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?


当前回答

import tensorflow as tf
import numpy as np

def softmax(x):
    return (np.exp(x).T / np.exp(x).sum(axis=-1)).T

logits = np.array([[1, 2, 3], [3, 10, 1], [1, 2, 5], [4, 6.5, 1.2], [3, 6, 1]])

sess = tf.Session()
print(softmax(logits))
print(sess.run(tf.nn.softmax(logits)))
sess.close()

其他回答

为了保持数值的稳定性,应减去max(x)。下面是softmax函数的代码;

def softmax (x):

if len(x.shape) > 1:
    tmp = np.max(x, axis = 1)
    x -= tmp.reshape((x.shape[0], 1))
    x = np.exp(x)
    tmp = np.sum(x, axis = 1)
    x /= tmp.reshape((x.shape[0], 1))
else:
    tmp = np.max(x)
    x -= tmp
    x = np.exp(x)
    tmp = np.sum(x)
    x /= tmp


return x

这将泛化并假设您正在规范化尾随维度。

def softmax(x: np.ndarray) -> np.ndarray:
    e_x = np.exp(x - np.max(x, axis=-1)[..., None])
    e_y = e_x.sum(axis=-1)[..., None]
    return e_x / e_y

所以,这实际上是对desertnaut的回答的一个评论,但由于我的声誉,我还不能评论它。正如他所指出的,只有当输入包含单个样本时,你的版本才是正确的。如果您的输入包含多个样本,则是错误的。然而,沙漠探险者的解决方案也是错误的。问题是一旦他得到一个一维的输入然后他又得到一个二维的输入。让我给你们看看这个。

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()

# desertnaut solution (copied from his answer): 
def desertnaut_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

# my (correct) solution:
def softmax(z):
    assert len(z.shape) == 2
    s = np.max(z, axis=1)
    s = s[:, np.newaxis] # necessary step to do broadcasting
    e_x = np.exp(z - s)
    div = np.sum(e_x, axis=1)
    div = div[:, np.newaxis] # dito
    return e_x / div

让我们以沙漠探险者为例:

x1 = np.array([[1, 2, 3, 6]]) # notice that we put the data into 2 dimensions(!)

输出如下:

your_softmax(x1)
array([[ 0.00626879,  0.01704033,  0.04632042,  0.93037047]])

desertnaut_softmax(x1)
array([[ 1.,  1.,  1.,  1.]])

softmax(x1)
array([[ 0.00626879,  0.01704033,  0.04632042,  0.93037047]])

你可以看到沙漠版本在这种情况下会失败。(如果输入是一维的,就不会像np那样。数组([1,2,3,6])。

现在让我们使用3个样本,因为这就是为什么我们使用二维输入的原因。下面的x2和沙漠例子中的x2不一样。

x2 = np.array([[1, 2, 3, 6],  # sample 1
               [2, 4, 5, 6],  # sample 2
               [1, 2, 3, 6]]) # sample 1 again(!)

该输入由一个有3个样本的批次组成。但样本一和样本三本质上是一样的。我们现在期望3行softmax激活,其中第一行应该与第三行相同,也与x1的激活相同!

your_softmax(x2)
array([[ 0.00183535,  0.00498899,  0.01356148,  0.27238963],
       [ 0.00498899,  0.03686393,  0.10020655,  0.27238963],
       [ 0.00183535,  0.00498899,  0.01356148,  0.27238963]])


desertnaut_softmax(x2)
array([[ 0.21194156,  0.10650698,  0.10650698,  0.33333333],
       [ 0.57611688,  0.78698604,  0.78698604,  0.33333333],
       [ 0.21194156,  0.10650698,  0.10650698,  0.33333333]])

softmax(x2)
array([[ 0.00626879,  0.01704033,  0.04632042,  0.93037047],
       [ 0.01203764,  0.08894682,  0.24178252,  0.65723302],
       [ 0.00626879,  0.01704033,  0.04632042,  0.93037047]])

我希望你能明白,这只是我的解的情况。

softmax(x1) == softmax(x2)[0]
array([[ True,  True,  True,  True]], dtype=bool)

softmax(x1) == softmax(x2)[2]
array([[ True,  True,  True,  True]], dtype=bool)

另外,下面是TensorFlows softmax实现的结果:

import tensorflow as tf
import numpy as np
batch = np.asarray([[1,2,3,6],[2,4,5,6],[1,2,3,6]])
x = tf.placeholder(tf.float32, shape=[None, 4])
y = tf.nn.softmax(x)
init = tf.initialize_all_variables()
sess = tf.Session()
sess.run(y, feed_dict={x: batch})

结果是:

array([[ 0.00626879,  0.01704033,  0.04632042,  0.93037045],
       [ 0.01203764,  0.08894681,  0.24178252,  0.657233  ],
       [ 0.00626879,  0.01704033,  0.04632042,  0.93037045]], dtype=float32)

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。

目标是使用Numpy和Tensorflow实现类似的结果。与原始答案的唯一变化是np的轴参数。和api。

初始方法:axis=0 -然而,当维度为N时,这并不能提供预期的结果。

修改方法:axis=len(e_x.shape)-1 -总是在最后一个维度上求和。这提供了与tensorflow的softmax函数类似的结果。

def softmax_fn(input_array):
    """
    | **@author**: Prathyush SP
    |
    | Calculate Softmax for a given array
    :param input_array: Input Array
    :return: Softmax Score
    """
    e_x = np.exp(input_array - np.max(input_array))
    return e_x / e_x.sum(axis=len(e_x.shape)-1)