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Python math_ops.floor_div函数代码示例

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

本文整理汇总了Python中tensorflow.python.ops.math_ops.floor_div函数的典型用法代码示例。如果您正苦于以下问题:Python floor_div函数的具体用法?Python floor_div怎么用?Python floor_div使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。



在下文中一共展示了floor_div函数的5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,您的评价将有助于我们的系统推荐出更棒的Python代码示例。

示例1: testConsistent

 def testConsistent(self):
   nums, divs = self.intTestData()
   with self.test_session():
     tf_result = (
         math_ops.floor_div(nums, divs) * divs + math_ops.floormod(nums, divs)
     ).eval()
     tf_nums = array_ops.constant(nums)
     tf_divs = array_ops.constant(divs)
     tf2_result = (tf_nums // tf_divs * tf_divs + tf_nums % tf_divs).eval()
     np_result = (nums // divs) * divs + (nums % divs)
     # consistentcy with numpy
     self.assertAllEqual(tf_result, np_result)
     # consistentcy with two forms of divide
     self.assertAllEqual(tf_result, tf2_result)
     # consistency for truncation form
     tf3_result = (
         math_ops.truncatediv(nums, divs) * divs
         + math_ops.truncatemod(nums, divs)
     ).eval()
     expanded_nums = np.reshape(np.tile(nums, divs.shape[1]),
                                (nums.shape[0], divs.shape[1]))
     # Consistent with desire to get numerator
     self.assertAllEqual(tf3_result, expanded_nums)
     # Consistent with desire to get numerator
     self.assertAllEqual(tf_result, expanded_nums)
开发者ID:Jackhuang945,项目名称:tensorflow,代码行数:25,代码来源:math_ops_test.py


示例2: testConsistent

 def testConsistent(self):
   nums, divs = self.intTestData()
   with self.test_session():
     tf_result = (
         math_ops.floor_div(nums, divs) * divs + math_ops.floor_mod(nums, divs)
     ).eval()
     tf_nums = array_ops.constant(nums)
     tf_divs = array_ops.constant(divs)
     tf2_result = (tf_nums // tf_divs * tf_divs + tf_nums % tf_divs).eval()
     np_result = (nums // divs) * divs + (nums % divs)
     self.assertAllEqual(tf_result, np_result)
     self.assertAllEqual(tf_result, tf2_result)
开发者ID:821760408-sp,项目名称:tensorflow,代码行数:12,代码来源:math_ops_test.py


示例3: _FloorModGrad

def _FloorModGrad(op, grad):
  """Returns grad * (1, -floor(x/y))."""
  x = math_ops.conj(op.inputs[0])
  y = math_ops.conj(op.inputs[1])

  sx = array_ops.shape(x)
  sy = array_ops.shape(y)
  rx, ry = gen_array_ops.broadcast_gradient_args(sx, sy)
  floor_xy = math_ops.floor_div(x, y)
  gx = array_ops.reshape(math_ops.reduce_sum(grad, rx), sx)
  gy = array_ops.reshape(
      math_ops.reduce_sum(grad * math_ops.negative(floor_xy), ry), sy)
  return gx, gy
开发者ID:PuchatekwSzortach,项目名称:tensorflow,代码行数:13,代码来源:math_grad.py


示例4: testDivideInt

 def testDivideInt(self):
   nums, divs = self.intTestData()
   with self.test_session():
     tf_result = math_ops.floor_div(nums, divs).eval()
     np_result = nums // divs
     self.assertAllEqual(tf_result, np_result)
开发者ID:LongJun123456,项目名称:tensorflow,代码行数:6,代码来源:math_ops_test.py


示例5: sample


#.........这里部分代码省略.........
      sequence. The entries index into the Halton sequence starting with 0 and
      hence, must be whole numbers. For example, sequence_indices=[0, 5, 6] will
      produce the first, sixth and seventh elements of the sequence. If this
      parameter is None, then the `num_results` parameter must be specified
      which gives the number of desired samples starting from the first sample.
    dtype: (Optional) The dtype of the sample. One of `float32` or `float64`.
      Default is `float32`.
    randomized: (Optional) bool indicating whether to produce a randomized
      Halton sequence. If True, applies the randomization described in
      Owen (2017) [arXiv:1706.02808].
    seed: (Optional) Python integer to seed the random number generator. Only
      used if `randomized` is True. If not supplied and `randomized` is True,
      no seed is set.
    name:  (Optional) Python `str` describing ops managed by this function. If
    not supplied the name of this function is used.

  Returns:
    halton_elements: Elements of the Halton sequence. `Tensor` of supplied dtype
    and `shape` `[num_results, dim]` if `num_results` was specified or shape
    `[s, dim]` where s is the size of `sequence_indices` if `sequence_indices`
    were specified.

  Raises:
    ValueError: if both `sequence_indices` and `num_results` were specified or
    if dimension `dim` is less than 1 or greater than 1000.
  """
  if dim < 1 or dim > _MAX_DIMENSION:
    raise ValueError(
        'Dimension must be between 1 and {}. Supplied {}'.format(_MAX_DIMENSION,
                                                                 dim))
  if (num_results is None) == (sequence_indices is None):
    raise ValueError('Either `num_results` or `sequence_indices` must be'
                     ' specified but not both.')

  dtype = dtype or dtypes.float32
  if not dtype.is_floating:
    raise ValueError('dtype must be of `float`-type')

  with ops.name_scope(name, 'sample', values=[sequence_indices]):
    # Here and in the following, the shape layout is as follows:
    # [sample dimension, event dimension, coefficient dimension].
    # The coefficient dimension is an intermediate axes which will hold the
    # weights of the starting integer when expressed in the (prime) base for
    # an event dimension.
    indices = _get_indices(num_results, sequence_indices, dtype)
    radixes = array_ops.constant(_PRIMES[0:dim], dtype=dtype, shape=[dim, 1])

    max_sizes_by_axes = _base_expansion_size(math_ops.reduce_max(indices),
                                             radixes)

    max_size = math_ops.reduce_max(max_sizes_by_axes)

    # The powers of the radixes that we will need. Note that there is a bit
    # of an excess here. Suppose we need the place value coefficients of 7
    # in base 2 and 3. For 2, we will have 3 digits but we only need 2 digits
    # for base 3. However, we can only create rectangular tensors so we
    # store both expansions in a [2, 3] tensor. This leads to the problem that
    # we might end up attempting to raise large numbers to large powers. For
    # example, base 2 expansion of 1024 has 10 digits. If we were in 10
    # dimensions, then the 10th prime (29) we will end up computing 29^10 even
    # though we don't need it. We avoid this by setting the exponents for each
    # axes to 0 beyond the maximum value needed for that dimension.
    exponents_by_axes = array_ops.tile([math_ops.range(max_size)], [dim, 1])

    # The mask is true for those coefficients that are irrelevant.
    weight_mask = exponents_by_axes >= max_sizes_by_axes
    capped_exponents = array_ops.where(
        weight_mask, array_ops.zeros_like(exponents_by_axes), exponents_by_axes)
    weights = radixes ** capped_exponents
    # The following computes the base b expansion of the indices. Suppose,
    # x = a0 + a1*b + a2*b^2 + ... Then, performing a floor div of x with
    # the vector (1, b, b^2, b^3, ...) will produce
    # (a0 + s1 * b, a1 + s2 * b, ...) where s_i are coefficients we don't care
    # about. Noting that all a_i < b by definition of place value expansion,
    # we see that taking the elements mod b of the above vector produces the
    # place value expansion coefficients.
    coeffs = math_ops.floor_div(indices, weights)
    coeffs *= 1 - math_ops.cast(weight_mask, dtype)
    coeffs %= radixes
    if not randomized:
      coeffs /= radixes
      return math_ops.reduce_sum(coeffs / weights, axis=-1)
    coeffs = _randomize(coeffs, radixes, seed=seed)
    # Remove the contribution from randomizing the trailing zero for the
    # axes where max_size_by_axes < max_size. This will be accounted
    # for separately below (using zero_correction).
    coeffs *= 1 - math_ops.cast(weight_mask, dtype)
    coeffs /= radixes
    base_values = math_ops.reduce_sum(coeffs / weights, axis=-1)

    # The randomization used in Owen (2017) does not leave 0 invariant. While
    # we have accounted for the randomization of the first `max_size_by_axes`
    # coefficients, we still need to correct for the trailing zeros. Luckily,
    # this is equivalent to adding a uniform random value scaled so the first
    # `max_size_by_axes` coefficients are zero. The following statements perform
    # this correction.
    zero_correction = random_ops.random_uniform([dim, 1], seed=seed,
                                                dtype=dtype)
    zero_correction /= (radixes ** max_sizes_by_axes)
    return base_values + array_ops.reshape(zero_correction, [-1])
开发者ID:QiangCai,项目名称:tensorflow,代码行数:101,代码来源:halton_sequence_impl.py



注:本文中的tensorflow.python.ops.math_ops.floor_div函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。


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Python math_ops.floormod函数代码示例发布时间:2022-05-27
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