infix-notation

问题 I have been looking without much luck for an implementation of Python that converts infix to prefix that ranges on a sufficient amount of arithmetic and logic operators and care about its properties on a good python implementation. More specifically I am interested on the operators that would appear on a conditional clause of a C program . (e.g. it would transform a > 0 && b > 1 in prefix. Since I am still newbie to Python I would appreciate if anyone could offer me the implementation or some

问题 I already found different solutions on Stackoverflow, but there were some things I didn´t understand. Whats the best method to calculate the Output of e.g.: ((1+(4*(2+3)))+((2+3)*(4*5))) ? My method looks as following, but I know there are lots of mistakes in it: public static int ComputeInfix(string infix) { Stack<char> operatorstack = new Stack<char>(); Stack<int> operandstack = new Stack<int>(); for(int j = 0; j < infix.Length; j++) { char c = infix[j]; if (c => 0 && c <= 9) { operandstack

问题 My lecturer gave me an assignment to create a program to convert an infix expression to postfix using Stack. I've made the stack classes and some functions to read the infix expression. But this one function, called inToPos(char string[]) which is responsible to convert the inFix expression in the string inFix to the post fix expression in the string postFix using stacks, is creating a breakpoint. Can you guys help me out and tell me what I'm doing wrong? These are my codes that badly needs

问题 Methods taking a single argument can be written as an infix operators in Scal. I.e. adding *(other:C) = foo(this, other) to class C, will allow us to write c1 * c2 instead of foo(c1,c2). But is there a way to define infix operators on existing classes that you cannot modify? E.g. if I wanted to write c1 + c2 instead of xor(c1,c2) , where c1,c2:Array[Byte] , I obviously cannot modify the Array-Class. I found this and tried implicit class Bytearray(a1:Array[Byte]) extends Anyval { def +(a2

问题 I am looking for algorithm postfix to infix notation which will produce the minimum number of the parentheses. I have found that but it will produce many, many parentheses: http://tajendrasengar.blogspot.com/2011/09/postfix-to-infix-algorithm.html For example The input: <ONP>abcd*/+~ The result: <INF>~(a+b/(c*d)) 回答1: What you need to do if you really want as few parentheses as possible, is similar to what the algorithm you linked to says. However... You should store an operator for each

问题 I'm trying to replicate, roughly, the dplyr package from R using Python/Pandas (as a learning exercise). Something I'm stuck on is the "piping" functionality. In R/dplyr, this is done using the pipe-operator %>% , where x %>% f(y) is equivalent to f(x, y) . If possible, I would like to replicate this using infix syntax (see here). To illustrate, consider the two functions below. import pandas as pd def select(df, *args): cols = [x for x in args] df = df[cols] return df def rename(df, **kwargs

问题 I want to know if there is a way to solve infix expressions in a single pass using 2 stacks? The stacks can be one for operator and the other for operands... The standard way to solve by shunt-yard algorithm is to convert the infix expression to postfix(reverse polish) and then solve. I don't want to convert the expression first to postfix. If the expression is like 2*3-(6+5)+8 , how to solve? 回答1: Quite late, but here is the answer. Take two stacks: operator stack { for operators and

问题 I want to create an expression tree given expression in infix form. Is it necessary to convert the expression to postfix first and then create the tree? I understand that it somehow depends on the problem itself. But assume that it is simple expression of mathematical function with unknowns and operators like: / * ^ + -. 回答1: No. If you're going to build an expression tree, then it's not necessary to convert the expression to postfix first. It will be simpler just to build the expression tree

问题 I'm currently working on a project to convert from postfix to infix using a stack in the form of a singly linked list. I've managed to convert expressions such as ab+ to (a+b) however when the expression gets longer such as ab+cd*- . It doesn't work. I'm considering pushing the previously converted expression back onto the stack however the stack is of type char and the expression is a string and it complains when I try to push it back. Should I make it a template and if so how would I do it

问题 My task is to evaluate a fully parenthesized infix expression using a stack. A Stack class has been written for me and I must not change or modify the Stack class. Here are the step by step directions for how to evaluate the infix expression: Just scan the expression from left to right. If it is anything other than a ), push it onto the stack. When you encounter a ), pop from the stack 4 times, do the math and push the value onto the stack. At the end you will have just one value in the stack