Question

为了分析数据，我需要每个输出维的损失，而我只得到一个损失，我怀疑这是所有输出维的损失的平均值。

任何帮助您了解我的损失是什么，以及如何为每个输出获得单独的损失：

2019-08-30 21:16:57 [scrapy.core.engine] INFO: Spider opened
2019-08-30 21:16:57 [scrapy.extensions.logstats] INFO: Crawled 0 pages (at 0 pages/min), scraped 0 items (at 0 items/min)
2019-08-30 21:16:57 [scrapy.extensions.telnet] INFO: Telnet console listening on 127.0.0.1:6023
2019-08-30 21:16:57 [scrapy.core.engine] DEBUG: Crawled (200) <GET https://www.zooplus.es/robots.txt> (referer: None)
2019-08-30 21:16:57 [scrapy.downloadermiddlewares.redirect] DEBUG: Redirecting (301) to <GET https://www.zooplus.es/shop/tienda_perros/pienso_perros/pienso_hipoalergenico> from <GET https://www.zooplus.es/shop/tienda_perros/pienso_perros/pienso_hipoalergenico/>
2019-08-30 21:16:58 [scrapy.core.engine] DEBUG: Crawled (200) <GET https://www.zooplus.es/shop/tienda_perros/pienso_perros/pienso_hipoalergenico> (referer: None)
2019-08-30 21:16:58 [scrapy.core.scraper] ERROR: Spider error processing <GET https://www.zooplus.es/shop/tienda_perros/pienso_perros/pienso_hipoalergenico> (referer: None)

我得到：

import numpy as np
import matplotlib.pyplot as plt
import tensorflow as tf
from scipy import stats
from keras import models 
from keras.models import Sequential
from keras.layers import Dense, Dropout
from keras import optimizers
from sklearn.model_selection import KFold

siz=100000
inp0=np.random.randint(100, 1000000 , size=(siz,3))
rand0=np.random.randint(-100, 100 , size=(siz,2))
a1=0.2;a2=0.8;a3=2.5;a4=2.6;a5=1.2;a6=0.3
oup1=np.dot(inp0[:,0],a1)+np.dot(inp0[:,1],a2)+np.dot(inp0[:,2],a3)\
+rand0[:,0]
oup2=np.dot(inp0[:,0],a4)+np.dot(inp0[:,1],a5)+np.dot(inp0[:,2],a6)\
 +rand0[:,1]
 oup_tot=np.concatenate((oup1.reshape(siz,1), oup2.reshape(siz,1)),\
                   axis=1)
normzer_inp = MinMaxScaler()
 inp_norm = normzer_inp.fit_transform(inp0)
 normzer_oup = MinMaxScaler()
 oup_norm = normzer_oup.fit_transform(oup_tot)
 X=inp_norm
Y=oup_norm
kfold = KFold(n_splits=2, random_state=None, shuffle=False)
opti_SGD = SGD(lr=0.01, momentum=0.9)
model1 = Sequential()
for train, test in kfold.split(X, Y):
    model = Sequential()
    model.add(Dense(64, input_dim=X.shape[1], activation='relu'))
    model.add(Dense(64, activation='relu'))
    model.add(Dense(64, activation='relu'))
    model.add(Dense(Y.shape[1], activation='linear'))
    model.compile(loss='mean_squared_error', optimizer=opti_SGD)

    history = model.fit(X[train], Y[train], \
            validation_data=(X[test], Y[test]), \
            epochs=100,batch_size=2048, verbose=2)

我想知道我现在所遭受的损失以及如何在每个输出维度上获得损失。

Answer 1

将函数列表传递给compile函数中的metrics参数。看到这里：https://keras.io/metrics/#custom-metrics

import keras.backend as K

...

def loss_first_dim(y_true, y_pred):
    return K.mean(K.square(y_pred[:, 0] - y_true[:, 0]))

model.compile(optimizer='rmsprop',
              loss='binary_crossentropy',
              metrics=[loss_first_dim])

...

如何在多输出回归中获得每个输出的损失？

1 个答案: