用于时间序列的 Keras Conv1D

Question

我只是深度学习领域的新手。

我对 Keras Conv1D 进行了第一次基本尝试。不确定我做了什么以及我是否做对了。我的输入数据只是美国各地商店每周的总销售额（总共 313 周），时间步长为 1。

这是我的代码：

from pandas import read_csv
import matplotlib.pyplot as plt
import numpy
from keras.datasets import imdb
from keras.models import Sequential
from keras.layers import Dense
from keras.layers import Flatten
from keras.layers.convolutional import Conv1D
from keras.layers.convolutional import MaxPooling1D
from keras.layers.embeddings import Embedding
from keras.preprocessing import sequence
from sklearn.preprocessing import MinMaxScaler
from sklearn.metrics import mean_squared_error

def create_dataset(dataset, look_back=1):
    dataX, dataY = [], []
    for i in range(len(dataset)-look_back):
        a = dataset[i:(i+look_back), 0]
        dataX.append(a)
        dataY.append(dataset[i + look_back, 0])
    return numpy.array(dataX), numpy.array(dataY)
seed = 7
numpy.random.seed(seed)
dataframe = read_csv('D:/MIS793/Dataset/Academic Dataset External 2/Python scripts/totalsale _byweek.csv', usecols=[1], engine='python')
plt.plot(dataframe)
plt.show()
dataset = dataframe.values
dataset = dataset.astype('float32')
# normalize the dataset
scaler = MinMaxScaler(feature_range=(0, 1))
dataset = scaler.fit_transform(dataset)
train_size = int(len(dataset) * 0.67)
test_size = len(dataset) - train_size
train, test = dataset[0:train_size,:], dataset[train_size:len(dataset),:]
# reshape into X=t and Y=t+1
look_back = 1
trainX, trainY = create_dataset(train, look_back)
testX, testY = create_dataset(test, look_back)
trainX = trainX.reshape(trainX.shape[0], trainX.shape[1], 1).astype('float32')
testX = testX.reshape(testX.shape[0], testX.shape[1], 1).astype('float32')
model = Sequential()
model.add(Conv1D(filters=10, kernel_size=1, padding='same', strides=1, activation='relu',input_shape=(1,1)))
model.add(MaxPooling1D(pool_size=1))
model.add(Flatten())
model.add(Dense(250, activation='relu'))
model.add(Dense(1, activation='linear'))
model.compile(loss='mse', optimizer='adam', metrics=['mae'])
print(model.summary())
model.fit(trainX, trainY, validation_data=(testX, testY), epochs=10, batch_size=100)
scores = model.evaluate(testX, testY, verbose=0)
print("Accuracy: %.2f%%" % (scores[1]*100))

这里有几件事不确定：

重塑 trainX 和 testX。
kernel_size 和 input_shape

我的想法是它只是销售价值的一个向量。 10 个过滤器，每个大小为 1，从一个值移动到另一个值。输入形状的格式为时间步长、维度。

我的准确率只有10.91%！所以我的第一个问题是我是否输入了正确的参数。

谢谢

升序

Answer 1

使用 model.metrics_names 您可以获得 scores 变量的标签。

在你的情况下它将是 ['loss', 'mean_absolute_error'].

所以你打印的不是 accuracy，而是 mae，乘以 100。

Answer 2

我尝试使用准确性而不是 mae。但是我得到的准确度是 0%。只是想知道这是关于预测数值，我真的应该使用准确性吗？这是我最新的代码。

from pandas import read_csv
import matplotlib.pyplot as plt
import numpy
from keras.datasets import imdb
from keras.models import Sequential
from keras.layers import Dense
from keras.layers import Flatten
from keras.layers import Dropout
from keras.layers.convolutional import Conv1D
from keras.layers.convolutional import MaxPooling1D
from keras.layers.embeddings import Embedding
from keras.preprocessing import sequence
from sklearn.preprocessing import MinMaxScaler
from sklearn.metrics import mean_squared_error


def create_dataset(dataset, look_back=1):
    dataX, dataY = [], []
    for i in range(len(dataset)-look_back):
        a = dataset[i:(i+look_back), 0]
        dataX.append(a)
        dataY.append(dataset[i + look_back, 0])
    return numpy.array(dataX), numpy.array(dataY)

# fix random seed for reproducibility
seed = 7
numpy.random.seed(seed)

dataframe = read_csv('D:/MIS793/Dataset/Academic Dataset External 2/Python scripts/totalsale _byweek.csv', usecols=[1], engine='python')
plt.plot(dataframe)
plt.show()

dataset = dataframe.values
dataset = dataset.astype('float32')
# normalize the dataset
scaler = MinMaxScaler(feature_range=(0, 1))
dataset = scaler.fit_transform(dataset)
train_size = int(len(dataset) * 0.67)
test_size = len(dataset) - train_size
train, test = dataset[0:train_size,:], dataset[train_size:len(dataset),:]

# reshape into X=t and Y=t+1
look_back = 1
trainX, trainY = create_dataset(train, look_back)
testX, testY = create_dataset(test, look_back)

trainX = trainX.reshape(trainX.shape[0], trainX.shape[1],1).astype('float32')
testX = testX.reshape(testX.shape[0], testX.shape[1],1).astype('float32')

model = Sequential()
model.add(Conv1D(filters=20, kernel_size=1, padding='same', strides=1, activation='relu',input_shape=(1,1)))
model.add(MaxPooling1D(pool_size=1))
model.add(Conv1D(filters=10, kernel_size=1, padding='same', strides=1, activation='relu'))
model.add(MaxPooling1D(pool_size=1))
model.add(Flatten())
model.add(Dense(4, activation='relu'))
model.add(Dense(1, activation='linear'))
model.compile(loss='mse', optimizer='adam', metrics=['accuracy'])
print(model.summary())
model.fit(trainX, trainY, validation_data=(testX, testY), epochs=10, batch_size=100)
scores = model.evaluate(testX, testY, verbose=0)
print("Accuracy: %.2f%%" % (scores[1]*100))

或者我应该和 MAE 一起去？

如果我选择 MAE，我的分数将如下所示：

[0.12740663779013364, 0.31208728355111426]

第一个是loss，第二个是MAE。在这种情况下，这不是更好的指标吗？

最后一行是这样的：

print("MAE: %.2f%%" % (分数[1]))

谢谢阿宁迪亚

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