本文整理汇总了Python中sympy.combinatorics.permutations.Cycle类的典型用法代码示例。如果您正苦于以下问题：Python Cycle类的具体用法？Python Cycle怎么用？Python Cycle使用的例子？那么恭喜您, 这里精选的类代码示例或许可以为您提供帮助。
在下文中一共展示了Cycle类的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: _print_Permutation
def _print_Permutation(self, expr):
from sympy.combinatorics.permutations import Permutation, Cycle
if Permutation.print_cyclic:
if not expr.size:
return "()"
# before taking Cycle notation, see if the last element is
# a singleton and move it to the head of the string
s = Cycle(expr)(expr.size - 1).__repr__()[len("Cycle") :]
last = s.rfind("(")
if not last == 0 and "," not in s[last:]:
s = s[last:] + s[:last]
s = s.replace(",", "")
return s
else:
s = expr.support()
if not s:
if expr.size < 5:
return "Permutation(%s)" % str(expr.array_form)
return "Permutation([], size=%s)" % expr.size
trim = str(expr.array_form[: s[-1] + 1]) + ", size=%s" % expr.size
use = full = str(expr.array_form)
if len(trim) < len(full):
use = trim
return "Permutation(%s)" % use示例2: _print_Permutation
def _print_Permutation(self, expr):
from sympy.combinatorics.permutations import Permutation, Cycle
if Permutation.print_cyclic:
if not expr.size:
return 'Permutation()'
# before taking Cycle notation, see if the last element is
# a singleton and move it to the head of the string
s = Cycle(expr)(expr.size - 1).__repr__()[len('Cycle'):]
last = s.rfind('(')
if not last == 0 and ',' not in s[last:]:
s = s[last:] + s[:last]
return 'Permutation%s' % s
else:
s = expr.support()
if not s:
if expr.size < 5:
return 'Permutation(%s)' % str(expr.array_form)
return 'Permutation([], size=%s)' % expr.size
trim = str(expr.array_form[:s[-1] + 1]) + ', size=%s' % expr.size
use = full = str(expr.array_form)
if len(trim) < len(full):
use = trim
return 'Permutation(%s)' % use