Azure ML调整模型超级参数 [英] Azure ML Tune Model Hyper Parameters

问题描述

这是在70-774考试参考书的本章结尾处提出的问题.

Here's question proposed at the end of the chapter in 70-774 exam reference book.

如果您将神经网络与已配置的Tune Model Hyperparameters模块连接 随机扫描且随机扫描的最大运行次数= 1，如何 在执行实验过程中训练了许多神经网络?为什么?如果你 将验证数据集连接到Tune Model超参数的第三个输入 模块，现在训练了多少个神经网络?

If you connect a neural network with a Tune Model Hyperparameters module configured with Random Sweep and Maximum number of runs on random sweep = 1, how many neural networks are trained during the execution of the experiment? Why? If you connect a validation dataset to the third input of the Tune Model Hyperparameters module, how many neural networks are trained now?

答案是:

没有验证数据集11(k倍交叉验证中的10 +用所有数据训练的1 与超参数的最佳组合).验证集只有1个神经 网络经过培训，因此如果您提供验证集，则不会使用最佳模型来培训最佳模型 它.

Without validation dataset 11 (10 of k-fold cross validation + 1 trained with all the data with the best combination of hyperparameters). With the validation set only 1 neural network is trained, so the best model is not trained using the validation set if you provide it.

10是哪里来的?据我了解，该数字应分别为2和1.它不应该创建n折，而n等于行程数吗?

Where does 10 come from? As far as I understand the number should be 2 and 1 respectively. Shouldn't it create n-folds where n is equal to the number of runs?

推荐答案

在不使用验证数据集的情况下使用调优模型超参数"模块时，这意味着，当您仅使用第二个输入数据端口时，该模块可以进行交叉验证模式.因此，可以通过对提供的数据集进行交叉验证来找到最佳参数模型，然后将数据集拆分为k倍.默认情况下，模块将数据分成10折.如果您想将数据拆分为不同的折数，则可以在第二个输入处连接分区和样本"模块，选择分配到折数"并指出所需的折数.在许多情况下，k = 5是一个合理的选择.

When you use the Tune Model Hyperparameters module without a validation dataset, this means, when you use only the 2nd input data port, the module works in cross-validation mode. So the best-parameters model is found by doing cross-validation over the provided dataset, and to do this, the dataset is splitted in k-folds. By default, the module splits the data in 10 folds. In case you want to split the data in a different number of folds, you can connect a Partition and Sample module at the 2nd input, selecting Assign to Folds and indicating the number of folds desired. In many cases k=5 is a reasonable option.

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