我目前在Tensorflow的机器学习项目中遇到麻烦。神经网络的输入是26x1的数字列表，而所需的输出是5x16的二进制数组。
我正在训练这些数据。这是我的testing sample和我的training sample。您可以通过右键单击网页并单击“另存为”从Git下载它。
此外，完整的代码是here，只是因为您不必从StackOverflow复制和粘贴多次。
我将检查我的代码。我认为没有必要了解其中的所有内容，但是无论如何，我都在为我的思考过程加注很多。问题的主要焦点是我是否在Tensorflow中正确地进行了机器学习。
'''Imports'''
import csv
import numpy as np
import tensorflow as tf
import tensorflow.keras as keras
np.set_printoptions(suppress=True)
''' Here we unpack the .csv files. I have chosen to put their contents
into lists. Do let me know if there exists a more efficient method. '''
distribution_train = []
probs_train = []
distribution_test = []
probs_test = []
with open('training_sample.csv') as csv_file:
csv_reader = csv.reader(csv_file, delimiter=',')
for row in csv_reader:
distribution_train.append(row[0])
probs_train.append(row[1])
with open('testing_sample.csv') as csv_file_1:
csv_reader_1 = csv.reader(csv_file_1, delimiter= ',')
for row in csv_reader_1:
distribution_test.append(row[0])
probs_test.append(row[1])
'''Get rid of the titles in the training_sample.csv file.'''
distribution_train.pop(0)
probs_train.pop(0)
我们的数据已从.csv文件中提取并存储在列表中，但是由于某些原因，数字以字符串形式存在。因此，我将定义一个提取所有数字的函数。
'''For some reason everything in my csv file is stored as strings. Or
maybe it's just because of the way I have unpacked it. The below function
is to convert it into floats so that TF can work with it.
It's crude, but it locates all the numbers and appends them to a list,
which then gets appended to a giant list called f.'''
def num_converter_flatten(csv_list):
f = []
for j in range(len(csv_list)):
append_this = []
for i in csv_list[j]:
if i == '1' or i == '2' or i == '3' or i == '4' or i == '5' or i == '6' or i == '7' or i == '8' or i =='9' or i =='0':
append_this.append(float(i))
f.append((append_this))
return f
'''Basically, this line is to convert the distribution_train and
probs_train which are currently strings
into numbers. And we normalize the training data.'''
x_train = num_converter_flatten(distribution_train)
y_train = num_converter_flatten(probs_train)
x_train = tf.keras.utils.normalize(x_train, axis=1)
现在，我们将其重塑为张量，并确保它们具有相同的浮点数。
'''This line we reshape x_train and y_train into tensors. The convertion
to float 32 is also necessary as I realised that A and B are different
floats for some reason.'''
A = tf.reshape(x_train, [-1,1*26])
B = tf.reshape(y_train, [-1,1*80])
A = tf.dtypes.cast(A, dtype = tf.float32)
B = tf.dtypes.cast(B, dtype = tf.float32)
'''Doing the same thing to x_test and y_test'''
x_test = num_converter_flatten(distribution_test)
y_test = num_converter_flatten(probs_test)
C = tf.reshape(x_test, [-1,1*26])
D = tf.reshape(y_test, [-1,1*80])
C = tf.dtypes.cast(C, dtype = tf.float32)
D = tf.dtypes.cast(D, dtype = tf.float32)
现在我们训练模型。
'''Model starts from here'''
model = tf.keras.models.Sequential()
'''I'm not too sure if relu is the right activation function to use here.
