以一个简单例子来说明各个 Loss 函数的使用
label_numpy = np.array([[0, 0, 0, 0], [0, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 1]], dtype=np.float) # 模拟 标签 out_numpy = np.array([[0, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]], dtype=np.float) # 模拟 预测 num_classes = 2
$l_n = -w_n[y_n * log{x_n} + (1 - y_n) * log(1 - x_n)]$
label = torch.from_numpy(label_numpy).unsqueeze(0) # N x C output = torch.from_numpy(out_numpy).unsqueeze(0) # N x C # ======================================================= # criterion = nn.BCELoss() loss = criterion(F.sigmoid(output), label) # 0.6219 # ======================================================= #
label = torch.from_numpy(label_numpy).unsqueeze(0) output = torch.from_numpy(out_numpy).unsqueeze(0) # ======================================================= # criterion = nn.BCEWithLogitsLoss() loss = criterion(output, label) # 0.6219 # ======================================================= #
这个损失将Sigmoid层和BCELoss合并在一个类中，且数值稳定性更好
具体计算过程如下
class BCEWithLogitsLoss(nn.Module): """ 这个版本在数值上比使用一个简单的Sigmoid和一个BCELoss as更稳定，通过将操作合并到一个层中，我们利用log-sum-exp技巧来实现数值稳定性。 """ def __init__(self): super(BCEWithLogitsLoss, self).__init__() def forward(self, input, target, weight=None, size_average=None, reduce=None, reduction=‘mean‘, pos_weight=None): if size_average is not None or reduce is not None: reduction = _Reduction.legacy_get_string(size_average, reduce) if not (target.size() == input.size()): raise ValueError("Target size ({}) must be the same as input size ({})".format(target.size(), input.size())) max_val = (-input).clamp(min=0) if pos_weight is None: loss = input - input * target + max_val + ((-max_val).exp() + (-input - max_val).exp()).log() else: log_weight = 1 + (pos_weight - 1) * target loss = input - input * target + log_weight * (max_val + ((-max_val).exp() + (-input - max_val).exp()).log()) if weight is not None: loss = loss * weight if reduction == ‘none‘: return loss elif reduction == ‘mean‘: return loss.mean() else: return loss.sum()
这里，输出为 one-hot 格式
$loss(x, class) = -log(\frac{e^{x[class]}}{\sum_j{e^{x[j]}}}) = -x[class] + log(\sum_j{e^{x[j]}})$
label_numpy = label_numpy.reshape((label_numpy.size)) out_numpy = out_numpy.reshape((label_numpy.size)) label = torch.from_numpy(label_numpy).long() onehot_output = np.eye(num_classes)[np.where(out_numpy>=0.5, 1, 0)] # convert to onehot format output = torch.from_numpy(onehot_output) # ======================================================= # criterion = nn.CrossEntropyLoss() loss = criterion(output, label) # 0.4383 # ======================================================= #
具体计算过程如下：
first = np.zeros(shape=(output.shape[0])) for i in range(output.shape[0]): first[i] = -output[i][label[i]] second = np.zeros(shape=(output.shape[0])) for i in range(output.shape[0]): for j in range(output.shape[1]): second[i] += np.exp(output[i][j]) res = (first + np.log(second)).mean()
原文：https://www.cnblogs.com/xuanyuyt/p/12923519.html
