如何在python中对嵌套字典进行排序？

Question

我在我的项目中有一个嵌套字典：

>>> dict = {
    "content": """<p>It is common for content in Arabic, Hebrew, and other languages that use right-to-left scripts to include numerals or include text from  other scripts. Both of these typically flow  left-to-right within the overall right-to-left  context. </p>""", 
    "name": "directory", 
    "decendent": [
         {
            "content": """<p>This article tells you how to write HTML where text with different writing directions is mixed <em>within a paragraph or other HTML block</em> (ie. <dfn id="term_inline">inline or phrasal</dfn> content). (A companion article <a href="/International/questions/qa-html-dir"><cite>Structural markup and right-to-left text in HTML</cite></a> tells you how to use HTML markup for  elements such as <code class="kw">html</code>, and structural markup such as <code class="kw">p</code> or <code class="kw">div</code> and forms.)</p>""", 
            "name": "subdirectory", 
            "decendent": None
        }, 
        {
            "content": """It tells you how to use HTML markup for  elements such as <code class="kw">html</code>, and structural markup such as <code class="kw">p</code> or <code class="kw">div</code> and forms.)""", 
            "name": "subdirectory_two", 
            "decendent": [
                {
                    "content": "Name 4", 
                    "name": "subsubdirectory", 
                    "decendent": None
                }
            ]
        }
    ]
}

以及生成它的函数：

def getname(dirpath):
    onlyfile = [entry for entry in os.listdir(dirpath) if os.path.isfile(os.path.join(dirpath, entry)) and entry.endswith(".txt")][0]
    onlyfilename = os.path.join(dirpath, onlyfile)

    decendent = []
    for entry in os.listdir(dirpath):
        entrypath = os.path.join(dirpath, entry)
        if os.path.isfile(entrypath):
            continue
        decendent.append(getname(entrypath))

    if len(decendent) == 0:
        decendent = None

    return {'name': onlyfilename.split("/")[-2],
            'path': onlyfilename.rsplit("/", 1)[-2] + "/index.html",
            'leaf': readFile(onlyfilename),
            'decendent': decendent}

问题是，它作为后代返回一组无序字典：

​

​

在我的情况下，当它们被嵌套时，如何排序？

Answer 1

您的descendent值只是一个列表。列表可以排序，无论它们被引用在哪里。它在字典里并不重要。

您的descendent列表顺序由os.listdir()返回的任何顺序设置。您可以对此进行排序，也可以简单地对descendent对象本身进行排序。例如，使用natural sort进行排序，或者使用任何其他条件进行排序。

例如，在natural_sort()结果上使用来自Mark Byer's answer的os.listdir()函数：

for entry in natural_sort(os.listdir(dirpath)):

按排序顺序列出目录条目，以便插入到descendant列表中。