I've tried different activation functions, but all run into the same
problem described below.'''
model.add(tf.keras.layers.Dense(180, activation=keras.activations.relu, input_shape=(26,)))
model.add(tf.keras.layers.Dense(2080, activation=keras.activations.relu))
model.add(tf.keras.layers.Dense(180, activation=keras.activations.relu))
'''I'm making the final layer 80 because I want TF to output the size of
the 'probs' list in the csv file'''
model.add(tf.keras.layers.Dense(80, activation=keras.activations.softplus))
'''Again I'm not sure if softplus is the best to use here. I've also
tested a number of activation functions for the last layer, and it also
runs to the same problem.'''
model.compile(optimizer='adam',
loss='binary_crossentropy',
metrics=['accuracy'])
model.fit(A,B, epochs=2, steps_per_epoch=16)
val_loss, val_acc = model.evaluate(C,D, steps = 128)
print (val_loss, val_acc)
'''Just saving the model'''
model.save('epic_equation_model_try1')
new_model = tf.keras.models.load_model('epic_equation_model_try1')
predictions = new_model.predict(C, steps = 1)
这是主要问题。如果运行模型（持续2个时间段），则精度会提高，但此后会降低，然后再次提高。最终，如果您再运行几个纪元，它将达到0.6221的峰值。为什么在这个数字达到顶峰？同样，这不是softplus激活所独有的。我测试过的其他几个激活函数（例如tanh或softmax或Sigmoid）也可以执行相同的操作。我是机器学习的新手，但我不认为应以如此一致且低的数字达到最高水平。
其次，如果我要求我的模型进行预测，它会为我提供一个非常小的数字列表。因此，例如，如果我希望我的模型预测我的testing_sample.csv的第一个数据样本，则可以这样做：
for i in range(1):
x = np.array(predictions[i]).reshape(5,16)
print (x)
我会得到这样的东西...
[[0.00000014 0.00000065 0.0000007 0.00000031 0.00000069 0.00000003
0.00000073 0.00000009 0.00000001 0.00000095 0.00000215 0.00000045
0.00000155 0.00000274 0.00000057 0.00053975]
[0.00000016 0.00000011 0.00000021 0.00000006 0.00000012 0.00000022
0.00000002 0.00000005 0.00000019 0.00000002 0.00000087 0.00000465
0.00000238 0.00000009 0.00003278 0.00001788]
[0.00000002 0.00000001 0.00000046 0.00000131 0.00000072 0.00000006
0.00000005 0.00000001 0.00000001 0.0000003 0.0000005 0.00000016
0.00000465 0.00000226 0.00000083 0.00002015]
[0.00000005 0.0000004 0.00000001 0.00000032 0.00000008 0.00000061
0.00000107 0.00000015 0.00000013 0.00000014 0.00000012 0.00000037
0.00000334 0.00000016 0.00000057 0.00018404]
[0.00000044 0.00000038 0.00000095 0.00000013 0.0000002 0.00000006
0.00000019 0.00000087 0.00000095 0.00000016 0.00000513 0.00000095
0.00000846 0.0000534 0.00000049 0.00000429]]
基本上是一个难看的16x5矩阵。很明显，有些数字明显大于其他数字，特别是如果您在顶部禁用了np.set_printoptions(suppress=True)。我不确定这些数字是什么。目前，我假设数字越大，表示该计算机更有信心将其设置为1。
当前，这就是我要将此输出矩阵转换为二进制的操作。但这是一种相当随意的方式。
'''This tests for only the first prediction. If you wwant to see more
predictions, change the range.'''
