Python| SymPy Permutation.cyclic_form() 方法
Permutation.cyclic_form() : cyclic_form()是一个 sympy Python库函数,它通过省略单例从规范符号返回循环符号。
Syntax :
sympy.combinatorics.permutations.Permutation.cyclic_form()
Return :
cyclic notation from the canonical notation
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]]