为什么平均平均百分比误差(mape)极高?

Sri*_*a G 6 python machine-learning neural-network keras tensorflow

我已经从machinelearningmastery获得了代码

我修改了model.compile()函数中添加MAPE指标,找出平均绝对误差百分比。运行代码后,每个时期的mape都非常大,将其视为百分比指标。我缺少明显的东西还是输出正确?输出如下:

Epoch 91/100
0s - loss: 0.0103 - mean_absolute_percentage_error: 1764997.4502
Epoch 92/100
0s - loss: 0.0103 - mean_absolute_percentage_error: 1765653.4924
Epoch 93/100
0s - loss: 0.0102 - mean_absolute_percentage_error: 1766505.5107
Epoch 94/100
0s - loss: 0.0102 - mean_absolute_percentage_error: 1766814.5450
Epoch 95/100
0s - loss: 0.0102 - mean_absolute_percentage_error: 1767510.8146
Epoch 96/100
0s - loss: 0.0101 - mean_absolute_percentage_error: 1767686.9054
Epoch 97/100
0s - loss: 0.0101 - mean_absolute_percentage_error: 1767076.2169
Epoch 98/100
0s - loss: 0.0100 - mean_absolute_percentage_error: 1767014.8481
Epoch 99/100
0s - loss: 0.0100 - mean_absolute_percentage_error: 1766592.8125
Epoch 100/100
0s - loss: 0.0100 - mean_absolute_percentage_error: 1766348.6332
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我运行的代码(省略了预测部分)如下:

import numpy
from numpy import array
import matplotlib.pyplot as plt
from pandas import read_csv
import math
from keras.models import Sequential
from keras.layers import Dense
from keras.layers import LSTM
from sklearn.preprocessing import MinMaxScaler
from sklearn.metrics import mean_squared_error

# convert an array of values into a dataset matrix
def create_dataset(dataset, look_back=1):
        dataX, dataY = [], []
        for i in range(len(dataset)-look_back-1):
                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
numpy.random.seed(7)
# load the dataset
dataframe = read_csv('airlinepassdata.csv', usecols=[1], engine='python', skipfooter=3)
dataset = dataframe.values

#dataset = array([0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0])
dataset = dataset.astype('float32')
# normalize the dataset
scaler = MinMaxScaler(feature_range=(0, 1))
dataset = scaler.fit_transform(dataset)
# split into train and test sets
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)
# reshape input to be [samples, time steps, features]
trainX = numpy.reshape(trainX, (trainX.shape[0], 1, trainX.shape[1]))
testX = numpy.reshape(testX, (testX.shape[0], 1, testX.shape[1]))
# create and fit the LSTM network
model = Sequential()
model.add(LSTM(4, input_shape=(1, look_back)))
model.add(Dense(1))
model.compile(loss='mse', optimizer='adam', metrics=['mape'])
model.fit(trainX, trainY, nb_epoch=100, batch_size=50, verbose=2)
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小智 11

我通过keras.backend.set_epsilon(1)在调用编译之前将模糊因子 epsilon 设置为 1 来解决这个问题。

提示在源代码中

def mean_absolute_percentage_error(y_true, y_pred):
diff = K.abs((y_true - y_pred) / K.clip(K.abs(y_true),
                                        K.epsilon(),
                                        None))
return 100. * K.mean(diff, axis=-1)
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这意味着,由于某种未知原因,K.abs(y_true)训练集上 MAPE 计算中的术语低于模糊默认值 (1e-7),因此它使用该默认值代替,因此是巨大的数字。

  • 将 K.epsilon 设置为 1 可确保分母始终为 1。在这种情况下,您实际上不是在执行平均绝对百分比误差,而是在执行平均绝对误差。也就是说,你可以通过传递 `mae` 作为损失函数而不是 `mape` 来完成同样的事情。对我来说似乎有点奇怪的是,keras 在这里使用 K.abs() 而不是 np.linalg.norm() 或其他一些查找标签向量长度的方法。他们不是采用预测与真实的向量差异,而是采用每个组件的差异,因此任何 0 都有无限损失。这对于 1-hot 失败 (3认同)
  • 在您的情况下,我认为这是因为您正在以 0-1 的范围调用 `MinMaxScaler`: `scaler = MinMaxScaler(feature_range=(0, 1)) dataset = scaler.fit_transform(dataset)` 这保证了最小值y_true 的值为 0。因为它不能除以 0,所以必须将 0 视为 `K.epsilon`,返回一个大数而不是无穷大。 (2认同)