感谢指导家伙们,至少我拿出了我自己的解决方案。由于这是我第一次做数学方程解析,请原谅我,如果我这样做是错误的或无效的,或者帮我找到这个错误:
基本上,这里是我做的步骤,这一点:
- 之前解析,总是验证模式。出现错误时抛出错误。
- 一旦经过验证,我们会为前缀符号转换做一个中缀表示法。此步骤要求“和”优先于“或”。
- 反转给定的模式
- 是否将中缀转换为后缀表示法转换。I dumb, I learn from this
- 做反向再次
- 中缀到前缀应该在这个阶段
- 构建从前缀符号,使得
- 节点总是有一棵树来完成,并且最大,有两个分支机构
- 遍历,直到它达到全叶
- 优化树,使得合并同类运营商一起(如多发性
$and
运营子$and
可以合并形成一个较短的树)
- 混合与设定的给定的标准,并且全部完成!
工作实例可以在这里找到:http://jsfiddle.net/chaoszcat/uGKYj/3/
工作如下代码:
(function() {
/**
* This is a source example of my original question on
* http://stackoverflow.com/questions/20986255/converting-conditional-equation-from-infix-to-prefix-notation
*
* This is my solution and use it at your own risk
* @author Lionel Chan <chaoszcat[at]gmail.com>
*/
/**
* isNumeric, from jQuery. Duplicated here to make this js code pure
* @param {mix} n Test subject
* @returns {boolean} true if it's numeric
*/
function isNumeric(n) {
return !isNaN(parseFloat(n))&&isFinite(n);
}
/**
* Node class - represent a operator or numeric node
* @param {string} token The token string, operator "and", "or", or numeric value
*/
function Node(token) {
this.parent = null;
this.children = []; //one node has two children at most
this.token = token;
this.is_operator = token === 'and' || token === 'or';
this.is_numeric = !this.is_operator;
this.destroyed = false;
}
Node.prototype = {
isOperator: function() { return this.is_operator;},
isNumeric: function() { return this.is_numeric;},
//While building tree, a node is full if there are two children
isFull: function() {
return this.children.length >= 2;
},
addChild: function(node) {
node.parent = this;
this.children.push(node);
},
hasParent: function() {
return this.parent !== null;
},
indexOfChild: function(node) {
for (var i = 0 ; i < this.children.length ; ++i) {
if (this.children[i] === node) {
return i;
}
}
return -1;
},
removeChild: function(node) {
var idx = this.indexOfChild(node);
if (idx >= 0) {
this.children[idx].parent = null; //remove parent relationship
this.children.splice(idx, 1); //splice it out
}
},
/**
* Pass my children to the target node, and destroy myself
*
* @param {Node} node A target node
*/
passChildrenTo: function(node) {
for (var i = 0 ; i < this.children.length ; ++i) {
node.addChild(this.children[i]);
}
this.destroy();
},
//Destroy this node
destroy: function() {
this.parent.removeChild(this);
this.children = null;
this.destroyed = true;
}
};
/**
* Tree class - node manipulation
* @param {array} prefixTokens The converted, prefix-notated tokens
*/
function Tree(prefixTokens) {
this.buildTree(prefixTokens);
//Optimize tree - so that the tree will merge multiple similar operators together
this.optimize(this.root);
}
Tree.prototype = {
root: null,
//Reference to the deepest operator node in the tree for next attachment point
deepestNode: null,
/**
* Render this tree with given criteria array
* @param {array} crits
* @returns {object} The built criteria
*/
render: function(crits) {
//After optimization, we build the criteria and that's all!
