Beispiel #1
0
def last_openleaf(request, project_id=None, skeleton_id=None):
    """ Return the ID of the nearest node (or itself), and its location string;
    or two nulls if none found. """
    tnid = int(request.POST['tnid'])
    cursor = connection.cursor()

    # Select all nodes and their tags
    cursor.execute('''
    SELECT t.id, t.parent_id, t.location, ci.name
    FROM treenode t LEFT OUTER JOIN (treenode_class_instance tci INNER JOIN class_instance ci ON tci.class_instance_id = ci.id) ON t.id = tci.treenode_id
    WHERE t.skeleton_id = %s
    ''' % int(skeleton_id))

    # Some entries repeated, when a node has more than one tag
    # Create a graph with edges from parent to child, and accumulate parents
    tree = nx.DiGraph()
    for row in cursor.fetchall():
        nodeID = row[0]
        if row[1]:
            # It is ok to add edges that already exist: DiGraph doesn't keep duplicates
            tree.add_edge(row[1], nodeID)
        else:
            tree.add_node(nodeID)
        tree.node[nodeID]['loc'] = row[2]
        if row[3]:
            props = tree.node[nodeID]
            tags = props.get('tags')
            if tags:
                tags.append(row[3])
            else:
                props['tags'] = [row[3]]

    if tnid not in tree:
        raise Exception("Could not find %s in skeleton %s" % (tnid, int(skeleton_id)))

    reroot(tree, tnid)
    distances = edge_count_to_root(tree, root_node=tnid)

    # Iterate end nodes, find closest
    nearest = None
    distance = tree.number_of_nodes() + 1
    loc = None
    other_tags = set(('uncertain continuation', 'not a branch', 'soma'))

    for nodeID, out_degree in tree.out_degree_iter():
        if 0 == out_degree:
            # Found an end node
            props = tree.node[nodeID]
            # Check if not tagged with a tag containing 'end'
            if not 'tags' in props and not [s for s in props if 'end' in s or s in other_tags]:
                # Found an open end
                d = distances[nodeID]
                if d < distance:
                    nearest = nodeID
                    distance = d
                    loc = props['loc']

    return HttpResponse(json.dumps((nearest, loc)))
Beispiel #2
0
def last_openleaf(request, project_id=None, skeleton_id=None):
    """ Return the ID of the nearest node (or itself), and its location string;
    or two nulls if none found. """
    tnid = int(request.POST['tnid'])
    cursor = connection.cursor()

    # Select all nodes and their tags
    cursor.execute('''
    SELECT t.id, t.parent_id, t.location, ci.name
    FROM treenode t LEFT OUTER JOIN (treenode_class_instance tci INNER JOIN class_instance ci ON tci.class_instance_id = ci.id) ON t.id = tci.treenode_id
    WHERE t.skeleton_id = %s
    ''' % int(skeleton_id))

    # Some entries repeated, when a node has more than one tag
    # Create a graph with edges from parent to child, and accumulate parents
    tree = nx.DiGraph()
    for row in cursor.fetchall():
        nodeID = row[0]
        if row[1]:
            # It is ok to add edges that already exist: DiGraph doesn't keep duplicates
            tree.add_edge(row[1], nodeID)
        else:
            tree.add_node(nodeID)
        tree.node[nodeID]['loc'] = row[2]
        if row[3]:
            props = tree.node[nodeID]
            tags = props.get('tags')
            if tags:
                tags.append(row[3])
            else:
                props['tags'] = [row[3]]

    if tnid not in tree:
        raise Exception("Could not find %s in skeleton %s" % (tnid, int(skeleton_id)))

    reroot(tree, tnid)
    distances = edge_count_to_root(tree, root_node=tnid)

    # Iterate end nodes, find closest
    nearest = None
    distance = tree.number_of_nodes() + 1
    loc = None
    other_tags = set(('uncertain continuation', 'not a branch', 'soma'))

    for nodeID, out_degree in tree.out_degree_iter():
        if 0 == out_degree:
            # Found an end node
            props = tree.node[nodeID]
            # Check if not tagged with a tag containing 'end'
            if not 'tags' in props and not [s for s in props if 'end' in s or s in other_tags]:
                # Found an open end
                d = distances[nodeID]
                if d < distance:
                    nearest = nodeID
                    distance = d
                    loc = props['loc']

    return HttpResponse(json.dumps((nearest, loc)))
Beispiel #3
0
def _node_centrality_by_synapse(tree, nodes, totalOutputs, totalInputs):
    """ tree: a DiGraph
        nodes: a dictionary of treenode ID vs Counts instance
        totalOutputs: the total number of output synapses of the tree
        totalInputs: the total number of input synapses of the tree
        Returns nothing, the results are an update to the Counts instance of each treenode entry in nodes, namely the nPossibleIOPaths. """
    # 1. Ensure the root is an end by checking that it has only one child; otherwise reroot at the first end node found

    if 0 == totalOutputs:
        # Not computable
        for counts in nodes.itervalues():
            counts.synapse_centrality = -1
        return

    if len(tree.successors(find_root(tree))) > 1:
        # Reroot at the first end node found
        tree = tree.copy()
        endNode = (nodeID for nodeID in nodes.iterkeys() if not tree.successors(nodeID)).next()
        reroot(tree, endNode)

    # 2. Partition into sequences, sorted from small to large
    sequences = sorted(partition(tree), key=len)

    # 3. Traverse all partitions counting synapses seen
    for seq in sequences:
        # Each seq runs from an end node towards the root or a branch node
        seenI = 0
        seenO = 0
        for nodeID in seq:
            counts = nodes[nodeID]
            seenI += counts.inputs + counts.seenInputs
            seenO += counts.outputs + counts.seenOutputs
            counts.seenInputs = seenI
            counts.seenOutputs = seenO
            counts.nPossibleIOPaths = counts.seenInputs * (totalOutputs - counts.seenOutputs) + counts.seenOutputs * (totalInputs - counts.seenInputs)
            counts.synapse_centrality = counts.nPossibleIOPaths / float(totalOutputs)