Python Graphlib Module done(*nodes)
In the previous article, we have seen get_ready( ) method in Graphlib
module.
In this article, we discuss about done(*nodes) in Graphlib module.
done(*nodes):
*This method marks a set of nodes returned by TopologicalSorter.get_ready( ) as processed ,unlocking an sucessor of each node in nodes being returned in the future by a call to TopologicalSorter.get_ready( )
codesnippet:
from graphlib import TopologicalSorter
graph = {1 : {6},2: {3}, 3: {1}, 4: {0, 1}, 5: {0, 2}}
t=TopologicalSorter(graph)
t.prepare()
r=t.get_ready( )
t.done(*r)
->It will raises ValueError-
*if any node in nodes has already been marked as processed by a previous call to this method.
*If a node was not added to the graph by using TopologicalSorter.add( )
*If called without calling prepare( ) or if node has not yet returned by get_ready()
Example:
from graphlib import TopologicalSorter
graph = {1 : {6},2: {3}, 3: {1}, 4: {0, 1}, 5: {0, 2}}
t=TopologicalSorter(graph)
r=t.get_ready()
t.done(*r)
Error:
Traceback (most recent call last):
File "C:\Users\findm\AppData\Local\Programs\Python\Python39\graphlib.py", line 8, in <module>
r=t.get_ready()
File "C:\Users\findm\AppData\Local\Programs\Python\Python39\lib\graphlib.py", line 117, in get_ready
raise ValueError("prepare() must be called first")
ValueError: prepare() must be called first
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