Question

我必须使用Keras在R中创建自定义丢失函数。我在R语言中的损失函数是微不足道的：

lossTradingReturns = function(y_true, y_pred) {
  y_true_diff = na.trim(diff(log(y_true)))
  y_pred_diff = na.trim(diff(log(y_pred)))
  sum( -(sign(y_pred_diff) * y_true_diff) )
}

我在Keras for R中翻译了如下：

lossTradingReturns = function(y_true, y_pred) {
   y_true_log = k_log(y_true)
   y_pred_log = k_log(y_pred)
   y_true_diff = y_true_log[2:batch_size] - y_true_log[1:(batch_size-1)]
   y_pred_diff = y_pred_log[2:batch_size] - y_pred_log[1:(batch_size-1)]
   y_true_diff = k_reshape(y_true_diff, (batch_size-1))
   y_pred_diff = k_reshape(y_pred_diff, (batch_size-1))
   return (k_sum( -(k_sign(y_pred_diff) * y_true_diff) ))
}

我的函数进行差分（y_t - y_t0）所以我从1024个元素（batch_size）开始，但最后我只有1023个元素来计算回报。错误消息表明我需要1024，我不明白为什么：该函数必须只返回一个标量...

无论如何，如果我错了，并且函数输出必须是1024张量是正确的，我怎样才能扩展1023张量，增加零值？

提前致谢

运行时

错误消息：

Error in py_call_impl(callable, dots$args, dots$keywords) :
InvalidArgumentError: Input to reshape is a tensor with 1024 values, but the requested shape has 1023
[[Node: loss_19/dense_2_loss/Reshape = Reshape[T=DT_FLOAT, Tshape=DT_INT32, _device="/job:localhost/replica:0/task:0/device:CPU:0"](loss_19/dense_2_loss/Sub, loss_19/dense_2_loss/Reshape_1/shape)]]

Caused by op u'loss_19/dense_2_loss/Reshape', defined at:
File "/home/peroni/.virtualenvs/r-tensorflow/lib/python2.7/site-packages/keras/models.py", line 863, in compile
**kwargs)
File "/home/peroni/.virtualenvs/r-tensorflow/lib/python2.7/site-packages/keras/engine/training.py", line 830, in compile
sample_weight, mask)

在收到评论后澄清我的策略：批量样本在设计上是连续的，但我正在检查以确保这一点！（感谢您的建议）。这是我使用的函数（shuffle = FALSE）来选择它们。也许你可以向我证实。

# data — The original array of floating-point data, which you normalized in listing 6.32.
# lookback — How many timesteps back the input data should go.
# delay — How many timesteps in the future the target should be.
# min_index and max_index — Indices in the data array that delimit which timesteps to draw from. This is useful for keeping a segment of the data for validation and another for testing.
# shuffle — Whether to shuffle the samples or draw them in chronological order.
# batch_size — The number of samples per batch.
# step — The period, in timesteps, at which you sample data. You’ll set it 6 in order to draw one data point every hour.
generator = function(data, Y.column=1, lookback, delay, min_index, max_index, shuffle = FALSE, batch_size = 128, step = 6, is_test = FALSE) {
  if (is.null(max_index))
    max_index <- nrow(data) - delay - 1
  i <- min_index + lookback
  function() {
    if (shuffle) {
      rows <- sample(c((min_index+lookback):max_index), size = batch_size)
    } else {
      if (i + batch_size >= max_index)
        i <<- min_index + lookback
      rows <- c(i:min(i+batch_size-1, max_index))
  i <<- i + length(rows)
    }
    samples <- array(0, dim = c(length(rows), 
                                lookback / step,
                                dim(data)[[-1]]-1))
    targets <- array(0, dim = c(length(rows)))

    for (j in 1:length(rows)) {
      indices <- seq(rows[[j]] - lookback, rows[[j]]-1, 
                     length.out = dim(samples)[[2]])
      samples[j,,] <- data[indices, -Y.column]
      targets[[j]] <- data[rows[[j]] + delay, Y.column]
    }            

    if (!is_test)
      return (list(samples, targets))
    else
      return (list(samples))
  }
}

我评估了之前区分信号的假设（使它们静止不动）但这改变了我的NN工作策略并导致非常低的性能......

Answer 1

错误发生在k_reshape。但是，您的损失函数不需要此重塑步骤，因为如果您离开k_sum，axis=NULL将取代张量的所有元素的均值。

以下损失功能对我来说很好：

lossTradingReturns = function(y_true, y_pred) {
   y_true_log = k_log(y_true)
   y_pred_log = k_log(y_pred)
   y_true_diff = y_true_log[2:batch_size] - y_true_log[1:(batch_size-1)]
   y_pred_diff = y_pred_log[2:batch_size] - y_pred_log[1:(batch_size-1)]
   return (k_sum( -(k_sign(y_pred_diff) * y_true_diff) ))
}

然而，这种损失功能对我来说看起来很奇怪。数据集在训练期间进行了混洗，因此每个时期的迷你批次不同。因此，在损失函数中取y的差异是没有意义的，因为减去的观察是完全随机的。在训练模型之前，您是不是希望立即在整个数据集中区分y变量？

Keras的自定义损失函数（R语言）

1 个答案: