来自 ANTLR 解析树的 Python AST?

标签 python antlr4 abstract-syntax-tree parse-tree

我找到了一个 ANTLRv4 Python3 grammer , 但它会生成一个解析树,它通常有很多无用的节点。

我正在寻找一个已知的包来从该解析树中获取 Python AST。

有这样的东西吗?

编辑:关于使用 Python ast 包的说明:我的项目是用 Java 编写的,我需要解析 Python 文件。

编辑 2:“AST”是指 http://docs.python.org/2/library/ast.html#abstract-grammar ,而“解析树”是指 http://docs.python.org/2/reference/grammar.html .

最佳答案

以下可能是一个开始:

public class AST {

    private final Object payload;

    private final List<AST> children;

    public AST(ParseTree tree) {
        this(null, tree);
    }

    private AST(AST ast, ParseTree tree) {
        this(ast, tree, new ArrayList<AST>());
    }

    private AST(AST parent, ParseTree tree, List<AST> children) {

        this.payload = getPayload(tree);
        this.children = children;

        if (parent == null) {
            walk(tree, this);
        }
        else {
            parent.children.add(this);
        }
    }

    public Object getPayload() {
        return payload;
    }

    public List<AST> getChildren() {
        return new ArrayList<>(children);
    }

    private Object getPayload(ParseTree tree) {
        if (tree.getChildCount() == 0) {
            return tree.getPayload();
        }
        else {
            String ruleName = tree.getClass().getSimpleName().replace("Context", "");
            return Character.toLowerCase(ruleName.charAt(0)) + ruleName.substring(1);
        }
    }

    private static void walk(ParseTree tree, AST ast) {

        if (tree.getChildCount() == 0) {
            new AST(ast, tree);
        }
        else if (tree.getChildCount() == 1) {
            walk(tree.getChild(0), ast);
        }
        else if (tree.getChildCount() > 1) {

            for (int i = 0; i < tree.getChildCount(); i++) {

                AST temp = new AST(ast, tree.getChild(i));

                if (!(temp.payload instanceof Token)) {
                    walk(tree.getChild(i), temp);
                }
            }
        }
    }

    @Override
    public String toString() {

        StringBuilder builder = new StringBuilder();

        AST ast = this;
        List<AST> firstStack = new ArrayList<>();
        firstStack.add(ast);

        List<List<AST>> childListStack = new ArrayList<>();
        childListStack.add(firstStack);

        while (!childListStack.isEmpty()) {

            List<AST> childStack = childListStack.get(childListStack.size() - 1);

            if (childStack.isEmpty()) {
                childListStack.remove(childListStack.size() - 1);
            }
            else {
                ast = childStack.remove(0);
                String caption;

                if (ast.payload instanceof Token) {
                    Token token = (Token) ast.payload;
                    caption = String.format("TOKEN[type: %s, text: %s]",
                            token.getType(), token.getText().replace("\n", "\\n"));
                }
                else {
                    caption = String.valueOf(ast.payload);
                }

                String indent = "";

                for (int i = 0; i < childListStack.size() - 1; i++) {
                    indent += (childListStack.get(i).size() > 0) ? "|  " : "   ";
                }

                builder.append(indent)
                        .append(childStack.isEmpty() ? "'- " : "|- ")
                        .append(caption)
                        .append("\n");

                if (ast.children.size() > 0) {
                    List<AST> children = new ArrayList<>();
                    for (int i = 0; i < ast.children.size(); i++) {
                        children.add(ast.children.get(i));
                    }
                    childListStack.add(children);
                }
            }
        }

        return builder.toString();
    }
}

并可用于为输入 "f(arg1='1')\n" 创建 AST,如下所示:

public static void main(String[] args) {

    Python3Lexer lexer = new Python3Lexer(new ANTLRInputStream("f(arg1='1')\n"));
    Python3Parser parser = new Python3Parser(new CommonTokenStream(lexer));

    ParseTree tree = parser.file_input();
    AST ast = new AST(tree);

    System.out.println(ast);
}

这将打印:

'- file_input
   |- stmt
   |  |- small_stmt
   |  |  |- atom
   |  |  |  '- TOKEN[type: 35, text: f]
   |  |  '- trailer
   |  |     |- TOKEN[type: 47, text: (]
   |  |     |- arglist
   |  |     |  |- test
   |  |     |  |  '- TOKEN[type: 35, text: arg1]
   |  |     |  |- TOKEN[type: 53, text: =]
   |  |     |  '- test
   |  |     |     '- TOKEN[type: 36, text: '1']
   |  |     '- TOKEN[type: 48, text: )]
   |  '- TOKEN[type: 34, text: \n]
   '- TOKEN[type: -1, text: ]

我意识到这仍然包含您可能不想要的节点,但您甚至可以添加一组您想要排除的标记类型。随意破解!

Here is a Gist包含上面代码的一个版本,带有正确的导入语句和一些 JavaDocs 和内联注释。

关于来自 ANTLR 解析树的 Python AST?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24766537/

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