def infix_to_postfix(self): prec = {'∧': 1, '∨': 1, '¬': 2,'*':4,'/':4,'+':3,'-':3} stack1 = Stack() lst = [] spl = self.split() for i in spl: if i.isdigit(): lst.append(i) elif i == '(': stack1.push(i) elif i == ')': top = stack1.pop() while top != '(': lst.append(top) top= stack1.pop() elif (i== '∧' or i=='∨' or i=='¬' or i=='+'or i=='-' or i=='*' or i=='/'): if stack1.peek() == '(': stack1.push(i) elif not stack1.empty() and (prec[stack1.peek()]>=prec[i]): lst.append(stack1.pop()) stack1.push(i) elif not stack1.empty() and (prec[stack1.peek()]<prec[i]): stack1.push(i) elif stack1.empty(): stack1.push(i) else: lst.append(i) while not stack1.empty(): lst.append(stack1.pop()) return lst

def infix_evaluator(infix_expression: str): token_list = infix_expression.split() print(token_list) pre_dict = {'': 2, '/': 2, '+': 1, '-': 1} ope_stack = [] num_stack = [] for i in token_list: if i.isdecimal() or '.' in i: num_stack.append(float(i)) elif i in '+-/': while ope_stack and pre_dict[ope_stack[-1]] >= pre_dict[i]: top = ope_stack.pop() op2 = num_stack.pop() op1 = num_stack.pop() num_stack.append(get_value(top, op1, op2)) ope_stack.append(i) while ope_stack: top = ope_stack.pop() op2 = num_stack.pop() op1 = num_stack.pop() num_stack.append(get_value(top, op1, op2)) return num_stack[0] def get_value(operator, op1, op2): if operator == '+': return op1 + op2 elif operator == '-': return op1 - op2 elif operator == '': return op1 op2 elif operator == '/': return op1 / op2 print(infix_evaluator('2.5 - 3 * 4'))请修改这个代码

def get_value(operator, op1, op2): if operator == '+': return op1 + op2 elif operator == '-': return op1 - op2 elif operator == '*': return op1 * op2 elif operator == '/': return op1 / op2 ...

这是上题的代码：def infix_to_postfix(expression): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} op_stack = [] postfix_list = [] token_list = expression.split() for token in token_list: if token.isalnum(): postfix_list.append(token) elif token == '(': op_stack.append(token) elif token == ')': top_token = op_stack.pop() while top_token != '(': postfix_list.append(top_token) top_token = op_stack.pop() else: # operator while op_stack and precedence[op_stack[-1]] >= precedence[token]: postfix_list.append(op_stack.pop()) op_stack.append(token) while op_stack: postfix_list.append(op_stack.pop()) return ' '.join(postfix_list) class Node: def init(self, value): self.value = value self.left_child = None self.right_child = None def build_expression_tree(postfix_expr): operator_stack = [] token_list = postfix_expr.split() for token in token_list: if token.isalnum(): node = Node(token) operator_stack.append(node) else: right_node = operator_stack.pop() left_node = operator_stack.pop() node = Node(token) node.left_child = left_node node.right_child = right_node operator_stack.append(node) return operator_stack.pop() def evaluate_expression_tree(node, variable_values): if node.value.isalnum(): return variable_values[node.value] else: left_value = evaluate_expression_tree(node.left_child, variable_values) right_value = evaluate_expression_tree(node.right_child, variable_values) if node.value == '!': return not left_value elif node.value == '&': return left_value and right_value elif node.value == '|': return left_value or right_value expression = "!a & (b | c)" postfix_expression = infix_to_postfix(expression) expression_tree = build_expression_tree(postfix_expression) variable_values = {'a': True, 'b': False, 'c': True} result = evaluate_expression_tree(expression_tree, variable_values) print(result)

def infix_to_postfix(expression): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} op_stack = [] postfix_list = [] token_list = expression.split() for token in token_list: if token.isalnum(): ...

