proj_value = self.value_conv(x).view(m_batchsize,-1,width*height) # B X C X N out = torch.bmm...-1).permute(0, 2, 1) key = key.view(b, c, -1) value = value.view(b, c, -1).permute(0, 2, 1) att = torch.bmm...(query, key) if self.use_scale: att = att.div(c**0.5) att = self.softmax(att) x = torch.bmm(att,..., c * h * w) p = p.view(b, 1, c * h * w) g = g.view(b, c * h * w, 1) att = torch.bmm...(p, g) if self.use_scale: att = att.div((c*h*w)**0.5) x = torch.bmm(att
同理,由于 torch.bmm 函数不支持广播,相对应的输入的两个张量必须为 3D。...import torch input = torch.randn(10, 3, 4) other = torch.randn(10, 4, 2) result = torch.bmm(input,
h = self.h(x).view(m_batchsize, -1, width * height) # B * C * (W * H) attention = torch.bmm...) # B * (W * H) * (W * H) attention = self.softmax(attention) self_attetion = torch.bmm
proj_key = self.key_conv(x).view(m_batchsize,-1,width*height) # B X C x (*W*H) energy = torch.bmm...proj_value = self.value_conv(x).view(m_batchsize,-1,width*height) # B X C X N out = torch.bmm...然后我们用torch.bmm()来做矩阵的乘法:(N,Channel//8)和(Channel//8,N)两个矩阵相乘,得到一个(N,N)的矩阵。
x.view(*size[:2],-1) f,g,h = self.query(x),self.key(x),self.value(x) beta = F.softmax(torch.bmm...(f.transpose(1,2), g), dim=1) o = self.gamma * torch.bmm(h, beta) + x return o.view
(N) g1 = self.g(x).view(m_batchsize, -1, width * height) # B X C x (*W*H) energy = torch.bmm...(N) X (N) h1 = self.h(x).view(m_batchsize, -1, width * height) # B X C X N out = torch.bmm
break a = random.randint(-10,10,size=(8,8)) 然而,让我们思考一个问题, 4.5 本节源码 3 83 apple 57345 uni4E00 torch.bmm
triton在TFLOPS这个指标层面是能够超过cublas的实现,但是后面我通过nsight system对每个kernel的具体执行时间进行了profiling,发现在torch.matmul或者torch.bmm...None, :] < N) tl.store(C_ptr, c, mask=c_mask) 然后写一个简单的单元测试,确保通过triton写出来的kernel能够和torch.matmul/torch.bmm...dtype=torch.float16) b = torch.randn((4, 512, 512), device='cuda', dtype=torch.float16) torch_output = torch.bmm...16x4096x4096, 16x4096x4096) 通过nsight system + nvtx就可以看到每个kernel的具体实现情况: img 添加图片注释,不超过 140 字(可选) 使用torch.bmm
input): matrix3x3 = self.input_transform(input) # batch matrix multiplication xb = torch.bmm...xb = F.relu(self.bn1(self.conv1(xb))) matrix64x64 = self.feature_transform(xb) xb = torch.bmm...if outputs.is_cuda: id3x3 = id3x3.cuda() id64x64 = id64x64.cuda() diff3x3 = id3x3 - torch.bmm...(m3x3, m3x3.transpose(1, 2)) diff64x64 = id64x64 - torch.bmm(m64x64, m64x64.transpose(1, 2)) return
input_embedding = input_embedding.unsqueeze(2) # [batch_size, embed_size, 1] pos_dot = torch.bmm...2), 1] pos_dot = pos_dot.squeeze(2) # [batch_size, (window * 2)] neg_dot = torch.bmm...tensor的第一个维度必须相同,后面两个维度必须满足矩阵乘法的要求 batch1 = torch.randn(10, 3, 4) batch2 = torch.randn(10, 4, 5) res = torch.bmm
torch.bmm(torch.ones((2,1,3), dtype = torch.float), torch.ones((2,3,2), dtype = torch.float)) tensor(...-1] # set transpose_b=True to swap the last two dimensions of key scores = torch.bmm...masked_softmax(scores, valid_length)) print("attention_weight\n",attention_weights) return torch.bmm...(-1) attention_weights = self.dropout(masked_softmax(scores, valid_length)) return torch.bmm
proj_key = self.key_conv(x).view(m_batchsize,-1,width*height) # B X C x (*W*H) energy = torch.bmm...proj_value = self.value_conv(x).view(m_batchsize,-1,width*height) # B X C X N out = torch.bmm...步骤二: energy = torch.bmm(proj_query,proj_key) 这一步是将batch_size中的每一对proj_query和proj_key分别进行矩阵相乘,输出为B×(W...out = torch.bmm(proj_value,attention.permute(0,2,1) ) out = out.view(m_batchsize,C,width,height) 在对proj_value
self.key(x).view(n_batch, C, -1) v = self.value(x).view(n_batch, C, -1) content_content = torch.bmm...energy = content_content + content_position attention = self.softmax(energy) out = torch.bmm
encoder_outputs, hidden): seq_len = encoder_outputs.size(1) # 计算注意力权重 attn_weights = torch.bmm...2)) attn_weights = torch.softmax(attn_weights, dim=2) # 加权求和得到上下文向量 context = torch.bmm
n_pts = input.size()[2] matrix3x3 = self.input_transform(input) input_transform_output = torch.bmm...input_transform_output) matrix64x64 = self.feature_transform(x) feature_transform_output = torch.bmm...id3x3.cuda() id64x64 = id64x64.cuda() # Calculate matrix differences diff3x3 = id3x3 - torch.bmm...(m3x3, m3x3.transpose(1, 2)) diff64x64 = id64x64 - torch.bmm(m64x64, m64x64.transpose(1, 2))
self.embedding # 特征进行cross interaction for i in range(self.cross_num): emb_tmp = torch.bmm
V = self.v(x) # V: batch_size * seq_len * dim_v atten = nn.Softmax(dim=-1)(torch.bmm..._norm_fact # Q * K.T() # batch_size * seq_len * seq_len output = torch.bmm(atten,V)
注意事项 torch.bmm()函数执行了一个批量矩阵-矩阵乘积,简化了注意力分数的计算,其中查询和键向量的形状为[batch_size, seq_len, hidden_dim]。...由于我们想独立地对批次中的所有序列做这件事,我们使用torch.bmm(),它需要两批矩阵并将第一批中的每个矩阵与第二批中的相应矩阵相乘。....]], grad_fn=) 最后一步是将注意力权重与数值相乘: attn_outputs = torch.bmm(weights, value) attn_outputs.shape...,我们以后可以使用: def scaled_dot_product_attention(query, key, value): dim_k = query.size(-1) scores = torch.bmm...(query, key.transpose(1, 2)) / sqrt(dim_k) weights = F.softmax(scores, dim=-1) return torch.bmm(weights
raw_weights = torch.bmm(x, x.transpose(1, 2)) # - torch.bmm is a batched matrix multiplication....y = torch.bmm(weights, x) 以上就是通过两个矩阵乘法和一个softmax实现的self-attention。...这确保我们可以像以前一样使用torch.bmm(),并且键、查询和值的整个集合将被视为稍微大一些的batch。 由于head和batch的维度不是彼此相邻,我们需要在重塑之前进行转置。...# apply the self attention to the values out = torch.bmm(dot, values).view(b, h, t, k) 为了统一attention...下面是 pytorch 中的实现: dot = torch.bmm(queries, keys.transpose(1, 2)) indices = torch.triu_indices(k, k,
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