根據(jù)文檔，該Reduction參數(shù)有 3 個值 -?SUM_OVER_BATCH_SIZE、SUM和NONE。y_true = [[0., 2.], [0., 0.]]y_pred = [[3., 1.], [2., 5.]]mae = tf.keras.losses.MeanAbsoluteError(reduction=tf.keras.losses.Reduction.SUM)mae(y_true, y_pred).numpy()> 5.5mae = tf.keras.losses.MeanAbsoluteError()mae(y_true, y_pred).numpy()> 2.75經(jīng)過各種試驗后我可以推斷出的計算結(jié)果是：-當REDUCTION = SUM,Loss = Sum over all samples {(Sum of differences between y_pred and y_target vector of each sample / No of element in y_target of the sample )} = { (abs(3-0) + abs(1-2))/2 } + { (abs(2-0) + abs(5-0))/2 } = {4/2} + {7/2} = 5.5.當REDUCTION = SUM_OVER_BATCH_SIZE,Loss = [Sum over all samples {(Sum of differences between y_pred and y_target vector of each sample / No of element in y_target of the sample )}] / Batch_size or No of Samples ?= [ { (abs(3-0)} + abs(1-2))/2 } + { (abs(2-0) + abs(5-0))/2 } ]/2 = [ {4/2} + {7/2} ]/2 = [5.5]/2 = 2.75.結(jié)果，SUM_OVER_BATCH_SIZE無非是SUM/batch_size。那么，為什么要調(diào)用它呢SUM_OVER_BATCH_SIZE？實際上是SUM將整個批次的損失相加，同時SUM_OVER_BATCH_SIZE計算該批次的平均損失。SUM_OVER_BATCH_SIZE我關(guān)于和的運作的假設(shè)是否SUM正確？
1 回答
犯罪嫌疑人X
TA貢獻2080條經(jīng)驗 獲得超4個贊
據(jù)我了解，您的假設(shè)是正確的。
如果您檢查 github [keras/losses_utils.py][1] 第 260-269 行，您將看到它確實按預期執(zhí)行。 SUM將總結(jié)批量維度中的損失，并SUM_OVER_BATCH_SIZE除以總SUM損失數(shù)（批量大?。?/p>
def reduce_weighted_loss(weighted_losses,
reduction=ReductionV2.SUM_OVER_BATCH_SIZE):
if reduction == ReductionV2.NONE:
loss = weighted_losses
else:
loss = tf.reduce_sum(weighted_losses)
if reduction == ReductionV2.SUM_OVER_BATCH_SIZE:
loss = _safe_mean(loss, _num_elements(weighted_losses))
return loss
您只需添加一對損失為零的輸出即可對前面的示例進行簡單檢查。
y_true = [[0., 2.], [0., 0.],[1.,1.]]
y_pred = [[3., 1.], [2., 5.],[1.,1.]]
mae = tf.keras.losses.MeanAbsoluteError(reduction=tf.keras.losses.Reduction.SUM)
mae(y_true, y_pred).numpy()
> 5.5
mae = tf.keras.losses.MeanAbsoluteError()
mae(y_true, y_pred).numpy()
> 1.8333
所以，你的假設(shè)是正確的。[1]：https://github.com/keras-team/keras/blob/v2.7.0/keras/utils/losses_utils.py#L25-L84
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