Python Graphlib Module get_ready( )
In the previous two articles, we have seen about static_order( ) in Graphlib module.
In this article, we are going to see about another method get_ready( )
get_ready( )
This method returns a tuple with all the nodes that are ready.
*Initially get_ready( ) returns all nodes with no predecessors and after by calling TopologicalSorter.done( ),mark them as processed.
*Further calls of get_ready( ) will return all new nodes that have all their predecessors already processed.
Codesnippet:
from graphlib import TopologicalSorter
graph = {1 : {6},2: {3}, 3: {1}, 4: {0, 1}, 5: {0, 2}}
t=TopologicalSorter(graph)
t.prepare( )
resulttuple=t.get_ready( )
print(resulttuple)
Output:
(6, 0)
->If empty tuples are returned that indicates that no progress can be made
->get_ready( ) raises ValueError if called without calling prepare( )
previously.
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( )
print(r)
Error:
Traceback (most recent call last):
File "C:\Users\findm\AppData\Local\Programs\Python\Python39\graphlib.py", line 9, 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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