class TreeNode: def init(self, val=None, left=None, right=None): self.val = val self.left = left self.right = right def infix_to_postfix(infix): operators = {'(': 0, ')': 0, 'NOT': 1, 'AND': 2, 'OR': 3} stack = [] postfix = [] for token in infix: if token in operators: if token == '(': stack.append(token) elif token == ')': while stack[-1] != '(': postfix.append(stack.pop()) stack.pop() else: while stack and operators[stack[-1]] >= operators[token]: postfix.append(stack.pop()) stack.append(token) else: postfix.append(token) while stack: postfix.append(stack.pop()) return postfix def postfix_to_tree(postfix): stack = [] for token in postfix: if token in {'NOT', 'AND', 'OR'}: right = stack.pop() if token == 'NOT': stack.append(TreeNode('NOT', None, right)) else: left = stack.pop() stack.append(TreeNode(token, left, right)) else: stack.append(TreeNode(token)) return stack.pop() def evaluate(root, values): if root.val in values: return values[root.val] elif root.val == 'NOT': return not evaluate(root.right, values) elif root.val == 'AND': return evaluate(root.left, values) and evaluate(root.right, values) elif root.val == 'OR': return evaluate(root.left, values) or evaluate(root.right, values) def print_tree(root, level=0): if root: print_tree(root.right, level + 1) print(' ' * 4 * level + '->', root.val) print_tree(root.left, level + 1) infix = input('请输入命题演算公式：').split() postfix = infix_to_postfix(infix) root = postfix_to_tree(postfix) print('后缀表达式：', postfix) print('二叉树构造过程：') print_tree(root) print('真值表：') variables = list(set(filter(lambda x: x not in {'NOT', 'AND', 'OR'}, infix))) for values in itertools.product([True, False], repeat=len(variables)): values = dict(zip(variables, values)) result = evaluate(root, values) print(values, '->', result)其中有错误NameError: name 'itertools' is not defined。请修改

def infix_to_postfix(infix): operators = {'(': 0, ')': 0, 'NOT': 1, 'AND': 2, 'OR': 3} stack = [] postfix = [] for token in infix: if token in operators: if token == '(': stack.append(token) ...

AttributeError: 'Stack' object has no attribute 'infix_to_postfix'

这个错误意味着在Stack对象上调用了名为infix_to_postfix的属性，但该属性在Stack对象中不存在。这可能是因为你正在使用一个自定义的Stack类，但是在该类中没有定义infix_to_postfix方法。请检查你的代码，确保你...

根据以下代码：class Node: def init(self, value): self.value = value self.left = None self.right = None def is_operator(c): return c in ['&', '|', '!'] def infix_to_postfix(infix): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} stack = [] postfix = [] for c in infix: if c.isalpha(): postfix.append(c) elif c == '(': stack.append(c) elif c == ')': while stack and stack[-1] != '(': postfix.append(stack.pop()) stack.pop() elif is_operator(c): while stack and precedence[c] <= precedence.get(stack[-1], 0): postfix.append(stack.pop()) stack.append(c) while stack: postfix.append(stack.pop()) return postfix def build_tree(postfix): stack = [] for c in postfix: if c.isalpha(): node = Node(c) stack.append(node) elif is_operator(c): node = Node(c) node.right = stack.pop() node.left = stack.pop() stack.append(node) return stack[-1] def evaluate(node, values): if node.value.isalpha(): return values[node.value] elif node.value == '!': return not evaluate(node.right, values) elif node.value == '&': return evaluate(node.left, values) and evaluate(node.right, values) elif node.value == '|': return evaluate(node.left, values) or evaluate(node.right, values) def calculate(formula, values): postfix = infix_to_postfix(formula) tree = build_tree(postfix) return evaluate(tree, values) 在该代码基础上，使用python语言，以菜单形式完成下面几个的输出：1.打印二叉树的构造过程；2.打印公式的后缀形式；3.二叉树的后序遍历序列；4.输入每个变量的值，计算并显示公式的真值，打印二叉树的评估过程；5.显示公式的真值表

def infix_to_postfix(infix): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} stack = [] postfix = [] for c in infix: if c.isalpha(): postfix.append(c) elif c == '(': stack.append(c) elif c == '...

