I have a basic function to parse a lisp expression. It's using a while
loop, but as an exercise I'd like to convert it into a recursive function. However, it's a bit tricky for me to do. Here is what I have thus far:
def build_ast(self, tokens=None):
# next two lines example input to make self-contained
LEFT_PAREN, RIGHT_PAREN = '(', ')'
tokens = ['(', '+', '2', '(', '*', '3', '4', ')', ')']
while RIGHT_PAREN in tokens:
right_idx = tokens.index(RIGHT_PAREN)
left_idx = right_idx - tokens[:right_idx][::-1].index(LEFT_PAREN)-1
extraction = [tokens[left_idx+1:right_idx],]
tokens = tokens[:left_idx] + extraction + tokens[right_idx+1:]
ast = tokens
return ast
And so it would parse something like this:
(+ 2 (* 3 4))
Into this:
[['+', '2', ['*', '3', '4']]]
What would be an example of how I could make the above function recursive? So far I've started with something like:
def build_ast(self, ast=None):
if ast is None: ast=self.lexed_tokens
if RIGHT_PAREN not in ast:
return ast
else:
right_idx = ast.index(RIGHT_PAREN)
left_idx = right_idx - ast[:right_idx][::-1].index(LEFT_PAREN)-1
ast = ast[:left_idx] + [ast[left_idx+1:right_idx],] + ast[right_idx+1:]
return self.build_ast(ast)
But it just comes across as a bit strange (as if the recursion isn't helpful here). What would be a better way to construct this? Or perhaps a better/more elegant algorithm to build this simple ast?
You can use a recursive generator function:
def _build_ast(tokens):
LEFT_PAREN, RIGHT_PAREN = '(', ')'
#consume the iterator until it is empty or a right paren occurs
while (n:=next(tokens, None)) is not None and n != RIGHT_PAREN:
#recursively call _build_ast if we encounter a left paren
yield n if n != LEFT_PAREN else list(_build_ast(tokens))
def build_ast(tokens):
#pass tokens as an iterator to _build_ast
return list(_build_ast(iter(tokens)))
tokens = ['(', '+', '2', '(', '*', '3', '4', ')', ')']
print(build_ast(tokens))
Output:
[['+', '2', ['*', '3', '4']]]