mindspore.ops.truncate_mod

View Source On Gitee
mindspore.ops.truncate_mod(x, y)[source]

Returns the remainder of division element-wise.

Inputs of x and y comply with the implicit type conversion rules to make the data types consistent. The inputs must be two tensors or one tensor and one scalar. When the inputs are two tensors, dtypes of them cannot be bool at the same time, and the shapes of them could be broadcast. When the inputs are one tensor and one scalar, the scalar could only be a constant.

Warning

  • The input data does not support 0.

  • When the elements of input exceed 2048 , the accuracy of operator cannot guarantee the requirement of double thousandths in the mini form.

  • Due to different architectures, the calculation results of this operator on NPU and CPU may be inconsistent.

  • If shape is expressed as (D1,D2… ,Dn), then D1*D2… *DN<=1000000,n<=8.

Parameters
  • x (Union[Tensor, numbers.Number, bool]) – The first input is a number, or a bool, or a tensor whose data type is number or bool.

  • y (Union[Tensor, numbers.Number, bool]) – The second input is a number, or a bool when the first input is a tensor, or a tensor whose data type is number or bool.

Returns

Tensor, the shape is the same as the one after broadcasting, and the data type is the one with higher precision among the two inputs.

Raises
  • TypeError – If neither x nor y is one of the following: Tensor, number, bool.

  • TypeError – If neither x nor y is a Tensor.

  • ValueError – If the shape x and y cannot be broadcasted to each other.

Supported Platforms:

Ascend GPU CPU

Examples

>>> import mindspore
>>> import numpy as np
>>> from mindspore import Tensor, ops
>>> x = Tensor(np.array([2, 4, -1]), mindspore.int32)
>>> y = Tensor(np.array([3, 3, 3]), mindspore.int32)
>>> output = ops.truncate_mod(x, y)
>>> print(output)
[ 2  1 -1]