🐛 Describe the bug When the devisor is -1 and input is a large negative integer, torch.floor_divide will throw a floating point exception. import torch input = torch.tensor([[-9223372036854775808]]) other = torch.tensor([-1]) torch.floor...
1.api add,minus,multiply,divide:对应位置的处理 matmul:mm,matmul,@,矩阵相乘 pow sqrt,rsqrt(平方根的导数) round:四舍五入 exp:e的对数 log:以e为底 trunc:取整数 frac:取小数 clamp:可以(min),(min,max) 2.api basic 矩阵相乘 多维上处理 次方: e: 小数: clamp: 3.程序 importtorch#ADDa = to...
import torch s = dir(torch) for i in s: print(i) 1. 2. 3. 4. 输出有一千多个结果 AI检测代码解析 AVG AggregationType AnyType Argument ArgumentSpec BFloat16Storage BFloat16Tensor BenchmarkConfig BenchmarkExecutionStats Block BoolStorage BoolTensor BoolType BufferDict ByteStorage ByteTensor CONV...
div() == divide() digamma() erf() erfc() erfinv() exp() exp2() expm1() fake_quantize_per_channel_affine() fake_quantize_per_tensor_affine() fix() == trunc() float_power() floor() floor_divide() fmod() frac() imag() ldexp() lerp() lgamma() log() log10() log1p() log...
#如果有gpu则使用gpu,此时device='cuda',否则使用cpu device="cuda" if torch.cuda.is_available() else "cpu" print(device) cuda #requires_grad:是否可被求导 #一般来说,神经网络学习的权重是可导的(requires_grad=True) my_tensor=torch.tensor([[1,2,3],[4,5,6]],dtype=torch.float32,device='...
Add/minus/multiply/divide Matmul(矩阵式相乘) Pow Sqrt/rsqrt Round basic(+ - * / add sub mul div) 建议直接使用运算符 AI检测代码解析 >>> a=torch.rand(3,4) >>> b=torch.rand(4) #broadingcast机制 >>> a+b tensor([[0.2349, 1.7635, 1.4385, 0.5826], ...
1 (1) 12 BINARY_ADD 14 BINARY_TRUE_DIVIDE 16 STORE_FAST 2 (x) 15 1...
Add/minus/multiply/divide Matmul(矩阵式相乘) Pow Sqrt/rsqrt Round basic(+ - * / add sub mul div) 建议直接使用运算符 view code matmul Torch.mm(only for 2d 不推荐) Torch.matmul(推荐) @ 注意:①*是element-wise,对应元素相乘;②.matmul是矩阵相乘 ...
torch.true_divide 是 支持bf16,fp16,fp32,uint8,int8,int16,int32,int64,bool torch.trunc 是 支持fp16,fp32 torch.xlogy 是 支持fp16,fp32,uint8,int8,int16,int32,int64,bool torch.argmax 是 支持bf16,fp16,fp32,fp64,uint8,int8,int16,int32,int64 torch.argmin 是 支持fp...
As a side note, if you find yourself wishing for the behavior of option 1, and it's true that the tensor you will call .backward() on (probably loss) is indeed a scalar which has the same value on every worker node, then you can just divide your tensor-to-be-differentiated by the...