return this.buildCriteria(this.root, crits);
},
/**
* Build criteria from root node. Recursive
*
* @param {Node} node
* @param {array} crits
* @returns {object} of criteria
*/
buildCriteria: function(node, crits) {
var output = {},
label = '$'+node.token;
output[label] = []; //cpnditions array
for (var i = 0 ; i < node.children.length ; ++i) {
if (node.children[i].isOperator()) {
output[label].push(this.buildCriteria(node.children[i], crits));
}else{
output[label].push(crits[node.children[i].token-1]);
}
}
return output;
},
/**
* Optimize the tree, we can simplify nodes with same operator. Recursive
*
* @param {Node} node
* @void
*/
optimize: function(node) {
//note that node.children.length will keep changing since the swapping children will occur midway. Rescan is required
for (var i = 0 ; i < node.children.length ; ++i) {
if (node.children[i].isOperator()) {
this.optimize(node.children[i]);
if (node.children[i].token === node.token) {
node.children[i].passChildrenTo(node);
i = 0; //rescan this level whenever a swap occured
}
}
}
},
/**
* Build tree from raw tokens
* @param {array} tokens
*/
buildTree: function(tokens) {
for (var i = 0 ; i < tokens.length ; ++i) {
this.addNode(new Node(tokens[i]));
}
},
/**
* Add node into tree
*
* @param {Node} node
*/
addNode: function(node) {
//If no root? The first node is root
if (this.root === null) {
this.root = node;
this.deepestNode = node;
return;
}
//if deepestNode is full, traverse up until we find a node with capacity
while(this.deepestNode && this.deepestNode.isFull()) {
this.deepestNode = this.deepestNode.parent;
}
if (this.deepestNode) {
this.deepestNode.addChild(node);
}
//If the current node is an operator, we move the deepestNode cursor to it
if (node.isOperator()) {
this.deepestNode = node;
}
}
};
/**
* Main criteria parser
*/
var CriteriaParser = {
/**
* Convert raw string of pattern (1 and 2 or 3) into the object of criteria pattern
*
* @param {string} str The raw pattern
* @param {array} crits The raw list of criteria
* @returns {String|Boolean}
*/
parse: function(str, crits) {
var tokens = this.tokenize(str),
validationResult = this.validate(tokens, crits),
prefixNotation = '';
//Once succeded, we proceed to convert it to prefix notation
if (validationResult === true) {
prefixNotation = this.infixToPrefix(tokens);
return (new Tree(prefixNotation)).render(crits);
}else{
return validationResult;
}
},
/**
* Convert the infix notation of the pattern (1 and 2 or 3) into prefix notation "or and 1 2 3"
*
* Note:
* - and has higher precedence than or
*
* Steps:
* 1. Reverse the tokens array
* 2. Do infix -> postfix conversion (http://www.cs.arizona.edu/classes/cs227/spring12/infix.pdf, http://scriptasylum.com/tutorials/infix_postfix/algorithms/infix-postfix/index.htm)
* 3. Reverse the result
*
* @param {array} tokens The tokenized tokens
* @returns {array} prefix notation of pattern
*/
infixToPrefix: function(tokens) {
var reversedTokens = tokens.slice(0).reverse(), //slice to clone, so not to disturb the original array
stack = [],
output = [];
//And since it's reversed, please regard "(" as closing bracket, and ")" as opening bracket
do {
var stackTop = stack.length > 0 ? stack[stack.length-1] : null,
token = reversedTokens.shift();
if (token === 'and') {
while(stackTop === 'and') {
output.push(stack.pop());
stackTop = stack.length > 0 ? stack[stack.length-1] : null;
}
stack.push(token);
stackTop = token;
}else if (token === 'or') {
while(stackTop === 'and' || stackTop === 'or') { //and has higher precedence, so it will be popped out
output.push(stack.pop());
stackTop = stack.length > 0 ? stack[stack.length-1] : null;
}
stack.push(token);
stackTop = token;
}else if (token === '(') { //'(' is closing bracket in reversed tokens
while(stackTop !== ')' && stackTop !== undefined) { //keep looping until found a "open -)" bracket
