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Python SymPy Permutation.cyclic_form()用法及代碼示例

Permutation.cyclic_form():cyclic_form()是一個sympy Python庫函數,通過省略單例,從規範符號返回循環符號。

用法:
sympy.combinatorics.permutations.Permutation.cyclic_form()

返回:
規範表示法中的循環表示法


代碼1:cyclic_form()示例

# Python code explaining 
# SymPy.Permutation.cyclic_form() 
  
# importing SymPy libraries 
from sympy.combinatorics.partitions import Partition 
from sympy.combinatorics.permutations import Permutation 
  
# Using from sympy.combinatorics.permutations.Permutation.cyclic_form() method  
  
# creating Permutation 
a = Permutation([2, 0, 3, 1, 5, 4]) 
  
b = Permutation([3, 1, 2, 5, 4, 0]) 
  
  
print ("Permutation a - cyclic_form form : ", a.cyclic_form) 
print ("Permutation b - cyclic_form form : ", b.cyclic_form)

輸出:

Permutation a – cyclic_form form : [[0, 2, 3, 1], [4, 5]]
Permutation b – cyclic_form form : [[0, 3, 5]]

代碼2:cyclic_form()示例– 2D排列

# Python code explaining 
# SymPy.Permutation.cyclic_form() 
  
# importing SymPy libraries 
from sympy.combinatorics.partitions import Partition 
from sympy.combinatorics.permutations import Permutation 
  
# Using from  
# sympy.combinatorics.permutations.Permutation.cyclic_form() method  
  
# creating Permutation 
a = Permutation([[2, 4, 0],  
                 [3, 1, 2], 
                 [1, 5, 6]]) 
  
# SELF COMMUTATING     
print ("Permutation a - cyclic_form form : ", a.cyclic_form)

輸出:

Permutation a – cyclic_form form : [[0, 3, 5, 6, 1, 2, 4]]



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