python:list()函数的困惑

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我不明白list函数是如何工作的。

以下是我所做的研究:

我正在看的文件:

文档

我特别要看这一段:

class list([iterable]) Return a list whose items are the same and in
the same order as iterable’s items. iterable may be either a sequence,
a container that supports iteration, or an iterator object. If
iterable is already a list, a copy is made and returned, similar to
iterable[:]. For instance, list('abc') returns ['a', 'b', 'c'] and
list( (1, 2, 3) ) returns [1, 2, 3]. If no argument is given, returns
a new empty list, [].

list is a mutable sequence type, as documented in Sequence Types —
str, unicode, list, tuple, bytearray, buffer, xrange. For other
containers see the built in dict, set, and tuple classes, and the
collections module.

这里是另一个帖子:

另一篇关于list函数的文章

在这个帖子上，有人发布了以下内容:

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>>> myList = ['hello']
>>> myList=list(myList[0])
>>> myList
['h', 'e', 'l', 'l', 'o']

但当我这样做的时候:

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for root, dirs, files in os.walk(os.getcwd()):
path_files.append(files)

path_files
[['combinedPdfs.py', 'meetingminutes.pdf', 'meetingminutes_encrypted.pdf', 'pdf_intro.py', 'pdf_paranoia.py', 'readDocx.py']]

>>> path_files_2 = list(path_files[0])
>>> path_files_2
['combinedPdfs.py', 'meetingminutes.pdf', 'meetingminutes_encrypted.pdf', 'pdf_intro.py', 'pdf_paranoia.py', 'readDocx.py']
>>> path_files_2[0]
'combinedPdfs.py'
>>> path_files_2[1]
'meetingminutes.pdf'

为什么我所做的工作与其他帖子的用户不同?

编辑# 1:

如果我这样运行:

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>>> myList2 = ['hello', 'goodbye']
>>> myList2[0]
'hello'
>>> myList2 = list(myList2)
>>> myList2
['hello', 'goodbye']
>>> myList2 = list(myList2[0])
>>> myList2
['h', 'e', 'l', 'l', 'o']

如果我这样运行:

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>>> myList4 = [['Hello', 'goodbye']]
>>> myList4 = list(myList4)
>>> myList4
[['Hello', 'goodbye']]
>>> myList4 = list(myList4[0])
>>> myList4
['Hello', 'goodbye']

我看到了这个定义，但我希望有一个更"门外汉"的方式来解释它。