for i in range(1):
MUTI = 500000
x = np.array(predictions[i]).reshape(5,16)
# print (x)
PX = MUTI*x
PX = np.round(PX, 2)
PX[PX<0.1] = 0
PX[PX>0.1] = 1
PX[PX==0.1] = 1
print (PX)
那么我在这里做错了什么？最好的选择是让输出矩阵对应于0和1，例如[0.999998，0.000002，0.000003，0.099995 ...]，但是到目前为止，我在Tensorflow中没有找到允许我执行此操作的函数。
此外，我是否正确执行了机器学习程序？我觉得所有激活功能的准确性都不应达到0.6221的峰值。
编辑：将最后一层的功能更改为S型后：
model.add(tf.keras.layers.Dense(80, activation=keras.activations.sigmoid))
这是模型现在的训练方式。
# Model Training
1/16 [>.............................] - ETA: 27s - loss: 0.6932 - acc: 0.4845
2/16 [==>...........................] - ETA: 20s - loss: 0.6919 - acc: 0.5277
3/16 [====>.........................] - ETA: 18s - loss: 0.6905 - acc: 0.5470
4/16 [======>.......................] - ETA: 15s - loss: 0.6889 - acc: 0.5577
5/16 [========>.....................] - ETA: 14s - loss: 0.6871 - acc: 0.5654
6/16 [==========>...................] - ETA: 12s - loss: 0.6851 - acc: 0.5720
7/16 [============>.................] - ETA: 11s - loss: 0.6831 - acc: 0.5769
8/16 [==============>...............] - ETA: 9s - loss: 0.6813 - acc: 0.5813
9/16 [===============>..............] - ETA: 8s - loss: 0.6799 - acc: 0.5851
10/16 [=================>............] - ETA: 7s - loss: 0.6787 - acc: 0.5883
11/16 [===================>..........] - ETA: 6s - loss: 0.6776 - acc: 0.5912
12/16 [=====================>........] - ETA: 4s - loss: 0.6766 - acc: 0.5936
13/16 [=======================>......] - ETA: 3s - loss: 0.6756 - acc: 0.5957
14/16 [=========================>....] - ETA: 2s - loss: 0.6746 - acc: 0.5975
15/16 [===========================>..] - ETA: 1s - loss: 0.6738 - acc: 0.5991
16/16 [==============================] - 19s 1s/step - loss: 0.6731 - acc: 0.6005
Epoch 2/2
1/16 [>.............................] - ETA: 16s - loss: 0.6626 - acc: 0.6221
2/16 [==>...........................] - ETA: 15s - loss: 0.6625 - acc: 0.6221
3/16 [====>.........................] - ETA: 14s - loss: 0.6624 - acc: 0.6221
4/16 [======>.......................] - ETA: 13s - loss: 0.6623 - acc: 0.6221
5/16 [========>.....................] - ETA: 12s - loss: 0.6622 - acc: 0.6221
6/16 [==========>...................] - ETA: 11s - loss: 0.6621 - acc: 0.6221
7/16 [============>.................] - ETA: 10s - loss: 0.6621 - acc: 0.6221
8/16 [==============>...............] - ETA: 9s - loss: 0.6621 - acc: 0.6221
9/16 [===============>..............] - ETA: 8s - loss: 0.6621 - acc: 0.6221
10/16 [=================>............] - ETA: 6s - loss: 0.6621 - acc: 0.6221
11/16 [===================>..........] - ETA: 5s - loss: 0.6620 - acc: 0.6221
12/16 [=====================>........] - ETA: 4s - loss: 0.6620 - acc: 0.6221
13/16 [=======================>......] - ETA: 3s - loss: 0.6620 - acc: 0.6221
14/16 [=========================>....] - ETA: 2s - loss: 0.6620 - acc: 0.6221
15/16 [===========================>..] - ETA: 1s - loss: 0.6619 - acc: 0.6221
16/16 [==============================] - 18s 1s/step - loss: 0.6619 - acc: 0.6221
# Testing Sample
1/128 [..............................] - ETA: 11s
6/128 [>.............................] - ETA: 3s
10/128 [=>............................] - ETA: 2s
14/128 [==>...........................] - ETA: 2s
19/128 [===>..........................] - ETA: 1s
