3

只是想知道是否有一种方法可以在一个时期的中间保存最高的准确率和最低的损失，并将其用作下一个时期的得分。通常我的数据以 43.56% 的准确率达到最大值，但我已经看到它在一个时期的中间一直上升到 46% 以上。有没有办法让我在那个时候停止时代并用它来让比分继续前进？

这是我现在正在运行的代码

import pandas as pd
import numpy as np
import pickle
import random
from skopt import BayesSearchCV
from sklearn.neural_network import MLPRegressor
from tensorflow.keras.models import Sequential
from tensorflow.keras.layers import Dense, Dropout, LSTM, Bidirectional, SimpleRNN, GRU
from tensorflow.keras.wrappers.scikit_learn import KerasRegressor
from tensorflow.keras import layers
import tensorflow_docs as tfdocs
import tensorflow as tf
from tensorflow.keras.callbacks import EarlyStopping, ModelCheckpoint
import tensorflow_docs.modeling
from tensorflow import keras
import warnings
warnings.filterwarnings("ignore")
warnings.filterwarnings('ignore', category=DeprecationWarning)

train_df = avenues[["LFA's", "Spend"]].sample(frac=0.8,random_state=0)
test_df = avenues[["LFA's", "Spend"]].drop(train_df.index)
train_df = clean_dataset(train_df)
test_df = clean_dataset(test_df)
train_df = train_df.reset_index(drop=True)
test_df = test_df.reset_index(drop=True)
train_stats = train_df.describe()
train_stats = train_stats.pop("LFA's")
train_stats = train_stats.transpose()
train_labels = train_df.pop("LFA's").values
test_labels = test_df.pop("LFA's").values
normed_train_data = np.array(norm(train_df)).reshape((train_df.shape[0], 1, 1))
normed_test_data = np.array(norm(test_df)).reshape((test_df.shape[0], 1, 1))
model = KerasRegressor(build_fn=build_model, epochs=25, 
                                   batch_size=1, verbose=0)
gs = BayesSearchCV(model, param_grid, cv=3, n_iter=25, n_jobs=1,
                               optimizer_kwargs={'base_estimator': 'RF'},
                               fit_params={"callbacks": [es_acc, es_loss, tfdocs.modeling.EpochDots()]})
try:
     gs.fit(normed_train_data, train_labels)
except Exception as e:
     print(e)
4

1 回答 1

1

尝试使用train_on_batch而不是fit. 通过这种方式，您可以控制您的 epoch 在您想要的每批之后停止（尽管从机器学习的角度来看，我怀疑这是否是一个好主意，最后您会得到不太通用的模型）。

于 2020-01-14T01:02:16.367 回答