Python 将函数转换为OOP
我有一个类似的json口述:Python 将函数转换为OOP,python,arrays,oop,recursion,initialization,Python,Arrays,Oop,Recursion,Initialization,我有一个类似的json口述: { "name": "Root Folder", "type": "folder", "children": [ { "name": "Folder A", "type": "folder", "children": [ { "name": "Folder A-A", "type": "folder", "children": [
{
"name": "Root Folder",
"type": "folder",
"children": [
{
"name": "Folder A",
"type": "folder",
"children": [
{
"name": "Folder A-A",
"type": "folder",
"children": [
{
"name": "File A-A-1",
"type": "file",
"children": []
}
]
},
{
"name": "Folder A-B",
"type": "folder",
"children": []
}
]
},
{
"name": "Folder B",
"type": "folder",
"children": [
{
"name": "Folder B-A",
"type": "folder",
"children": [
{
"name": "Folder B-A-A",
"type": "folder",
"children": []
}
]
},
{
"name": "Folder B-B",
"type": "folder",
"children": []
}
]
},
{
"name": "File Root",
"type": "file",
"children": []
}
]
}
从这个命令中,我想自动获取每个成员以及级别的值“Path”、“Name”和“type”,因此root是level0,Folder A level 1,Folder AA level 2等等
为此,我创建了以下Python代码,该代码采用json字典并将其中的值放入数组中:
import json
import os
from jinja2 import Template
with open('Folder_structure.json') as f:
data = json.load(f)
def arr4tmpl(data,arr_path,arr_name,arr_type,arr_level):
for child in data['children']:
arr_path.extend([child['path']])
arr_name.extend([child['name']])
arr_type.extend([child['type']])
arr_level.extend([arr4tmpl.level])
arr4tmpl.level+=1
arr4tmpl(child,arr_path,arr_name,arr_type,arr_level)
arr4tmpl.level-=1
return(arr_path,arr_name,arr_type,arr_level)
def main():
arr4tmpl.level = 0
arr=[]
arr_path=[]
arr_name=[]
arr_type=[]
arr_level=[]
(arr_path,arr_name,arr_type,arr_level)=arr4tmpl(data,arr_path,arr_name,arr_type,arr_level)
这很好,但现在必须将其转变为面向对象的解决方案。因此我想到了以下内容(根据答案编辑,谢谢):
其思想是通过init方法初始化参数,然后在arr4tmpl方法中扩展它们。
据我所知,创建对象应该执行init方法,但在执行代码时,我得到错误:
File "Folder_structure.py", line 34, in <module>
template.arr4tmpl(data,level)
NameError: name 'level' is not defined
文件“Folder_structure.py”,第34行,在
模板arr4tmpl(数据,级别)
NameError:未定义名称“级别”
当您想要修改对象变量时,必须使用self引用它们
def arr4tmpl(self,data,level):
for child in data['children']:
self.arr_path.extend([child['path']])
self.arr_name.extend([child['name']])
self.arr_type.extend([child['type']])
self.arr_level.extend([level])
self.level+=1
self.arr4tmpl(self,child,level) #BUT this line will cause all sorts of recursion issues, since you don't have a stop criteria
self.level-=1
return(arr_path,arr_name,arr_type,arr_level)
啊,是的,我真的找到了一个方法:
import json
import os
from jinja2 import Template
class CreateTemplate(object):
arr_path=[]
arr_name=[]
arr_type=[]
arr_level=[]
level=0
def arr4tmpl(self,data):
for child in data['children']:
self.arr_path.extend([child['path']])
self.arr_name.extend([child['name']])
self.arr_type.extend([child['type']])
self.arr_level.extend([self.level])
self.level+=1
self.arr4tmpl(child)
self.level-=1
return(self.arr_path,self.arr_name,self.arr_type,self.arr_level)
with open('Folder_structure.json') as f:
data = json.load(f)
template=CreateTemplate()
template.arr4tmpl(data)
我不确定init为什么不能工作,因为我认为它会在创建对象时初始化参数,然后这些变量可以用于方法arr4tmpl
也不确定arr4tmpl方法是否可以改进,但至少它现在确实创建了这些阵列。谢谢 您没有调用类的成员不要忘记,您仍然需要在
arr4tmpl
方法中引用self
——因此arr\u path.extend(…)
应该是self.arr\u path.extend(…)
,等等。欢迎使用StackOverflow。请按照您创建此帐户时的建议,阅读并遵循帮助文档中的发布指南。适用于这里。在您发布MCVE代码并准确描述问题之前,我们无法有效地帮助您。我们应该能够将您发布的代码粘贴到文本文件中,并重现您描述的问题。“不工作”不是一个问题规范。
import json
import os
from jinja2 import Template
class CreateTemplate(object):
arr_path=[]
arr_name=[]
arr_type=[]
arr_level=[]
level=0
def arr4tmpl(self,data):
for child in data['children']:
self.arr_path.extend([child['path']])
self.arr_name.extend([child['name']])
self.arr_type.extend([child['type']])
self.arr_level.extend([self.level])
self.level+=1
self.arr4tmpl(child)
self.level-=1
return(self.arr_path,self.arr_name,self.arr_type,self.arr_level)
with open('Folder_structure.json') as f:
data = json.load(f)
template=CreateTemplate()
template.arr4tmpl(data)