Seit TensorFlow herauskam, habe ich versucht, etwas anderes als MNIST zu lernen. Wir werden eine Regression durchführen, um eine Korrelation zu erstellen. (Addition) Da die Ausgabe der Zwischenschicht für die Anzahl der Daten zu groß war, wurde der Wert der Zwischenschicht korrigiert.
--Die verwendeten Daten sind Diaveten
import sklearn
import tensorflow as tf
from sklearn import datasets
import numpy as np
diabetes = datasets.load_diabetes()
#Lade Daten
print "load diabetes data"
data = diabetes["data"].astype(np.float32)
target = diabetes['target'].astype(np.float32).reshape(len(diabetes['target']), 1)
#Unterteilt in Trainingsdaten und Testdaten
N=342
x_train, x_test = np.vsplit(data, [N])
y_train, y_test = np.vsplit(target, [N])
N_test = y_test.size
x= tf.placeholder("float",shape=[None,10])
#1. Schicht Eingang 10 Ausgang 256
with tf.name_scope('l1') as scope:
weightl1 = tf.Variable(tf.truncated_normal([10, 256], stddev=0.1),name="weightl1")
biasel1 = tf.Variable(tf.constant(1.0, shape=[256]), name="biasel1")
outputl1=tf.nn.relu(tf.matmul(x,weightl1) + biasel1)
#2. Schicht Eingang 256 Ausgang 1
with tf.name_scope('l2') as scope:
weightl2 = tf.Variable(tf.truncated_normal([256, 1], stddev=0.1),name="weightl2")
biasel2 = tf.Variable(tf.constant(1.0, shape=[1]), name="biasel2")
outputl2=tf.nn.relu(tf.matmul(outputl1,weightl2) + biasel2)
"""
Funktion zur Fehlerberechnung
Berechnen Sie den Fehler mit MSE
"""
def loss(output):
with tf.name_scope('loss') as scope:
loss = tf.reduce_mean(tf.square(output - y_train))
return loss
loss_op = loss(outputl2)
optimizer = tf.train.AdagradOptimizer(0.04)
train_step = optimizer.minimize(loss_op)
#Aufnahmefehler
best = float("inf")
#Initialisieren
init_op = tf.initialize_all_variables()
with tf.Session() as sess:
#drin
sess.run(init_op)
for i in range(20001):
loss_train = sess.run(loss_op, feed_dict={x:x_train})
sess.run(train_step, feed_dict={x:x_train})
if loss_train < best:
best = loss_train
best_match = sess.run(outputl2, feed_dict={x:x_test})
if i %1000 == 0:
print "step {}".format(i)
pearson = np.corrcoef(best_match.flatten(), y_test.flatten())
print 'train loss = {} ,test corrcoef={}'.format(best,pearson[0][1])
load diabetes data
step 0
train loss = 29000.1777344 ,test corrcoef=0.169487254139
step 1000
train loss = 3080.2097168 ,test corrcoef=0.717823972634
step 2000
train loss = 2969.1887207 ,test corrcoef=0.72972180486
step 3000
train loss = 2938.4609375 ,test corrcoef=0.73486349373
step 4000
train loss = 2915.63330078 ,test corrcoef=0.737497869454
step 5000
train loss = 2896.14331055 ,test corrcoef=0.739029181368
step 6000
train loss = 2875.51708984 ,test corrcoef=0.74006814362
step 7000
train loss = 2856.36816406 ,test corrcoef=0.741115477047
step 8000
train loss = 2838.77026367 ,test corrcoef=0.742113966068
step 9000
train loss = 2822.453125 ,test corrcoef=0.743066699589
step 10000
train loss = 2807.88916016 ,test corrcoef=0.743988699821
step 11000
train loss = 2795.09057617 ,test corrcoef=0.744917437376
step 12000
train loss = 2783.8828125 ,test corrcoef=0.745871358086
step 13000
train loss = 2773.68457031 ,test corrcoef=0.747112534114
step 14000
train loss = 2764.80224609 ,test corrcoef=0.748115829411
step 15000
train loss = 2756.6628418 ,test corrcoef=0.748800330555
step 16000
train loss = 2749.1340332 ,test corrcoef=0.749471871992
step 17000
train loss = 2741.78881836 ,test corrcoef=0.750184567587
step 18000
train loss = 2734.56054688 ,test corrcoef=0.750722087518
step 19000
train loss = 2727.18579102 ,test corrcoef=0.751146409281
step 20000
train loss = 2719.29101562 ,test corrcoef=0.751330770654
Es scheint, dass Sie es ganz frei einstellen können. Es scheint, dass es einige Zeit dauern wird, es zu meistern.
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