def infix_evaluator(infix_expression: str): token_list = infix_expression.split() print(token_list) pre_dict = {'': 2, '/': 2, '+': 1, '-': 1} ope_stack = [] num_stack = [] for i in token_list: if i.isdecimal() or '.' in i: num_stack.append(float(i)) elif i in '+-/': while ope_stack and pre_dict[ope_stack[-1]] >= pre_dict[i]: top = ope_stack.pop() op2 = num_stack.pop() op1 = num_stack.pop() num_stack.append(get_value(top, op1, op2)) ope_stack.append(i) while ope_stack: top = ope_stack.pop() op2 = ope_stack.pop() op1 = ope_stack.pop() ope_stack.append(get_value(top, op1, op2)) return ope_stack[0] def get_value(operator, op1, op2): if operator == '+': return op1 + op2 elif operator == '-': return op1 - op2 elif operator == '': return op1 op2 elif operator == '/': return op1 / op2 print(infix_evaluator('2.5 - 3 * 4'))

这段代码实现了一个中缀表达式求值的函数。函数的输入是一个中缀表达式字符串 infix_expression，输出是表达式...在本例中，我们调用 infix_evaluator() 函数，传入参数 '2.5 - 3 * 4'，函数输出结果为 -9.5。

这段代码有什么问题class Node: def init(self, value): self.value = value self.left_child = None self.right_child = None def build_tree(postfix_expr): opstack = [] token_list = postfix_expr.split() for token in token_list: if token in "ABCDEFGHIJKLMNOPQRSTUVWXYZ" : opstack.append(Node(token)) else: right_child = opstack.pop() left_child = opstack.pop() new_node = Node(token) new_node.left_child = left_child new_node.right_child = right_child opstack.append(new_node) return opstack.pop() def evaluate_tree(tree, var_dict): if tree.left_child and tree.right_child: left_value = evaluate_tree(tree.left_child, var_dict) right_value = evaluate_tree(tree.right_child, var_dict) return evaluate_op(tree.value, left_value, right_value) else: return var_dict[tree.value] def evaluate_op(op, left_value, right_value): if op == "&": return left_value and right_value elif op == "|": return left_value or right_value else: return not left_value def calculate(expr, var_dict): postfix_expr = infix_to_postfix(expr) tree = build_tree(postfix_expr) result = evaluate_tree(tree, var_dict) return result

这段代码的问题在于没有提供 infix_to_postfix 函数的实现。因此，我们无法确定整个程序的正确性。另外，我们也无法确定 var_dict 字典包含的键和值，因为该字典在代码中没有被定义。需要注意的是，如果 var_dict ...

std::string infixToPostfix(const std::string& infix)

std::string infixToPostfix(const std::string& infix)是一个用于将中缀表达式转换为后缀表达式的函数。中缀表达式是我们通常使用的表达式形式，例如：2 + 3 * 4。而后缀表达式是一种更加方便计算机处理的表达式...

class Node: def init(self, value=None, left=None, right=None): self.value = value self.left = left self.right = right class Stack: def init(self): self.items = [] def push(self, item): self.items.append(item) def pop(self): return self.items.pop() def peek(self): return self.items[-1] def is_empty(self): return len(self.items) == 0 def infix_to_postfix(infix): precedence = {'(': 0, 'AND': 1, 'OR': 1, 'NOT': 2} # 运算符优先级 postfix = [] stack = Stack() tokens = infix.split() for token in tokens: if token.isalnum(): postfix.append(token) elif token == '(': stack.push(token) elif token == ')': while stack.peek() != '(': postfix.append(stack.pop()) stack.pop() else: while not stack.is_empty() and precedence[stack.peek()] >= precedence[token]: postfix.append(stack.pop()) stack.push(token) while not stack.is_empty(): postfix.append(stack.pop()) return postfix def build_tree(postfix): stack = Stack() for token in postfix: if token.isalnum(): stack.push(Node(token)) else: right = stack.pop() left = stack.pop() stack.push(Node(token, left, right)) return stack.pop() def evaluate(node, values): if node.value.isalnum(): return values[node.value] else: left_value = evaluate(node.left, values) right_value = evaluate(node.right, values) if node.value == 'AND': return left_value and right_value elif node.value == 'OR': return left_value or right_value else: return not right_value def print_tree(node, indent=0): if node: print(' ' * indent + node.value) print_tree(node.left, indent + 2) print_tree(node.right, indent + 2) infix = 'A AND (B OR C) AND NOT D' postfix = infix_to_postfix(infix) print('Infix:', infix) print('Postfix:', postfix) tree = build_tree(postfix) print('Tree:') print_tree(tree) values = {'A': True, 'B': False, 'C': True, 'D': True} result = evaluate(tree, values) print('Result:', result)一句一句解释这段代码

