Question

我有一个高度不平衡的3D数据集，其中约80％的体积是背景数据，我只对前景元素感兴趣，这些元素在随机位置约占总体积的20％。这些位置记录在给网络的标签张量中。目标张量是二进制的，其中0表示背景，而1表示我们感兴趣或要分割的区域。

每个卷的大小为[30,512,1024]。我正在使用大小为[30,64,64]的块遍历每个卷。因此，我的大多数块在目标张量中只有0值。

我了解到DiceLoss非常适合此类问题，已成功用于3D MRI扫描分割中。一个简单的实现从这里开始：https://github.com/pytorch/pytorch/issues/1249#issuecomment-305088398

def dice_loss(input, target):
    smooth = 1.

    iflat = input.view(-1)
    tflat = target.view(-1)
    intersection = (iflat * tflat).sum()

    return 1 - ((2. * intersection + smooth) /
              (iflat.sum() + tflat.sum() + smooth))

这对我不起作用，我的意思是对于一个补丁，其中我所有的背景都是tflat.sum()将是0。这也将使intersection 0，因此，对于我的大多数补丁或块，我将得到1的回报。

这是对的吗？这不是应该如何工作的。但是我为此感到苦恼，因为这是我的网络输出：

idx:  0 of  312 - Training Loss:  1.0 - Training Accuracy:  3.204042239857152e-11
idx:  5 of  312 - Training Loss:  0.9876335859298706 - Training Accuracy:  0.0119545953348279
idx:  10 of  312 - Training Loss:  1.0 - Training Accuracy:  7.269467666715101e-11
idx:  15 of  312 - Training Loss:  0.7320756912231445 - Training Accuracy:  0.22638492286205292
idx:  20 of  312 - Training Loss:  0.3599294424057007 - Training Accuracy:  0.49074622988700867
idx:  25 of  312 - Training Loss:  1.0 - Training Accuracy:  1.0720428988975073e-09
idx:  30 of  312 - Training Loss:  1.0 - Training Accuracy:  1.19782361807097e-09
idx:  35 of  312 - Training Loss:  1.0 - Training Accuracy:  1.956790285362331e-09
idx:  40 of  312 - Training Loss:  1.0 - Training Accuracy:  1.6055999862985004e-09
idx:  45 of  312 - Training Loss:  1.0 - Training Accuracy:  7.580232552761856e-10
idx:  50 of  312 - Training Loss:  1.0 - Training Accuracy:  9.510597864803572e-10
idx:  55 of  312 - Training Loss:  1.0 - Training Accuracy:  1.341515676323013e-09
idx:  60 of  312 - Training Loss:  0.7165247797966003 - Training Accuracy:  0.02658153884112835
idx:  65 of  312 - Training Loss:  1.0 - Training Accuracy:  4.528208030762926e-09
idx:  70 of  312 - Training Loss:  0.3205708861351013 - Training Accuracy:  0.6673439145088196
idx:  75 of  312 - Training Loss:  0.9305377006530762 - Training Accuracy:  2.3437689378624782e-05
idx:  80 of  312 - Training Loss:  1.0 - Training Accuracy:  5.305786885401176e-07
idx:  85 of  312 - Training Loss:  1.0 - Training Accuracy:  4.0612556517771736e-07
idx:  90 of  312 - Training Loss:  0.8207412362098694 - Training Accuracy:  0.0344742126762867
idx:  95 of  312 - Training Loss:  0.7463213205337524 - Training Accuracy:  0.19459737837314606
idx:  100 of  312 - Training Loss:  1.0 - Training Accuracy:  4.863646818620282e-09
idx:  105 of  312 - Training Loss:  0.35790306329727173 - Training Accuracy:  0.608722984790802
idx:  110 of  312 - Training Loss:  1.0 - Training Accuracy:  3.3852198821904267e-09
idx:  115 of  312 - Training Loss:  1.0 - Training Accuracy:  1.5268487585373691e-09
idx:  120 of  312 - Training Loss:  1.0 - Training Accuracy:  3.46353523639209e-09
idx:  125 of  312 - Training Loss:  1.0 - Training Accuracy:  2.5878148582347826e-11
idx:  130 of  312 - Training Loss:  1.0 - Training Accuracy:  2.3601216467272756e-11
idx:  135 of  312 - Training Loss:  1.0 - Training Accuracy:  1.1504343033763575e-09
idx:  140 of  312 - Training Loss:  0.4516671299934387 - Training Accuracy:  0.13879922032356262

我不认为网络可以从中学到任何东西。

现在我很困惑，因为我的问题应该不会太复杂，因为我确信MRI扫描也具有目标张量，并且其中大多数都表示背景。我在做什么错了？

谢谢

Answer 1

如果您的算法预测所有背景体素的值都应该恰好为0，那么您将获得1的回报，但是如果预测任何正值（如果您使用S型激活，则肯定会得到），它仍然可以改善通过输出尽可能少的损耗。换句话说，分子不能超过smooth，但算法仍可以学习使分母尽可能小。

如果您对算法的行为不满意，则可以尝试增加批处理大小（这样，所有卷都不会出现前景下降的机会）或直接跳过此类批处理。它可能会或可能不会帮助学习。

话虽这么说，但我个人从未成功使用Dice / IoU作为损失函数来学习细分，并且通常选择二进制交叉熵或类似的损失，而将前者作为验证指标。

骰子丢失和样本中目标张量中没有数据？

1 个答案: