乙状结肠功能预测导出到DF

时间:2019-05-15 18:27:13

标签: python tensorflow machine-learning nlp sigmoid

我是tensorflow的新手,所以我试图通过解决kaggle上的二进制分类问题来弄脏我的手。我已经使用Sigmoid函数训练了模型,并在测试时获得了非常好的准确性,但是当我尝试将预测导出到df进行提交时,出现以下错误...我已经附上了代码以及预测和输出,请建议我做错了,我怀疑这与我的S型函数有关,谢谢。

This is output of the predictions....the expected is 1s and 0s

INFO:tensorflow:Restoring parameters from ./movie_review_variables
Prections are [[3.8743019e-07]
 [9.9999821e-01]
 [1.7650980e-01]
 ...
 [9.9997473e-01]
 [1.4901161e-07]
 [7.0333481e-06]]
#Importing tensorflow
import tensorflow as tf
#defining hyperparameters
learning_rate = 0.01
training_epochs = 1000
batch_size = 100
num_labels = 2
num_features = 5000
train_size = 20000

#defining the placeholders and encoding the y placeholder
X = tf.placeholder(tf.float32, shape=[None, num_features])
Y = tf.placeholder(tf.int32, shape=[None])
y_oneHot = tf.one_hot(Y, 1)

#defining the model parameters -- weight and bias
W = tf.Variable(tf.zeros([num_features, 1]))
b = tf.Variable(tf.zeros([1]))

#defining the sigmoid model and setting up the learning algorithm
y_model = tf.nn.sigmoid(tf.add(tf.matmul(X, W), b))
cost = tf.nn.sigmoid_cross_entropy_with_logits(logits=y_model, labels=y_oneHot)
train_optimizer = tf.train.GradientDescentOptimizer(learning_rate).minimize(cost)

#defining operation to measure success rate
correct_prediction = tf.equal(tf.argmax(y_model, 1), tf.argmax(y_oneHot, 1))
accuracy = tf.reduce_mean(tf.cast(correct_prediction, tf.float32))

#saving variables
saver = tf.train.Saver()

#executing the graph and saving the model variables
with tf.Session() as sess: #new session
    tf.global_variables_initializer().run()

    #Iteratively updating parameter batch by batch
    for step in range(training_epochs * train_size // batch_size):
        offset = (step * batch_size) % train_size
        batch_xs = x_train[offset:(offset + batch_size), :]
        batch_labels = y_train[offset:(offset + batch_size)]
        #run optimizer on batch
        err, _ = sess.run([cost, train_optimizer], feed_dict={X:batch_xs, Y:batch_labels})
        if step % 1000 ==0:
            print(step, err) #print ongoing result
    #Print final learned parameters
    w_val = sess.run(W)
    print('w', w_val)
    b_val = sess.run(b)
    print('b', b_val)
    print('Accuracy', accuracy.eval(feed_dict={X:x_test, Y:y_test}))
    save_path = saver.save(sess, './movie_review_variables')
    print('Model saved in path {}'.format(save_path))



#creating csv file for kaggle submission
with tf.Session() as sess:
    saver.restore(sess, './movie_review_variables')
    predictions = sess.run(y_model, feed_dict={X: test_data_features})
    subm2 = pd.DataFrame(data={'id':test['id'],'sentiment':predictions})
    subm2.to_csv('subm2nlp.csv', index=False, quoting=3)
    print("I am done predicting")
INFO:tensorflow:Restoring parameters from ./movie_review_variables
---------------------------------------------------------------------------
Exception                                 Traceback (most recent call last)
<ipython-input-85-fd74ed82109c> in <module>()
      5 #     print('Prections are {}'.format(predictions))
      6 
----> 7     subm2 = pd.DataFrame(data={'id':test['id'], 'sentiment':predictions})
      8     subm2.to_csv('subm2nlp.csv', index=False, quoting=3)
      9     print("I am done predicting")

Exception: Data must be 1-dimensional

2 个答案:

答案 0 :(得分:0)

您可以看到sigmoid function的定义。这将始终具有连续输出。如果要离散化输出,则需要确定一些阈值,将其设置为1以上,将其设置为0以下。

pred = tf.math.greater(y_model, tf.constant(0.5))

但是,您必须谨慎选择合适的阈值,因为这不能保证您的模型可以通过概率很好地进行校准。您可以基于对某些保留的验证集的最佳区分来选择合适的阈值。

此步骤仅用于评估很重要,因为您将无法通过此操作反向传播损耗信号。

答案 1 :(得分:0)

您需要为S型输出设置一些阈值。例如。将输出分成两个间​​距为0.5的垃圾箱:

>>> import numpy as np
>>> x = np.linspace(0, 10, 20)
>>> x
array([ 0.        ,  0.52631579,  1.05263158,  1.57894737,  2.10526316,
        2.63157895,  3.15789474,  3.68421053,  4.21052632,  4.73684211,
        5.26315789,  5.78947368,  6.31578947,  6.84210526,  7.36842105,
        7.89473684,  8.42105263,  8.94736842,  9.47368421, 10.        ])
>>> q = 0.5     # The continuous value between two discrete points
>>> y = q * np.round(x/q)
>>> y
array([ 0. ,  0.5,  1. ,  1.5,  2. ,  2.5,  3. ,  3.5,  4. ,  4.5,  5.5,
        6. ,  6.5,  7. ,  7.5,  8. ,  8.5,  9. ,  9.5, 10. ])