def is_unordered_exact_match()

in utils/trees.py [0:0]

• 23 lines of code
• 12 McCabe index (conditional complexity)

    def is_unordered_exact_match(self, otree):
        '''
        This method is used to check if `otree`, a tree object, is an exact match 
        with the tree object calling this method.

        :param otree: (TopSemanticTree/ExpressSemanticTree) Tree object to compare with.
        :return: (bool) If the two trees have an unordered exact match or not.  
        '''
        if not isinstance(otree, type(self)):
            raise TypeError(f"Expected both trees to be of type {type(self)} "
                f"but one of them is of type {type(otree)}")
        # check if the roots of trees are same. 
        # If not, its not an exact match.
        if self.root_symbol() != otree.root_symbol():
            return False
        # Check if the number of children of both trees are same.
        # If not, its not an exact match.
        if len(self.children()) != len(otree.children()):
            return False
        # Count the number of equal sub-trees, i.e. children
        matched_subtrees_idx_1, matched_subtrees_idx_2 = set(), set()
        for i,child_1 in enumerate(self.children()):
            for j,child_2 in enumerate(otree.children()):
                if (i not in matched_subtrees_idx_1) and \
                    (j not in matched_subtrees_idx_2):
                    if child_1.is_unordered_exact_match(child_2):
                        matched_subtrees_idx_1.add(i)
                        matched_subtrees_idx_2.add(j)
        # If the no. of matched sub-trees are equal to the 
        # number of children itself, that means all childdren are a match
        # and hence its an exact match, otherwise not. 
        return len(matched_subtrees_idx_1) == len(self.children())