Python can determine whether a file or folder exists,

Python can determine whether a file or folder exists,

Before reading or writing a file, you must determine whether the file or directory exists. Otherwise, some processing methods may cause errors in the program. Therefore, it is best to determine whether a file exists before performing any operations.

Here we will introduce three methods to determine whether a file or folder exists: OS module, Try statement, and pathlib module.

1. Use the OS Module

The OS. path. exists () method in the OS module is used to check whether a file exists.

• Determine whether a file exists

import osos.path.exists(test_file.txt)#Trueos.path.exists(no_exist_file.txt)#False

• Determine whether a folder exists

import osos.path.exists(test_dir)#Trueos.path.exists(no_exist_dir)#False

We can see thatos.path.exists()To determine whether the file and folder are the same.

In fact, this method still has a problem. Suppose you want to check whether the file "test_data" exists, but there is a folder named "test_data" in the current path, which may lead to misjudgment. To avoid this situation, you can:

• Only check files

import osos.path.isfile("test-data")

In this method, if the file "test-data" does not exist, False is returned, and True is returned.

That is, the file exists. You may also need to determine whether the file can be read and written.

Determine whether a file can be read/written.

Useos.access()Method to Determine whether a file can be read and written.

Syntax:

os.access(, )

Path is the file path and mode is the operation mode. There are several types of files:

• OS. F_ OK: Check whether the file exists;
• OS. R_ OK: Check whether the file is readable;
• OS. W_ OK: Check whether the file can be written;
• OS. X_ OK: Check whether the file can be executed

This method returns True or False by determining whether the file path exists and the permissions in various access modes.

import osif os.access("/file/path/foo.txt", os.F_OK):  print "Given file path is exist."if os.access("/file/path/foo.txt", os.R_OK):  print "File is accessible to read"if os.access("/file/path/foo.txt", os.W_OK):  print "File is accessible to write"if os.access("/file/path/foo.txt", os.X_OK):  print "File is accessible to execute"

2. Use the Try statement

You can directly use the open () method in the program to check whether the file exists and can be read and written.

Syntax:

open()

If your open file does not exist, the program will throw an error and use the try statement to capture the error.

The program may not be able to access files for many reasons:

• If your open file does not exist, a FileNotFoundError exception will be thrown;
• If the file exists but has no access permission, a PersmissionError error is thrown.

Therefore, you can use the following code to determine whether a file exists:

try:  f =open()  f.close()except FileNotFoundError:  print "File is not found."except PersmissionError:  print "You don't have permission to access this file."

In fact, there is no need to deal with every exception in such detail. The above two exceptions are subclasses of IOError. So you can simplify the program:

try:  f =open()  f.close()except IOError:  print "File is not accessible."

The try statement is simple and elegant to handle all exceptions. In addition, you do not need to introduce other external modules.

3. Use the pathlib Module

The pathlib module is a built-in module in Python 3, but the third-party module needs to be installed separately in Python 2.

To use pathlib, you must first use the file path to create a path object. This path can be a file name or directory path.

• Check whether the path exists

path = pathlib.Path("path/file")path.exist()

• Check whether the path is a file

path = pathlib.Path("path/file")path.is_file()

Summary

The above section describes three methods for Python to determine whether a file or folder exists. I hope this will be helpful to you. If you have any questions, please leave a message for me, the editor will reply to you in a timely manner. Thank you very much for your support for the help House website!