0

我正在使用 h2o 给出的示例进行心电图异常检测。当尝试手动计算 MSE 时，我得到了不同的结果。为了证明差异，我使用了最后一个测试用例，但所有 23 个用例都不同。附上完整代码：

谢谢，伊莱。

suppressMessages(library(h2o))
localH2O = h2o.init(max_mem_size = '6g', # use 6GB of RAM of *GB available
                nthreads = -1) # use all CPUs (8 on my personal computer :3)

# Download and import ECG train and test data into the H2O cluster
train_ecg <- h2o.importFile(path = "http://h2o-public-test-data.s3.amazonaws.com/smalldata/anomaly/ecg_discord_train.csv",
                          header = FALSE,
                          sep = ",")
test_ecg <- h2o.importFile(path = "http://h2o-public-test-data.s3.amazonaws.com/smalldata/anomaly/ecg_discord_test.csv",
                         header = FALSE,
                         sep = ",")
# Train deep autoencoder learning model on "normal"
# training data, y ignored
anomaly_model <- h2o.deeplearning(x = names(train_ecg),
                                 training_frame = train_ecg,
                                 activation = "Tanh",
                                 autoencoder = TRUE,
                                 hidden = c(50,20,50),
                                 l1 = 1e-4,
                                 epochs = 100)

# Compute reconstruction error with the Anomaly
# detection app (MSE between output layer and input layer)
recon_error <- h2o.anomaly(anomaly_model, test_ecg)

# Pull reconstruction error data into R and
# plot to find outliers (last 3 heartbeats)
recon_error <- as.data.frame(recon_error)
recon_error
plot.ts(recon_error)
test_recon <- h2o.predict(anomaly_model, test_ecg)

t <- as.vector(test_ecg[23,])
r <- as.vector(test_recon[23,])
mse.23 <- sum((t-r)^2)/length(t)
mse.23
recon_error[23,]

> mse.23
[1] 2.607374
> recon_error[23,]
[1] 8.264768
4

2 回答 2

1

这不是一个真正的答案，但我做了@Arno Candel 的建议。我尝试将测试和训练数据组合起来并规范化为 0 - 1。之后，我将组合和规范化的数据拆分回测试和训练数据，并运行 OP 生成的脚本。但是，我仍然使用手动计算得到不同的 MSE。当我分别对测试和训练数据进行归一化时，MSE 也有所不同。我可以做些什么来正确地进行手动计算吗？

suppressMessages(library(purrr))
suppressMessages(library(dplyr))
suppressMessages(library(h2o))

localH2O = h2o.init(max_mem_size = '6g', # use 6GB of RAM of *GB available
                nthreads = -1) # use all CPUs (8 on my personal computer :3)

# Download and import ECG train and test data into the H2O cluster
train_ecg <- h2o.importFile(path = "http://h2o-public-test-data.s3.amazonaws.com/smalldata/anomaly/ecg_discord_train.csv",
                          header = FALSE,
                          sep = ",")
test_ecg <- h2o.importFile(path = "http://h2o-public-test-data.s3.amazonaws.com/smalldata/anomaly/ecg_discord_test.csv",
                         header = FALSE,
                         sep = ",")
### adding this section
# normalize data 
train_ecg <- as.data.frame(train_ecg)
test_ecg <- as.data.frame(test_ecg)

dat <- rbind(train_ecg,test_ecg)

get_desc <- function(x) {
  map(x, ~list(
    min = min(.x),
    max = max(.x),
    mean = mean(.x),
    sd = sd(.x)
  ))
}

normalization_minmax <- function(x, desc) {
  map2_dfc(x, desc, ~(.x - .y$min)/(.y$max - .y$min))
}

desc <- dat %>%
  get_desc()

dat <- dat %>%
  normalization_minmax(desc)

train_ecg  <- as.matrix(dat[1:20,]) ; test_ecg <- as.matrix(dat[21:43,])

# Train deep autoencoder learning model on "normal"
# training data, y ignored
anomaly_model <- h2o.deeplearning(x = names(train_ecg),
                                 training_frame = train_ecg,
                                 activation = "Tanh",
                                 autoencoder = TRUE,
                                 hidden = c(50,20,50),
                                 l1 = 1e-4,
                                 epochs = 100)

# Compute reconstruction error with the Anomaly
# detection app (MSE between output layer and input layer)
recon_error <- h2o.anomaly(anomaly_model, test_ecg)

# Pull reconstruction error data into R and
# plot to find outliers (last 3 heartbeats)
recon_error <- as.data.frame(recon_error)
recon_error
plot.ts(recon_error)
test_recon <- h2o.predict(anomaly_model, test_ecg)

t <- as.vector(test_ecg[23,])
r <- as.vector(test_recon[23,])
mse.23 <- sum((t-r)^2)/length(t)
mse.23
recon_error[23,]


> mse.23
[1] 23.14947
> recon_error[23,]
[1] 8.076866
于 2019-04-10T03:28:25.230 回答
0

对于 H2O 中的自动编码器，MSE 数学在归一化空间中完成，以避免数值缩放问题。例如，如果您有分类特征或非常大的数字，则神经网络自动编码器不能直接对这些数字进行操作，而是首先对数字特征进行虚拟 one-hot 编码和归一化，然后进行 fwd/back重建误差的传播和计算（在归一化和扩展的空间中）。对于纯数字数据，您可以首先手动将每列除以其范围（最大值-最小值），并且您的结果应该匹配。

这是一个明确执行此检查的 JUnit（在该数据集上）： https://github.com/h2oai/h2o-3/blob/master/h2o-algos/src/test/java/hex/deeplearning/DeepLearningAutoEncoderTest。 java#L86-L104

您还可以查看https://0xdata.atlassian.net/browse/PUBDEV-2078了解更多信息。

于 2016-01-20T11:06:18.930 回答