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    <p>I ended up implementing most of this as an exercise.</p> <p>First, find the number of possible locations to split: the number of non-function nodes.</p> <pre><code>def count(obj): total = 0 for o in obj[1:]: # Add the node itself. total += 1 if isinstance(o, list): total += count(o) return total </code></pre> <p>Then, a helper: given an index in the above range, figure out where it is.</p> <pre><code>def find_idx(tree, idx): """ Return the node containing the idx'th function parameter, and the index of that parameter. If the tree contains fewer than idx parameters, return (None, None). """ if not isinstance(idx, list): # Stash this in a list, so recursive calls share the same value. idx = [idx] for i, o in enumerate(tree): # Skip the function itself. if i == 0: continue if idx[0] == 0: return tree, i idx[0] -= 1 if isinstance(o, list): container, result_index = find_idx(o, idx) if container is not None: return container, result_index return None, None </code></pre> <p>Doing the swap is pretty simple now:</p> <pre><code>def random_swap(tree1, tree2): from random import randrange pos_in_1 = randrange(0, count(tree1)) pos_in_2 = randrange(0, count(tree2)) parent1, idx1 = find_idx(tree1, pos_in_1) parent2, idx2 = find_idx(tree2, pos_in_2) # Swap: parent1[idx1], parent2[idx2] = parent2[idx2], parent1[idx1] c = 1 tree1 = ["f:2", c, ["f:1", c]] tree2 = ["f:2", ["f:2", ["f:2", c, c], ["f:2", c, c]], ["f:3", ["f:4", c, c, c, c], ["f:2", c, c], c]] while True: random_swap(tree1, tree2) print tree1 print tree2 </code></pre> <p>This doesn't implement a max depth, but it's a start.</p> <p>This will also never replace the root node, where a node in tree1 becomes the new tree2 and all of tree2 becomes a node in tree1. A workaround would be to wrap the whole thing in eg. [lambda a: a, tree], so editable nodes always have a parent node.</p> <p>This isn't very efficient. Maintaining node counts could make it faster, but then you'd need to store a reference to the parent, too, in order to update the counts efficiently. If you go that route, you'll really want to find or implement a real tree class.</p>
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