infix_to_postfix 函数接收一个中缀表达式 infix，通过运算符优先级将其转换为后缀表达式 postfix，并返回后缀表达式的列表。 build_tree 函数接收一个后缀表达式 postfix，通过栈的操作构建二叉树，并返回根节点。...

def infix_to_postfix(infix): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} postfix = '' stack = Stack() for ch in infix: if ch == '(': stack.push(ch) elif ch == ')': while stack.top() != '(': postfix += stack.pop() stack.pop() elif ch in precedence: while stack.is_empty() == False and stack.top() != '(' and precedence[ch] <= precedence[stack.top()]: postfix += stack.pop() stack.push(ch) else: postfix += ch while stack.is_empty() == False: postfix += stack.pop() return postfix解释这段代码

这段代码是一个中缀表达式转后缀表达式的实现。它使用了一个栈来辅助转换过程。具体来说，它遍历中缀表达式的每个字符，如果是左括号，就将它压入栈中；如果是右括号，则将栈中的元素弹出并加入后缀表达式中，直到...

用栈代码实现步骤1：将中缀表达式转换为后缀表达式步骤2：计算中缀表达式

def infix_to_postfix(expression): stack = [] postfix_expression = [] for char in expression: if char.isdigit(): postfix_expression.append(char) elif char == '(': stack.append(char) elif char =...

Accepted 2.05215ms 404KiB foo.cc: In function 'std::string infixToPostfix(std::string)': foo.cc:19:23: warning: comparison of integer expressions of different signedness: 'int' and 'std::__cxx11::basic_string<char>::size_type' {aka 'long unsigned int'} [-Wsign-compare] 19 | for (int i = 0; i < expression.length(); i++) { | ^~~~~~~~~~~~~~~~~~~

string postfix = infixToPostfix(expression); cout << postfix ; return 0; } 通过将 i 的类型更改为 std::__cxx11::basic_string<char>::size_type，可以消除警告。请再次运行代码，看看警告是否...

第1关：基于栈的中缀算术表达式求值

def infix_to_postfix(expression): # 优先级字典 priority = {'+': 1, '-': 1, '*': 2, '/': 2, '(': 0, ')': 0} # 初始化栈 stack = [] # 初始化后缀表达式 postfix = '' # 遍历中缀表达式 for item in ...

实现简单算术表示式到抽象语法树的转换。具体要求如下：（1）用户任意输入一个正确的算术表示式，程序将其转换成抽象语法树输出；（2）输入以#结束的中缀表达式（包括+-/(）数字)，例如：x=a+bc；（3）结果需以GUI界面展示。

def infix_to_postfix(infix_expr): # 定义运算符优先级 prec = {'*': 3, '/': 3, '+': 2, '-': 2, '(': 1} # 定义一个栈来存储运算符 op_stack = [] # 定义一个列表来存储后缀表达式 postfix_list = [] # 将...

TypeError: infix_to_postfix() missing 1 required positional argument: 'infix'

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TypeError: infix_to_postfix() missing 1 required positional argument: 'infix'

用栈代码实现步骤1：将中缀表达式转换为后缀表达式步骤2：计算中缀表达式