误差计算
目录
TensorFlow2教程完整教程目录（更有python、go、pytorch、tensorflow、爬虫、人工智能教学等着你）：https://www.cnblogs.com/nickchen121/p/10840284.html
Outline
- MSE
- Cross Entropy Loss
- Hinge Loss
MSE
- \(loss = \frac{1}{N}\sum(y-out)^2\)
- \(L_{2-norm} = \sqrt{\sum(y-out)}\)
import tensorflow as tf
y = tf.constant([1, 2, 3, 0, 2])
y = tf.one_hot(y, depth=4) # max_label=3种
y = tf.cast(y, dtype=tf.float32)
out = tf.random.normal([5, 4])
out
<tf.Tensor: id=117, shape=(5, 4), dtype=float32, numpy=
array([[ 0.8138832 , -1.1521571 , 0.05197939, 2.3684442 ],
[ 0.28827545, -0.35568208, -0.3952962 , -1.2576817 ],
[-0.4354525 , -1.9914867 , 0.37045303, -0.38287213],
[-0.7680094 , -0.98293644, 0.62572837, -0.5673917 ],
[ 1.5299634 , 0.38036177, -0.28049606, -0.708137 ]],
dtype=float32)>
loss1 = tf.reduce_mean(tf.square(y - out))
loss1
<tf.Tensor: id=122, shape=(), dtype=float32, numpy=1.5140966>
loss2 = tf.square(tf.norm(y - out)) / (5 * 4)
loss2
<tf.Tensor: id=99, shape=(), dtype=float32, numpy=1.3962512>
loss3 = tf.reduce_mean(tf.losses.MSE(y, out))
loss3
<tf.Tensor: id=105, shape=(), dtype=float32, numpy=1.3962513>
Entropy
- Uncertainty
- measure of surprise
- lower entropy --> more info.
\[\text{Entropy} = -\sum_{i}P(i)log\,P(i)
\]
a = tf.fill([4], 0.25)
a * tf.math.log(a) / tf.math.log(2.)
<tf.Tensor: id=134, shape=(4,), dtype=float32, numpy=array([-0.5, -0.5, -0.5, -0.5], dtype=float32)>
-tf.reduce_sum(a * tf.math.log(a) / tf.math.log(2.))
<tf.Tensor: id=143, shape=(), dtype=float32, numpy=2.0>
a = tf.constant([0.1, 0.1, 0.1, 0.7])
-tf.reduce_sum(a * tf.math.log(a) / tf.math.log(2.))
<tf.Tensor: id=157, shape=(), dtype=float32, numpy=1.3567797>
a = tf.constant([0.01, 0.01, 0.01, 0.97])
-tf.reduce_sum(a * tf.math.log(a) / tf.math.log(2.))
<tf.Tensor: id=167, shape=(), dtype=float32, numpy=0.24194068>
Cross Entropy
\[H(p,q) = -\sum{p(x)log\,q(x)} \\
H(p,q) = H(p) + D_{KL}(p|q)
\]
- for p = q
Minima: H(p,q) = H(p)
- for P: one-hot encodint
\(h(p:[0,1,0]) = -1log\,1=0\)
\(H([0,1,0],[p_0,p_1,p_2]) = 0 + D_{KL}(p|q) = -1log\,q_1\) # p,q即真实值和预测值相等的话交叉熵为0
Binary Classification
- Two cases（第二种格式只需要输出一种情况，节省计算，无意义）
Single output
\[H(P,Q) = -P(cat)log\,Q(cat) - (1-P(cat))log\,(1-Q(cat)) \\
P(dog) = (1-P(cat)) \\
\]
\[\begin{aligned}
H(P,Q) & = -\sum_{i=(cat,dog)}P(i)log\,Q(i)\\
& = -P(cat)log\,Q(cat) - P(dog)log\,Q(dog)-(ylog(p)+(1-y)log\,(1-p))
\end{aligned}
\]
Classification
- \(H([0,1,0],[p_0,p_1,p_2])=0+D_{KL}(p|q) = -1log\,q_1\)
\[\begin{aligned}
& P_1 = [1,0,0,0,0]\\
& Q_1=[0.4,0.3,0.05,0.05,0.2]
\end{aligned}
\]
\[\begin{aligned}
H(P_1,Q_1) & = -\sum{P_1(i)}log\,Q_1(i) \\
& = -(1log\,0.4+0log\,0.3+0log\,0.05+0log\,0.05+0log\,0.2) \\
& =-log\,0.4 \\
& \approx{0.916}
\end{aligned}
\]
\[\begin{aligned}
& P_1 = [1,0,0,0,0]\\
& Q_1=[0.98,0.01,0,0,0.01]
\end{aligned}
\]
\[\begin{aligned}
H(P_1,Q_1) & = -\sum{P_1(i)}log\,Q_1(i) \\
& =-log\,0.98 \\
& \approx{0.02}
\end{aligned}
\]
tf.losses.categorical_crossentropy([0, 1, 0, 0], [0.25, 0.25, 0.25, 0.25])
<tf.Tensor: id=186, shape=(), dtype=float32, numpy=1.3862944>
tf.losses.categorical_crossentropy([0, 1, 0, 0], [0.1, 0.1, 0.8, 0.1])
<tf.Tensor: id=205, shape=(), dtype=float32, numpy=2.3978953>
tf.losses.categorical_crossentropy([0, 1, 0, 0], [0.1, 0.7, 0.1, 0.1])
<tf.Tensor: id=243, shape=(), dtype=float32, numpy=0.35667497>
tf.losses.categorical_crossentropy([0, 1, 0, 0], [0.01, 0.97, 0.01, 0.01])
<tf.Tensor: id=262, shape=(), dtype=float32, numpy=0.030459179>
tf.losses.BinaryCrossentropy()([1],[0.1])
<tf.Tensor: id=306, shape=(), dtype=float32, numpy=2.3025842>
tf.losses.binary_crossentropy([1],[0.1])
<tf.Tensor: id=333, shape=(), dtype=float32, numpy=2.3025842>
Why not MSE？
- sigmoid + MSE
gradient vanish
- converge slower
- However
e.g. meta-learning
logits-->CrossEntropy
