Python 递归地修改字典
我创建了一个类,它应该将嵌套列表转换为字典。以下是我的意见:Python 递归地修改字典,python,algorithm,dictionary,recursion,Python,Algorithm,Dictionary,Recursion,我创建了一个类,它应该将嵌套列表转换为字典。以下是我的意见: ['function:and', ['variable:X', 'function:>=', 'value:13'], ['variable:Y', 'function:==', 'variable:W']] 输出应为以下形式的词典: { "function": "and", "args": [ { "function": ">=", "args": [
['function:and',
['variable:X', 'function:>=', 'value:13'],
['variable:Y', 'function:==', 'variable:W']]
输出应为以下形式的词典:
{
"function": "and",
"args": [
{
"function": ">=",
"args": [
{
"variable": "X"
},
{
"value": 13
}
]
},
{
"function": "==",
"args": [
{
"variable": "Y"
},
{
"variable": "W"
}
]
}
]
}
这是接收输入列表并应返回所需字典的类
class Tokenizer(object):
def __init__(self, tree):
self.tree = tree
self.filter = {}
def to_dict(self, triple):
my_dict = {}
try:
first = triple[0]
second = triple[1]
third = triple[2]
except KeyError:
return
if type(second) == str and type(third) == str:
my_dict['function'] = second.split(':')[-1]
my_dict['args'] = [
{first.split(':')[0]: first.split(':')[1]},
{third.split(':')[0]: third.split(':')[1]}]
# case recursive
if type(second) == list:
my_dict['function'] = first.split(':')[-1]
my_dict['args'] = [second, third]
return my_dict
def walk(self, args):
left = self.to_dict(args[0])
right = self.to_dict(args[1])
if isinstance(left, dict):
if 'args' in left.keys():
left = self.walk(left['args'])
if isinstance(right, dict):
if 'args' in right.keys():
right = self.walk(right['args'])
args = [left, right]
return args
def run(self):
self.filter.update(self.to_dict(self.tree))
if 'args' in self.filter.keys():
self.filter['args'] = self.walk(self.filter['args'])
tree = [
'function:and',
['variable:X', 'function:>=', 'value:13'],
['variable:Y', 'function:==', 'variable:W']
]
import pprint
pp = pprint.PrettyPrinter(indent=4)
t = Tokenizer(tree)
t.run()
pp.pprint(t.filter)
我的递归方法walk
没有做它应该做的事情,我完全是递归的吸盘,所以我不知道我做错了什么
我得到的结果是:
{ 'args': [[None, None], [None, None]], 'function': 'and'}
对于您的特定测试用例,您根本不需要进行递归。您可以对您的通话进行评论:
def walk(self, args):
left = self.to_dict(args[0])
right = self.to_dict(args[1])
#if isinstance(left, dict):
# if 'args' in left.keys():
# left = self.walk(left['args'])
#if isinstance(right, dict):
# if 'args' in right.keys():
# right = self.walk(right['args'])
args = [left, right]
return args
并获得所需的输出。
如果允许在输入中使用嵌套函数,则只需进入递归:
['function:and',
['variable:X', 'function:>=', 'value:13'],
['function:==',
['variable:R', 'function:>=', 'value:1'],
['variable:Z', 'function:==', 'variable:K']
]
]
然后必须检查基本情况,因此只有当args
键的值包含未处理的值时,才进入递归:
def walk(self, args):
left = self.to_dict(args[0])
right = self.to_dict(args[1])
if isinstance(left, dict):
if 'args' in left.keys() and isinstance(left['args'][0], list):
left = self.walk(left['args'])
if isinstance(right, dict):
if 'args' in right.keys() and isinstance(right['args'][0], list):
right = self.walk(right['args'])
args = [left, right]
return args
然后你会得到这个:
{ 'args': [ { 'args': [{ 'variable': 'X'}, { 'value': '13'}],
'function': '>='},
{ 'args': [ { 'args': [ { 'variable': 'R'},
{ 'value': '1'}],
'function': '>='},
{ 'args': [ { 'variable': 'Z'},
{ 'variable': 'K'}],
'function': '=='}],
'function': '=='}],
'function': 'and'}
另外,如果输入列表是一个规则结构,并且在函数名字段之后始终有参数字段,则会更容易。然后,您可以将
方法大大简化为dict
方法。对于您的特定测试用例,您根本不需要进行递归。您可以对您的通话进行评论:
def walk(self, args):
left = self.to_dict(args[0])
right = self.to_dict(args[1])
#if isinstance(left, dict):
# if 'args' in left.keys():
# left = self.walk(left['args'])
#if isinstance(right, dict):
# if 'args' in right.keys():
# right = self.walk(right['args'])
args = [left, right]
return args
并获得所需的输出。
如果允许在输入中使用嵌套函数,则只需进入递归:
['function:and',
['variable:X', 'function:>=', 'value:13'],
['function:==',
['variable:R', 'function:>=', 'value:1'],
['variable:Z', 'function:==', 'variable:K']
]
]
然后必须检查基本情况,因此只有当args
键的值包含未处理的值时,才进入递归:
def walk(self, args):
left = self.to_dict(args[0])
right = self.to_dict(args[1])
if isinstance(left, dict):
if 'args' in left.keys() and isinstance(left['args'][0], list):
left = self.walk(left['args'])
if isinstance(right, dict):
if 'args' in right.keys() and isinstance(right['args'][0], list):
right = self.walk(right['args'])
args = [left, right]
return args
然后你会得到这个:
{ 'args': [ { 'args': [{ 'variable': 'X'}, { 'value': '13'}],
'function': '>='},
{ 'args': [ { 'args': [ { 'variable': 'R'},
{ 'value': '1'}],
'function': '>='},
{ 'args': [ { 'variable': 'Z'},
{ 'variable': 'K'}],
'function': '=='}],
'function': '=='}],
'function': 'and'}
另外,如果输入列表是一个规则结构,并且在函数名字段之后始终有参数字段,则会更容易。然后,您可以将
方法大大简化为_dict
方法。我的方法没有做它应该做的事情,所以它在做什么?@JohnGordon:编辑以添加输出,谢谢!可能是@das-g的一个例子:修复了,谢谢。帕斯特不是你的朋友也许有了这个“开始”,你会更容易找到方向。我的方法没有做它应该做的,那么它在做什么?@JohnGordon:编辑以添加输出,谢谢!可能是@das-g的一个例子:修复了,谢谢。帕斯特不是你的朋友也许有了这个“开始”,你会更容易找到路。