class graphlib.TopologicalSorter(graph=None)  
Provides functionality to topologically sort a graph of hashable nodes. A topological order is a linear ordering of the vertices in a graph such that for every directed edge u -> v from vertex u to vertex v, vertex u comes before vertex v in the ordering. For instance, the vertices of the graph may represent tasks to be performed, and the edges may represent constraints that one task must be performed before another; in this example, a topological ordering is just a valid sequence for the tasks. A complete topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph. If the optional graph argument is provided it must be a dictionary representing a directed acyclic graph where the keys are nodes and the values are iterables of all predecessors of that node in the graph (the nodes that have edges that point to the value in the key). Additional nodes can be added to the graph using the add() method. In the general case, the steps required to perform the sorting of a given graph are as follows:  Create an instance of the TopologicalSorter with an optional initial graph. Add additional nodes to the graph. Call prepare() on the graph. While is_active() is True, iterate over the nodes returned by get_ready() and process them. Call done() on each node as it finishes processing.  In case just an immediate sorting of the nodes in the graph is required and no parallelism is involved, the convenience method TopologicalSorter.static_order() can be used directly: >>> graph = {"D": {"B", "C"}, "C": {"A"}, "B": {"A"}}
>>> ts = TopologicalSorter(graph)
>>> tuple(ts.static_order())
('A', 'C', 'B', 'D')
 The class is designed to easily support parallel processing of the nodes as they become ready. For instance: topological_sorter = TopologicalSorter()

# Add nodes to 'topological_sorter'...

topological_sorter.prepare()
while topological_sorter.is_active():
    for node in topological_sorter.get_ready():
        # Worker threads or processes take nodes to work on off the
        # 'task_queue' queue.
        task_queue.put(node)

    # When the work for a node is done, workers put the node in
    # 'finalized_tasks_queue' so we can get more nodes to work on.
    # The definition of 'is_active()' guarantees that, at this point, at
    # least one node has been placed on 'task_queue' that hasn't yet
    # been passed to 'done()', so this blocking 'get()' must (eventually)
    # succeed.  After calling 'done()', we loop back to call 'get_ready()'
    # again, so put newly freed nodes on 'task_queue' as soon as
    # logically possible.
    node = finalized_tasks_queue.get()
    topological_sorter.done(node)
  
add(node, *predecessors)  
Add a new node and its predecessors to the graph. Both the node and all elements in predecessors must be hashable. If called multiple times with the same node argument, the set of dependencies will be the union of all dependencies passed in. It is possible to add a node with no dependencies (predecessors is not provided) or to provide a dependency twice. If a node that has not been provided before is included among predecessors it will be automatically added to the graph with no predecessors of its own. Raises ValueError if called after prepare(). 
  
prepare()  
Mark the graph as finished and check for cycles in the graph. If any cycle is detected, CycleError will be raised, but get_ready() can still be used to obtain as many nodes as possible until cycles block more progress. After a call to this function, the graph cannot be modified, and therefore no more nodes can be added using add(). 
  
is_active()  
Returns True if more progress can be made and False otherwise. Progress can be made if cycles do not block the resolution and either there are still nodes ready that haven’t yet been returned by TopologicalSorter.get_ready() or the number of nodes marked TopologicalSorter.done() is less than the number that have been returned by TopologicalSorter.get_ready(). The __bool__() method of this class defers to this function, so instead of: if ts.is_active():
    ...
 it is possible to simply do: if ts:
    ...
 Raises ValueError if called without calling prepare() previously. 
  
done(*nodes)  
Marks a set of nodes returned by TopologicalSorter.get_ready() as processed, unblocking any successor of each node in nodes for being returned in the future by a call to TopologicalSorter.get_ready(). Raises ValueError if any node in nodes has already been marked as processed by a previous call to this method or if a node was not added to the graph by using TopologicalSorter.add(), if called without calling prepare() or if node has not yet been returned by get_ready(). 
  
get_ready()  
Returns a tuple with all the nodes that are ready. Initially it returns all nodes with no predecessors, and once those are marked as processed by calling TopologicalSorter.done(), further calls will return all new nodes that have all their predecessors already processed. Once no more progress can be made, empty tuples are returned. Raises ValueError if called without calling prepare() previously. 
  
static_order()  
Returns an iterable of nodes in a topological order. Using this method does not require to call TopologicalSorter.prepare() or TopologicalSorter.done(). This method is equivalent to: def static_order(self):
    self.prepare()
    while self.is_active():
        node_group = self.get_ready()
        yield from node_group
        self.done(*node_group)
 The particular order that is returned may depend on the specific order in which the items were inserted in the graph. For example: >>> ts = TopologicalSorter()
>>> ts.add(3, 2, 1)
>>> ts.add(1, 0)
>>> print([*ts.static_order()])
[2, 0, 1, 3]

>>> ts2 = TopologicalSorter()
>>> ts2.add(1, 0)
>>> ts2.add(3, 2, 1)
>>> print([*ts2.static_order()])
[0, 2, 1, 3]
 This is due to the fact that “0” and “2” are in the same level in the graph (they would have been returned in the same call to get_ready()) and the order between them is determined by the order of insertion. If any cycle is detected, CycleError will be raised. 
  New in version 3.9.