从文件层次结构创建嵌套字典

Question

我想知道是否有人可以指出正确的方向。我试图从文件路径列表创建一个嵌套字典，类似于下面的内容。这个列表会根据用户的输入而改变，所以我想它需要是递归的。关于从哪里开始的任何指示？

编辑： 此外，字典将被转换为 JSON 并用于使用 D3.js.

创建图表

fileDict = [
    {
        "name": "BaseLevel",
        "children": [
          {
            "name": "/etc/",
            "children": [
              {
                "name": "/etc/passwd",
              },
              {
                "name": "/etc/group"
              }
            ]
          },
          {
            "name": "/root/",
            "children": [
              {
                "name": "/root/test",
              }
            ]
          }
        ]
      }
    ]

我能得到的最接近的例子是这个

    records = ["base/images/graphs/one.png", "base/images/tikz/two.png",
"base/refs/images/three.png", "base/one.txt", "base/chapters/two.txt"]

recordsSplit = map(lambda x: x.split("/"), records)

for record in recordsSplit:
    here = result
    for item in record[:-1]:
        if not item in here:
            here[item] = {}
            here = here[item]
        if "###content###" not in here:
            here["###content###"] = []
            here["###content###"].append(record[-1])

print json.dumps(result, indent=4)

Answer 1

是否值得制作一个 class 而不是一个命令？写了一个应该做你想做的简短的

class FileSystem():
    
    def __init__(filePath=None):
        self.children = []
        if files != None:
            try:
                self.name, child = files.split("/", 2)
                self.children.append(FileSystem(filePath))
            except (ValueError):
                 pass
            
    def addChild(filePath):
        self.children.append(FileSystem(filePath))
    
    def getChildren():
        return self.children

    def printAllChildren():
        print "Name: "+ self.name
        print "{ Children:"
        for child in self.children:
            child.printAllChildren()
        print "}"

然后您可以输入第一个路径并保存对它的引用，例如

myFileSystem = FileSystem("base/pictures/whatever.png")

这个 myFileSystem 将是您对“基本”级别的参考，使用它和它的方法您应该能够做您想做的事。

然后当您有第二条路径要添加时，您必须通过在 myFileSystem 上使用 getChildren() 找到要添加的正确节点，直到发现差异，然后使用 addChild() 将文件路径的其余部分添加到该节点。然后使用 myFileSystem.printAllChildren() 将打印出整个文件系统。

--------编辑--------

对我写了一半的代码不太满意，喜欢挑战，所以这是一个易于使用的 class

class FileSystem():

    def __init__(self,filePath=None):
        self.children = []
        if filePath != None:
            try:
                self.name, child = filePath.split("/", 1)
                self.children.append(FileSystem(child))
            except (ValueError):
                self.name = filePath
            
    def addChild(self, filePath):
        try:
            thisLevel, nextLevel = filePath.split("/", 1)
            try:
                if thisLevel == self.name:
                    thisLevel, nextLevel = nextLevel.split("/", 1)
            except (ValueError):
                self.children.append(FileSystem(nextLevel))
                return
            for child in self.children:
                if thisLevel == child.name:
                    child.addChild(nextLevel)
                    return
            self.children.append(FileSystem(nextLevel))
        except (ValueError):
            self.children.append(FileSystem(filePath))

    def getChildren(self):
        return self.children
        
    def printAllChildren(self, depth = -1):
        depth += 1
        print "\t"*depth + "Name: "+ self.name
        if len(self.children) > 0:
            print "\t"*depth +"{ Children:"
            for child in self.children:
                child.printAllChildren(depth)
            print "\t"*depth + "}"
        
records = ["base/images/graphs/one.png", "base/images/tikz/two.png",
"base/refs/images/three.png", "base/one.txt", "base/chapters/two.txt"]

myFiles = FileSystem(records[0])
for record in records[1:]:
    myFiles.addChild(record)

myFiles.printAllChildren()

正如你在最后看到的那样，当我简单地执行 myFiles.addChild(record) 时，addChild 函数现在负责在树中找到正确的位置以使其进入。printAllChildren() 给出了正确的输出至少对于那些参数。

让我知道它是否有任何意义，就像我说的那样它没有经过全面测试，所以一些极端情况（例如试图添加另一个基地？）可能会使它变得奇怪。

EDIT2

class FileSystem():

    def __init__(self,filePath=None):
        self.children = []
        if filePath != None:
            try:
                self.name, child = filePath.split("/", 1)
                self.children.append(FileSystem(child))
            except (ValueError):
                self.name = filePath

    def addChild(self, filePath):
        try:
            thisLevel, nextLevel = filePath.split("/", 1)
            try:
                if thisLevel == self.name:
                    thisLevel, nextLevel = nextLevel.split("/", 1)
            except (ValueError):
                self.children.append(FileSystem(nextLevel))
                return
            for child in self.children:
                if thisLevel == child.name:
                    child.addChild(nextLevel)
                    return
            self.children.append(FileSystem(nextLevel))
        except (ValueError):
            self.children.append(FileSystem(filePath))

    def getChildren(self):
        return self.children

    def printAllChildren(self, depth = -1):
        depth += 1
        print "\t"*depth + "Name: "+ self.name
        if len(self.children) > 0:
            print "\t"*depth +"{ Children:"
            for child in self.children:
                child.printAllChildren(depth)
            print "\t"*depth + "}"
            
    def makeDict(self):
        if len(self.children) > 0:
            dictionary = {self.name:[]}
            for child in self.children:
                dictionary[self.name].append(child.makeDict())
            return dictionary
        else:
            return self.name
                

records = ["base/images/graphs/one.png", "base/images/tikz/two.png",
"base/refs/images/three.png", "base/one.txt", "base/chapters/two.txt"]

myFiles = FileSystem(records[0])
for record in records[1:]:
    myFiles.addChild(record)

print myFiles.makeDict()

Answer 2

当您有这样的文件时：

['testdata/hhohoho.mdf', 'testdata/dvojka/rerere.bdf', 'testdata/jedna/sss.txt']

你的输出结构如下：

Name: testdata
{ Children:
    Name: hhohoho.mdf
    Name: rerere.bdf
    Name: sss.txt
}

你有一个错误：

self.children.append(FileSystem(nextLevel))
    except (ValueError):
        self.children.append(FileSystem(filePath))

解决方式如下：

 self.children.append(FileSystem(thisLevel))
        for child in self.children:
            if thisLevel == child.name:
                child.addChild(nextLevel)
                return


Name: testdata
{ Children:
    Name: hhohoho.mdf
    Name: dvojka
    { Children:
            Name: rerere.bdf
    }
    Name: jedna
    { Children:
            Name: sss.txt
    }
}

从文件层次结构创建嵌套字典

Creating Nested Dictionary from File hierarchy

python

dictionary

nested

file

hierarchy

--------编辑--------