the file you want to access.
For example, if you want to access file.txt which is inside the Document, try opening the IDE from Document by right clicking in the directory and clicking "Open with "