wsp*_*irs 1 python numpy tensorflow softmax
我有以下基于MNIST示例的代码。它有两种修改方式:
1)我没有使用单热向量,所以我只使用 tf.equal(y, y_)
2)我的结果是二进制的:0或1
import tensorflow as tf
import numpy as np
# get the data
train_data, train_results = get_data(2000, 2014)
test_data, test_results = get_data(2014, 2015)
# setup a session
sess = tf.Session()
x_len = len(train_data[0])
y_len = len(train_results[0])
# make placeholders for inputs and outputs
x = tf.placeholder(tf.float32, shape=[None, x_len])
y_ = tf.placeholder(tf.float32, shape=[None, y_len])
# create the weights and bias
W = tf.Variable(tf.zeros([x_len, 1]))
b = tf.Variable(tf.zeros([1]))
# initialize everything
sess.run(tf.initialize_all_variables())
# create the "equation" for y in terms of x
y_prime = tf.matmul(x, W) + b
y = tf.nn.softmax(y_prime)
# construct the error function
cross_entropy = tf.nn.softmax_cross_entropy_with_logits(y_prime, y_)
# setup the training algorithm
train_step = tf.train.GradientDescentOptimizer(0.01).minimize(cross_entropy)
# train the thing
for i in range(1000):
rand_rows = np.random.choice(train_data.shape[0], 100, replace=False)
_, w_out, b_out, ce_out = sess.run([train_step, W, b, cross_entropy], feed_dict={x: train_data[rand_rows, :], y_: train_results[rand_rows, :]})
print("%d: %s %s %s" % (i, str(w_out), str(b_out), str(ce_out)))
# compute how many times it was correct
correct_prediction = tf.equal(y, y_)
# find the accuracy of the predictions
accuracy = tf.reduce_mean(tf.cast(correct_prediction, "float"))
print(sess.run(accuracy, feed_dict={x: test_data, y_: test_results}))
for i in range(0, len(test_data)):
res = sess.run(y, {x: [test_data[i]]})
print("RES: " + str(res) + " ACT: " + str(test_results[i]))
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精度始终为0.5(因为我的测试数据的1s与0s差不多)。的值W,并b似乎总是增加,可能是因为值cross_entropy总是全部为零的向量。
当我尝试将此模型用于预测时,预测始终为1:
RES: [[ 1.]] ACT: [ 0.]
RES: [[ 1.]] ACT: [ 1.]
RES: [[ 1.]] ACT: [ 0.]
RES: [[ 1.]] ACT: [ 1.]
RES: [[ 1.]] ACT: [ 0.]
RES: [[ 1.]] ACT: [ 1.]
RES: [[ 1.]] ACT: [ 0.]
RES: [[ 1.]] ACT: [ 0.]
RES: [[ 1.]] ACT: [ 1.]
RES: [[ 1.]] ACT: [ 0.]
RES: [[ 1.]] ACT: [ 1.]
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我在这里做错了什么?
您似乎在预测单个标量,而不是矢量。softmax op会为每个示例生成矢量值的预测。此向量必须总和为1。当向量仅包含一个元素时,该元素必须始终为1。如果要使用softmax解决此问题,可以将[1,0]用作当前输出的目标使用[0],并在当前使用[1]的地方使用[0,1]。另一种选择是您可以继续使用一个数字,但是将输出层更改为Sigmoid而不是softmax,并将cost函数也更改为基于Sigmoid的cost函数。
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