output.push(stack.pop());
stackTop = stack.length > 0 ? stack[stack.length-1] : null;
}
stack.pop(); //remove the open ")" bracket
stackTop = stack.length > 0 ? stack[stack.length-1] : null;
}else if (token === ')') { //')' is opening bracket in reversed tokens
stack.push(token);
}else if (isNumeric(token)) {
output.push(token);
}else if (token === undefined) {
// no more tokens. Just shift everything out from stack
while(stack.length) {
stackTop = stack.pop();
if (stackTop !== undefined && stackTop !== ')') {
output.push(stackTop);
}
}
}
}while(stack.length || reversedTokens.length);
//Reverse output and we are done
return output.reverse();
},
/**
* Tokenized the provided pattern
* @param {string} str The raw pattern from user
* @returns {array} A tokenized array
*/
tokenize: function(str) {
var pattern = str.replace(/\s/g, ''), //remove all the spaces :) not needed
tokens = pattern.split(''),
tokenized = [];
//Tokenize it and verify
var token = null,
next = null;
//attempts to concatenate the "and" and "or" and numerics
while (tokens.length > 0) {
token = tokens.shift();
next = tokens.length > 0 ? tokens[0] : null;
if (token === '(' || token === ')') {
tokenized.push(token);
}else if (token === 'a' && tokens.length >= 2 && tokens[0] === 'n' && tokens[1] === 'd') { //and
tokenized.push(token + tokens.shift() + tokens.shift());
}else if (token === 'o' && tokens.length >= 1 && next === 'r') { //or
tokenized.push(token + tokens.shift());
}else if (isNumeric(token)) {
while(isNumeric(next)) {
token += next;
tokens.shift(); //exhaust it
next = tokens.length > 0 ? tokens[0] : null;
}
tokenized.push(token);
}else{
tokenized.push(token);
}
}
return tokenized;
},
/**
* Attempt to validate tokenized tokens
*
* @param {array} tokens The tokenized tokens
* @param {array} crits The user provided criteria set
* @returns {Boolean|String} Returns boolean true if succeeded, string if error occured
*/
validate: function(tokens, crits) {
var valid = true,
token = null,
stack = [],
nextToken = null,
criteria_count = crits.length;
for (var i = 0 ; i < tokens.length ; ++i) {
token = tokens[i];
nextToken = i < tokens.length - 1 ? tokens[i+1] : null;
if (token === '(') {
stack.push('(');
if (!isNumeric(nextToken) && nextToken !== '(' && nextToken !== ')') {
throw 'Unexpected token "'+nextToken+'"';
}
}else if (token === ')') {
if (stack.length > 0) {
stack.pop();
}else{
throw 'Unexpected closing bracket';
}
if (nextToken !== ')' && nextToken !== 'and' && nextToken !== 'or' && nextToken !== null) {
throw 'Unexpected token "'+nextToken+'"';
}
}else if (token === 'and' || token === 'or') {
if (!isNumeric(nextToken) && nextToken !== '(') {
throw 'Unexpected token "'+nextToken+'"';
}
}else if (isNumeric(token) && token <= criteria_count) {
if (nextToken !== ')' && nextToken !== 'and' && nextToken !== 'or') {
throw 'Unexpected token "'+nextToken+'"';
}
}else{
//anything not recognized, die.
throw 'Unexpected token "'+token+'"';
}
}
//Last step - check if we have all brackets closed
if (valid && stack.length > 0) {
throw 'Missing '+stack.length+' closing bracket';
}
return valid;
}
};
//This is an example pattern and criteria set. Note that pattern numbers must match criteria numbers.
var pattern = '((1 or 3) and (2 or 4) or 5)',
crits = [
1, 2, 3, 4, 5
];
//lazy on the document on load. Just delay
setTimeout(function() {
var result;
try {
result = JSON.stringify(CriteriaParser.parse(pattern, crits), undefined, 4);
}catch(e) {
result = e;
}
var pre = document.createElement('pre');
pre.innerHTML = result;
document.body.appendChild(pre);
}, 10);
})();
http://blog.reverberate.org/2013/07/ll-and-lr-parsing -demystified.html – zerkms
谢谢@zerkms!我肯定在研究中需要这个:) –
PS:我认为你的第二种情况下的AST缺少第二个'$或'标记。它应该是这样的:http://sketchia.com/draw_G2HXgRv.html PPS:疯狂的绘画技巧,我知道) – zerkms