23/128 [====>.........................] - ETA: 1s
27/128 [=====>........................] - ETA: 1s
32/128 [======>.......................] - ETA: 1s
36/128 [=======>......................] - ETA: 1s
41/128 [========>.....................] - ETA: 1s
45/128 [=========>....................] - ETA: 1s
50/128 [==========>...................] - ETA: 1s
54/128 [===========>..................] - ETA: 1s
59/128 [============>.................] - ETA: 1s
63/128 [=============>................] - ETA: 0s
67/128 [==============>...............] - ETA: 0s
72/128 [===============>..............] - ETA: 0s
76/128 [================>.............] - ETA: 0s
81/128 [=================>............] - ETA: 0s
84/128 [==================>...........] - ETA: 0s
88/128 [===================>..........] - ETA: 0s
92/128 [====================>.........] - ETA: 0s
96/128 [=====================>........] - ETA: 0s
101/128 [======================>.......] - ETA: 0s
105/128 [=======================>......] - ETA: 0s
109/128 [========================>.....] - ETA: 0s
112/128 [=========================>....] - ETA: 0s
116/128 [==========================>...] - ETA: 0s
120/128 [===========================>..] - ETA: 0s
124/128 [============================>.] - ETA: 0s
128/128 [==============================] - 2s 15ms/step
# Val loss, val acc
3.21346378326416 0.6113420724868774
# Model prediction
[0.07243679 0.06086067 0.17687203 0.0424496 0.04154298]
[[3.17100697e-04 1.11687055e-04 1.74964574e-04 6.10632123e-05
9.15286364e-05 5.74094338e-05 1.06784762e-04 9.21065075e-05
3.97201220e-04 6.81500824e-05 2.94565433e-03 1.35827821e-03
2.81211367e-04 1.00520747e-02 6.01871812e-04 5.57196997e-02]
[4.75579909e-05 1.06101972e-04 6.00069507e-05 4.96198081e-05
1.41731420e-04 8.79359577e-05 7.46832447e-05 4.33949099e-05
6.03557055e-05 7.12051406e-05 8.45988281e-04 2.34248699e-03
2.74868420e-04 1.51112420e-03 7.97806482e-04 5.43457977e-02]
[1.06174128e-04 6.14731325e-05 2.77487037e-04 9.75391740e-05
6.97174910e-05 1.57534625e-04 1.94240944e-04 7.85565353e-05
8.36232939e-05 4.36835981e-05 4.04849125e-04 6.86666509e-03
3.01186665e-04 3.41302366e-04 2.24954495e-03 1.65538445e-01]
[4.85532328e-05 5.30333891e-05 1.46693186e-04 2.40156965e-04
6.14130186e-05 6.84155602e-05 1.68141501e-04 9.09807641e-05
2.58948799e-04 6.01471947e-05 1.68107694e-03 8.52260040e-04
1.04383787e-03 4.75410791e-03 4.93220054e-04 3.24286185e-02]
[2.70062083e-05 1.19853627e-04 2.31390568e-05 4.41022166e-05
6.90615489e-05 2.53524253e-04 1.09332977e-04 7.17278526e-05
1.98521622e-04 3.77545693e-05 1.86053314e-03 2.42721115e-04
1.85445603e-03 1.16664183e-03 1.00720196e-03 3.44574042e-02]]
该模型仍然产生接近零的值，并且其精度仍达到0.6221的峰值。我应该为此设置一个阈值使其成为二进制吗？
答案 0 :(得分：0)
您应该将输出层的激活更改为Sigmoid，这将为您提供[0，1]范围内的值。然后，您可以应用阈值来获取二进制值。
model.add(tf.keras.layers.Dense(80, activation=keras.activations.sigmoid))
请注意，keras通过将输出阈值设为0.5来计算二进制精度，并且它将分别计算每个输出的精度。
答案 1 :(得分：0)
因此，您想从（1,26）（实值？）矩阵中预测5 * 16二进制矩阵。
import keras.backend as K
from keras.metrics import binary_accuracy
def custom_metric(y_true, y_pred):
threshold =0.5
thresholded_values = K.greater(y_pred, threshold)
return binary_accuracy(y_true, thresholded_values)
PS我没有测试代码，也许您需要将布尔张量转换为一个真正的值...