mindspore.ops.floor_mod
- mindspore.ops.floor_mod(x, y)[source]
Computes the remainder of division element-wise. It's a flooring divide. E.g. \(floor(x / y) * y + mod(x, y) = x\).
Inputs of x and y comply with the implicit type conversion rules to make the data types consistent. When the inputs are two tensors, dtypes of them cannot be both bool, and the shapes of them could be broadcast. When the inputs are one tensor and one scalar, the scalar could only be a constant.
\[out_{i} =\text{floor}(x_{i} // y_{i})\]where the \(floor\) indicates the Floor operator, for more details, please refer to the
mindspore.ops.Floor
operator.Warning
Data of input y should not be 0, or the maximum value of its dtype will be returned.
When the elements of input exceeds 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
- Returns
Tensor, the shape is the same as the one after broadcasting, and the data type is the one with higher precision of the two inputs.
- Raises
TypeError – If neither x nor y is a Tensor, number.Number, or bool.
- 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.floor_mod(x, y) >>> print(output) [2 1 2]