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  • Python之文件路径名的操作

    使用 os.path 模块中的函数来操作路径名
    import os
    
    # 获取当前文件路径
    path=os.path.abspath(__file__)  # 获取绝对路径 /home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py
    print(path)
    print(os.path.realpath(__file__))  # 获取真实路径  /home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py
    
    # 文件基础名字
    print(os.path.basename(path))  # 文件与IO.py
    
    # 获取文件目录
    print(os.path.dirname(path))   # /home/zzy/PycharmProjects/MyTest/testPackage/secondPackage
    
    # 文件路径拼接
    print(os.path.join("new","zzy",os.path.basename(path)))  # new/zzy/文件与IO.py
    
    # 展开用户的主目录
    print(os.path.expanduser("~/PycharmProjects/MyTest/testPackage/secondPackage"))   # /home/zzy/PycharmProjects/MyTest/testPackage/secondPackage
    
    # 拆分文件扩展名
    print(os.path.splitext(path))  # ('/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO', '.py')
    
    # 拆分文件名和目录
    print(os.path.split(path))  # ('/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage', '文件与IO.py')
    对于任何的文件名的操作,你都应该使用 os.path 模块,特别是为了可移植性考虑的时候更应如此,
    因为 os.path 模块知道Unix和Windows系统之间的差异并且能够可靠地处理类似

    测试文件是否存在
    print(os.path.exists("MyTest/testPackage"))  # False
    print(os.path.exists("/home/zzy/PycharmProjects/MyTest/testPackage"))  # True
    判断文件类型
    print(os.path.isdir("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage"))  # True
    print(os.path.isdir("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py"))  # False
    
    print(os.path.isfile("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py"))  # True
    
    print(os.path.islink("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py"))  # False
    print(os.path.islink("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage"))  # False
     
     
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  • 原文地址:https://www.cnblogs.com/zzy-9318/p/10518284.html
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