filenames =os.listdir(excited)
时间: 2024-03-10 20:30:57 浏览: 56
As an AI language model, I don't have access to your file system or the current directory. However, assuming that `excited` is a string variable containing the path to a directory, the code `filenames = os.listdir(excited)` would list all the files in that directory and assign them to a list called `filenames`.
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filenames = os.listdir(d_path)
I'm sorry, but as an AI language model, I cannot provide a specific answer without additional context. However, I can explain the purpose of this line of code.
The code "os.listdir(d_path)" is used to retrieve a list of all the files and directories present in the directory path specified by the variable "d_path". The function "os.listdir" belongs to the "os" module in Python, which provides a way to interact with the operating system. This code can be used to retrieve a list of filenames in a specific directory, which can be useful for performing operations on multiple files at once.
79. 功能与下面程序段相同的是( ) import os path=os.getcwd() s=os.listdir(path) print(s) A. import os path=os.getcwd() s=os.walk(path) print(s) B. import os path=os.getcwd() s=os.walk(path) for dirpath,dirnames,filenames in s: print(filenames) C. import os path=os.getcwd() s=os.walk(path) for dirpath,dirnames,filenames in s: print(dirpath) D. import os path=os.getcwd() s=os.walk(path) for dirpath,dirnames,filenames in s: print(dirnames)
答案是 A。
程序段中,os.getcwd() 返回当前工作目录的路径,os.listdir(path) 返回指定目录下所有文件和子目录的名称列表,因此程序段的作用是列出当前工作目录下所有文件和子目录的名称列表。
选项 A 中的程序段与原程序段相同,os.walk(path) 也是递归地遍历指定目录及其子目录,并返回每个子目录中的文件和子目录的信息,但返回的信息形式不同,os.walk(path) 返回的是一个迭代器,可以通过遍历迭代器的方式获取每个子目录中的文件和子目录的详细信息。选项 B、C、D 中的程序段都是通过遍历 os.walk(path) 返回的迭代器来获取子目录中的文件和子目录的信息,只是打印的信息不同,因此这三个选项的功能与原程序段相同。
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