Python Sympy Permutation.next_trotterjohnson()用法及代碼示例

Permutation.next_trotterjohnson()：next_trotterjohnson()是一個sympy Python庫函數，它以Trotter-Johnson順序返回下一個排列。如果self是最後一個排列，則返回None。

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

返回：Trotter-Johnson順序中的下一個排列

代碼1：next_trotterjohnson()示例

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

輸出：

Permutation a – next_trotterjohnson form : (0 3 1 2)
Permutation b – next_trotterjohnson form : (0 1 5)(2 3 4)

代碼2：next_trotterjohnson()示例– 2D排列

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

輸出：

Permutation a – next_trotterjohnson form : (6)(0 3 5 1 2 4)

相關用法

注：本文由純淨天空篩選整理自noobestars101大神的英文原創作品 SymPy | Permutation.next_trotterjohnson() in Python。非經特殊聲明，原始代碼版權歸原作者所有，本譯文未經允許或授權，請勿轉載或複製。