Python SymPy Permutation.cyclic_form()用法及代码示例

Permutation.cyclic_form()：cyclic_form()是一个sympy Python库函数，通过省略单例，从规范符号返回循环符号。

用法：

sympy.combinatorics.permutations.Permutation.cyclic_form()

返回：
规范表示法中的循环表示法

代码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)

输出：

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

代码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)

输出：

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

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注：本文由纯净天空筛选整理自noobestars101大神的英文原创作品 Python | SymPy Permutation.cyclic_form() method。非经特殊声明，原始代码版权归原作者所有，本译文未经允许或授权，请勿转载或复制。