从Udacity的深度学习课程中，y_i的softmax仅仅是指数除以整个Y向量的指数之和:
其中S(y_i)是y_i的软最大函数e是指数函数j是no。输入向量Y中的列。
我试过以下几种方法:
import numpy as np
def softmax(x):
"""Compute softmax values for each sets of scores in x."""
e_x = np.exp(x - np.max(x))
return e_x / e_x.sum()
scores = [3.0, 1.0, 0.2]
print(softmax(scores))
返回:
[ 0.8360188 0.11314284 0.05083836]
但建议的解决方案是:
def softmax(x):
"""Compute softmax values for each sets of scores in x."""
return np.exp(x) / np.sum(np.exp(x), axis=0)
它产生与第一个实现相同的输出，尽管第一个实现显式地取每列与Max的差值，然后除以和。
有人能用数学方法解释一下吗?一个是对的，另一个是错的?
实现在代码和时间复杂度方面是否相似?哪个更有效率?
编辑。从1.2.0版本开始，scipy包含了softmax作为一个特殊函数:
https://scipy.github.io/devdocs/generated/scipy.special.softmax.html
我写了一个在任意轴上应用softmax的函数:
def softmax(X, theta = 1.0, axis = None):
"""
Compute the softmax of each element along an axis of X.
Parameters
----------
X: ND-Array. Probably should be floats.
theta (optional): float parameter, used as a multiplier
prior to exponentiation. Default = 1.0
axis (optional): axis to compute values along. Default is the
first non-singleton axis.
Returns an array the same size as X. The result will sum to 1
along the specified axis.
"""
# make X at least 2d
y = np.atleast_2d(X)
# find axis
if axis is None:
axis = next(j[0] for j in enumerate(y.shape) if j[1] > 1)
# multiply y against the theta parameter,
y = y * float(theta)
# subtract the max for numerical stability
y = y - np.expand_dims(np.max(y, axis = axis), axis)
# exponentiate y
y = np.exp(y)
# take the sum along the specified axis
ax_sum = np.expand_dims(np.sum(y, axis = axis), axis)
# finally: divide elementwise
p = y / ax_sum
# flatten if X was 1D
if len(X.shape) == 1: p = p.flatten()
return p
正如其他用户所描述的那样，减去最大值是很好的做法。我在这里写了一篇详细的文章。
softmax函数是一种激活函数，它将数字转换为和为1的概率。softmax函数输出一个向量，表示结果列表的概率分布。它也是深度学习分类任务中使用的核心元素。
当我们有多个类时，使用Softmax函数。
它对于找出有最大值的类很有用。概率。
Softmax函数理想地用于输出层，在那里我们实际上试图获得定义每个输入类的概率。
取值范围是0 ~ 1。
Softmax函数将对数[2.0,1.0,0.1]转换为概率[0.7,0.2,0.1]，概率和为1。Logits是神经网络最后一层输出的原始分数。在激活发生之前。为了理解softmax函数，我们必须看看第(n-1)层的输出。
softmax函数实际上是一个arg max函数。这意味着它不会返回输入中的最大值，而是返回最大值的位置。
例如:
softmax之前
X = [13, 31, 5]
softmax后
array([1.52299795e-08, 9.99999985e-01, 5.10908895e-12]
代码:
import numpy as np
# your solution:
def your_softmax(x):
"""Compute softmax values for each sets of scores in x."""
e_x = np.exp(x - np.max(x))
return e_x / e_x.sum()
# correct solution:
def softmax(x):
"""Compute softmax values for each sets of scores in x."""
e_x = np.exp(x - np.max(x))
return e_x / e_x.sum(axis=0)
# only difference
我很好奇它们之间的性能差异
import numpy as np
def softmax(x):
"""Compute softmax values for each sets of scores in x."""
return np.exp(x) / np.sum(np.exp(x), axis=0)
def softmaxv2(x):
"""Compute softmax values for each sets of scores in x."""
e_x = np.exp(x - np.max(x))
return e_x / e_x.sum()
def softmaxv3(x):
"""Compute softmax values for each sets of scores in x."""
e_x = np.exp(x - np.max(x))
return e_x / np.sum(e_x, axis=0)
def softmaxv4(x):
"""Compute softmax values for each sets of scores in x."""
return np.exp(x - np.max(x)) / np.sum(np.exp(x - np.max(x)), axis=0)
x=[10,10,18,9,15,3,1,2,1,10,10,10,8,15]
使用
print("----- softmax")
%timeit a=softmax(x)
print("----- softmaxv2")
%timeit a=softmaxv2(x)
print("----- softmaxv3")
%timeit a=softmaxv2(x)
print("----- softmaxv4")
%timeit a=softmaxv2(x)
增加x内部的值(+100 +200 +500…)我使用原始numpy版本得到的结果始终更好(这里只是一个测试)
----- softmax
The slowest run took 8.07 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 17.8 µs per loop
----- softmaxv2
The slowest run took 4.30 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 23 µs per loop
----- softmaxv3
The slowest run took 4.06 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 23 µs per loop
----- softmaxv4
10000 loops, best of 3: 23 µs per loop
直到……x内的值达到~800，则得到
----- softmax
/usr/local/lib/python3.6/dist-packages/ipykernel_launcher.py:4: RuntimeWarning: overflow encountered in exp
after removing the cwd from sys.path.
/usr/local/lib/python3.6/dist-packages/ipykernel_launcher.py:4: RuntimeWarning: invalid value encountered in true_divide
after removing the cwd from sys.path.
The slowest run took 18.41 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 23.6 µs per loop
----- softmaxv2
The slowest run took 4.18 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 22.8 µs per loop
----- softmaxv3
The slowest run took 19.44 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 23.6 µs per loop
----- softmaxv4
The slowest run took 16.82 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 22.7 µs per loop
就像一些人说的，你的版本在“大数字”上更稳定。对于小数字来说，情况可能正好相反。
