Python Graphlib module static_order( )-2
In the previous article, we have seen some information about static_order( )
In this article, we shall see more about static_order( ) method
*The order of elements returned by static_order( ) depends on how u inserted the elements in the graph.
Example:
Topologicalorder-1:
from graphlib import TopologicalSorter
ts = TopologicalSorter()
ts.add(3, 2, 1)
ts.add(1, 0)
print("TopologicalOrder1=",end=" ")
print([*ts.static_order()])
Output:
TopologicalOrder1= [2, 0, 1, 3]
Topologicalorder-2:
from graphlib import TopologicalSorter
ts2 = TopologicalSorter()
ts2.add(1, 0)
ts2.add(3, 2, 1)
print("TopologicalOrder2=",end=" ")
print([*ts2.static_order()])
Output:
TopologicalOrder2= [0, 2, 1, 3]
By observing TopologicalOrder1 and TopologicalOrder2, we conclude
that order of elements depends upon order of insertion.
*If any cycles is detected in the graph, CycleError will be raised.
For Example:
from graphlib import TopologicalSorter
tos = TopologicalSorter()
tos.add(1, 0)
tos.add(3, 2, 1)
tos.add(0, 1)
print([*tos.static_order()])
Output:
Traceback (most recent call last):
File "C:\Users\findm\AppData\Local\Programs\Python\Python39\graphlib1.py", line 15, in <module>
print([*tos.static_order()])
File "C:\Users\findm\AppData\Local\Programs\Python\Python39\lib\graphlib.py", line 242, in static_order
self.prepare()
File "C:\Users\findm\AppData\Local\Programs\Python\Python39\lib\graphlib.py", line 104, in prepare
raise CycleError(f"nodes are in a cycle", cycle)
graphlib.CycleError: ('nodes are in a cycle', [1, 0, 1])
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