python - Cython 浮点除法 PyExc_ZeroDivision 错误检查

Question

我正在做一些循环密集型计算并将代码转换为 Cython。我使用 cython -a 选项进行了分析，并检查了 .html 文件，似乎每当我进行浮点除法时，都会出现一些黄线，它会执行以下操作：

if (unlikely(__pyx_t_37 == 0)) {
        PyErr_Format(PyExc_ZeroDivisionError, "float division");
        {__pyx_filename = __pyx_f[0]; __pyx_lineno = 84; __pyx_clineno = __LINE__; goto __pyx_L1_error;}
      }

我想这是针对分频器为 0 的情况。我为此使用了一个常数，并且分频器不可能为 0，我想知道是否有什么办法可以让它更快。

Answer 1

您需要添加@cython.cdivision(True)以避免异常检查。

import cython

cdef double pydivision():
  cdef int i
  cdef double k, j
  k = 2.0
  j = 0.0
  for i in range(10):
    j += i/k
  # Generated code: Python exception checking
  # /* "checksum.pyx":9
  # *   j = 0.0
  # *   for i in range(10):
  # *     j += i/k             # <<<<<<<<<<<<<<
  # *   return j
  # *
  # */
  #    if (unlikely(__pyx_v_k == 0)) {
  #      PyErr_Format(PyExc_ZeroDivisionError, "float division");
  #      {__pyx_filename = __pyx_f[0]; __pyx_lineno = 9; __pyx_clineno = __LINE__; goto __pyx_L1_error;}
  #    }
  #    __pyx_v_j = (__pyx_v_j + (__pyx_v_i / __pyx_v_k));
  #  }
  return j

#This decorator works wonders
@cython.cdivision(True)
cdef double cdivision():
  cdef int i
  cdef double k, j
  k = 2.0
  j = 0.0
  for i in range(10):
    j += i/k
  # Generated code: no exception checking
  # /* "checksum.pyx":20
  # *   j = 0.0
  # *   for i in range(10):
  # *     j += i/k             # <<<<<<<<<<<<<<
  # *   return j
  # *
  # */
  #    __pyx_v_j = (__pyx_v_j + (__pyx_v_i / __pyx_v_k));
  #  }
  return j

Answer 2

1/divisor如果除数是常数，则可以乘以