Question

我试图在Keras中实现真正的积极指标：

def TP(y_true, y_pred):
    estimated = K.argmax(y_pred, axis=1)
    truth = K.argmax(y_true, axis=1)
    TP = K.sum(truth * estimated)
    return TP

基于我的最后一层输出形状：（batch，2）。该功能已经在numpy argmax等效项下进行了测试，并且运行良好。

我使用了一个cross_entropy损失函数，每个时期给我度量值。但是这个值怎么可能是十进制数？？我在做什么错呢？谢谢！

已编辑：以下是Keras模型的示例代码：

def TP(y_true, y_pred):
    estimated = K.argmax(y_pred, axis=1)
    truth = K.argmax(y_true, axis=1)
    TP = K.sum(truth * estimated)
    return TP

epochs = 10
batch_size = 2

model = Sequential([
        Dense(32, input_shape=(4,)),
        Activation('relu'),
        Dense(2),
        Activation('softmax'),
])
model.compile(optimizer='adam',
              loss='categorical_crossentropy',
              metrics=['accuracy', TP])

model.summary()

train = np.array([[17,0,1,0],[17,0,1,0],[17,0,1,0],[17,0,1,0],[17,0,1,0], [2,1,0,1],[0,1,0,1],[0,1,0,1],[0,1,0,1],[0,1,0,1]])
labels = np.array([ [1,0],[1,0],[1,0],[1,0],[1,0], [0,1],[0,1],[0,1],[0,1],[0,1] ])

model.fit(train, labels, epochs=epochs, batch_size=batch_size, verbose=2)

在这里，显示 TP 功能的测试似乎有效

def npTP(y_true, y_pred):
    estimated = np.argmax(y_pred, axis=1)
    truth = np.argmax(y_true, axis=1)
    TP = np.sum(truth * estimated)
    return TP

y_true = np.array([ [1,0],[1,0],[1,0],[1,0],[1,0], [0,1],[0,1],[0,1],[0,1],[0,1] ])
y_pred = np.array([ [0,1],[0,1],[0,1],[0,1],[0,1], [0,1],[0,1],[0,1],[0,1],[0,1]])
print("np check : ")
print(npTP(y_true, y_pred))

运行此代码将给出以下输出：

Using TensorFlow backend.

_________________________________________________________________
Layer (type)                 Output Shape              Param #   
=================================================================
dense_1 (Dense)              (None, 32)                160       
_________________________________________________________________
activation_1 (Activation)    (None, 32)                0         
_________________________________________________________________
dense_2 (Dense)              (None, 2)                 66        
_________________________________________________________________
activation_2 (Activation)    (None, 2)                 0         
=================================================================
Total params: 226
Trainable params: 226
Non-trainable params: 0
_________________________________________________________________
Epoch 1/10
 - 0s - loss: 0.3934 - acc: 0.6000 - TP: 0.2000
Epoch 2/10                           ^^^^^^^^^^ here are the decimal values
 - 0s - loss: 0.3736 - acc: 0.6000 - TP: 0.2000
Epoch 3/10                           ^^^^^^^^^^
 - 0s - loss: 0.3562 - acc: 0.6000 - TP: 0.2000
Epoch 4/10                           ^^^^^^^^^^
 - 0s - loss: 0.3416 - acc: 0.7000 - TP: 0.4000
Epoch 5/10                           ^^^^^^^^^^
 - 0s - loss: 0.3240 - acc: 1.0000 - TP: 1.0000
Epoch 6/10
 - 0s - loss: 0.3118 - acc: 1.0000 - TP: 1.0000
Epoch 7/10
 - 0s - loss: 0.2960 - acc: 1.0000 - TP: 1.0000
Epoch 8/10
 - 0s - loss: 0.2806 - acc: 1.0000 - TP: 1.0000
Epoch 9/10
 - 0s - loss: 0.2656 - acc: 1.0000 - TP: 1.0000
Epoch 10/10
 - 0s - loss: 0.2535 - acc: 1.0000 - TP: 1.0000

np check : 
5

谢谢！

Answer 1

正如desertnaut所指出的，in this thread解释了答案。

Keras在批次和纪元之间进行运行平均值。

这里有batch_size=2和10个样本，每个时期进行5次训练（10/2=5）。

要了解时期1的输出指标，在5次训练后TP的总数必须为1，因此该指标为1/5 = 0.2。纪元4在5次训练中有2个TP，给出了指标2/5 = 0.4。

了解Keras中的指标计算

1 个答案:

Keras在批次和纪元之间进行运行平均值。