如何在 Python 中计算逻辑 sigmoid 函数？
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How to calculate a logistic sigmoid function in Python?
提问by Richard Knop
This is a logistic sigmoid function:
这是一个逻辑 sigmoid 函数：
I know x. How can I calculate F(x) in Python now?
我知道 x。我现在如何在 Python 中计算 F(x)？
Let's say x = 0.458.
假设 x = 0.458。
F(x) = ?
F(x) = ?
采纳答案by unwind
This should do it:
这应该这样做：
import math
def sigmoid(x):
return 1 / (1 + math.exp(-x))
And now you can test it by calling:
现在你可以通过调用来测试它：
>>> sigmoid(0.458)
0.61253961344091512
Update: Note that the above was mainly intended as a straight one-to-one translation of the given expression into Python code. It is nottested or known to be a numerically sound implementation. If you know you need a very robust implementation, I'm sure there are others where people have actually given this problem some thought.
更新：请注意，上述内容主要是将给定表达式直接一对一翻译成 Python 代码。它没有经过测试或已知是数字上可靠的实现。如果你知道你需要一个非常健壮的实现，我相信还有其他人实际上已经考虑过这个问题。
回答by ghostdog74
another way
其它的办法
>>> def sigmoid(x):
... return 1 /(1+(math.e**-x))
...
>>> sigmoid(0.458)
回答by Théo T
It is also available in scipy: http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.logistic.html
它也可以在 scipy 中使用：http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.logistic.html
In [1]: from scipy.stats import logistic
In [2]: logistic.cdf(0.458)
Out[2]: 0.61253961344091512
which is only a costly wrapper (because it allows you to scale and translate the logistic function) of another scipy function:
这只是另一个 scipy 函数的昂贵包装器（因为它允许您缩放和转换逻辑函数）：
In [3]: from scipy.special import expit
In [4]: expit(0.458)
Out[4]: 0.61253961344091512
If you are concerned about performances continue reading, otherwise just use expit.
如果您担心性能，请继续阅读，否则只需使用expit.
Some benchmarking:
一些基准测试：
In [5]: def sigmoid(x):
....: return 1 / (1 + math.exp(-x))
....:
In [6]: %timeit -r 1 sigmoid(0.458)
1000000 loops, best of 1: 371 ns per loop
In [7]: %timeit -r 1 logistic.cdf(0.458)
10000 loops, best of 1: 72.2 μs per loop
In [8]: %timeit -r 1 expit(0.458)
100000 loops, best of 1: 2.98 μs per loop
As expected logistic.cdfis (much) slower than expit. expitis still slower than the python sigmoidfunction when called with a single value because it is a universal function written in C ( http://docs.scipy.org/doc/numpy/reference/ufuncs.html) and thus has a call overhead. This overhead is bigger than the computation speedup of expitgiven by its compiled nature when called with a single value. But it becomes negligible when it comes to big arrays:
正如预期的那样logistic.cdf（远）慢于expit. 使用单个值调用时expit仍然比 pythonsigmoid函数慢，因为它是用 C ( http://docs.scipy.org/doc/numpy/reference/ufuncs.html)编写的通用函数，因此有调用开销。expit当使用单个值调用时，此开销大于其编译性质给出的计算加速。但是当涉及到大数组时，它变得可以忽略不计：
In [9]: import numpy as np
In [10]: x = np.random.random(1000000)
In [11]: def sigmoid_array(x):
....: return 1 / (1 + np.exp(-x))
....:
(You'll notice the tiny change from math.expto np.exp(the first one does not support arrays, but is much faster if you have only one value to compute))
（您会注意到从math.expto的微小变化np.exp（第一个不支持数组，但如果您只有一个值要计算，速度会快得多））
In [12]: %timeit -r 1 -n 100 sigmoid_array(x)
100 loops, best of 1: 34.3 ms per loop
In [13]: %timeit -r 1 -n 100 expit(x)
100 loops, best of 1: 31 ms per loop
But when you really need performance, a common practice is to have a precomputed table of the the sigmoid function that hold in RAM, and trade some precision and memory for some speed (for example: http://radimrehurek.com/2013/09/word2vec-in-python-part-two-optimizing/)
但是当你真的需要性能时，一个常见的做法是在 RAM 中预先计算一个 sigmoid 函数的表，并用一些精度和内存来换取一些速度（例如：http: //radimrehurek.com/2013/09 /word2vec-in-python-part-two-optimizing/)
Also, note that expitimplementation is numerically stable since version 0.14.0: https://github.com/scipy/scipy/issues/3385
另外，请注意，expit自 0.14.0 版以来，实现在数值上是稳定的：https: //github.com/scipy/scipy/issues/3385
回答by Neil G
Here's how you would implement the logistic sigmoid in a numerically stable way (as described here):
这里是你将如何实现在数字上稳定的方式物流乙状结肠（如描述这里）：
def sigmoid(x):
"Numerically-stable sigmoid function."
if x >= 0:
z = exp(-x)
return 1 / (1 + z)
else:
z = exp(x)
return z / (1 + z)
Or perhaps this is more accurate:
或者这可能更准确：
import numpy as np
def sigmoid(x):
return math.exp(-np.logaddexp(0, -x))
Internally, it implements the same condition as above, but then uses log1p.
在内部，它实现与上述相同的条件，但随后使用log1p.
In general, the multinomial logistic sigmoid is:
一般来说，多项logistic sigmoid是：
def nat_to_exp(q):
max_q = max(0.0, np.max(q))
rebased_q = q - max_q
return np.exp(rebased_q - np.logaddexp(-max_q, np.logaddexp.reduce(rebased_q)))
回答by czxttkl
Good answer from @unwind. It however can't handle extreme negative number (throwing OverflowError).
来自@unwind 的好答案。但是它不能处理极端负数（抛出溢出错误）。
My improvement:
我的改进：
def sigmoid(x):
try:
res = 1 / (1 + math.exp(-x))
except OverflowError:
res = 0.0
return res
回答by dontloo
Another way by transforming the tanhfunction:
转换tanh函数的另一种方法：
sigmoid = lambda x: .5 * (math.tanh(.5 * x) + 1)
回答by Philipp Schwarz
I feel many might be interested in free parameters to alter the shape of the sigmoid function. Second for many applications you want to use a mirrored sigmoid function. Third you might want to do a simple normalization for example the output values are between 0 and 1.
我觉得很多人可能对改变 sigmoid 函数形状的自由参数感兴趣。其次，对于您想要使用镜像 sigmoid 函数的许多应用程序。第三，您可能想要进行简单的归一化，例如输出值介于 0 和 1 之间。
Try:
尝试：
def normalized_sigmoid_fkt(a, b, x):
'''
Returns array of a horizontal mirrored normalized sigmoid function
output between 0 and 1
Function parameters a = center; b = width
'''
s= 1/(1+np.exp(b*(x-a)))
return 1*(s-min(s))/(max(s)-min(s)) # normalize function to 0-1
And to draw and compare:
并绘制和比较：
def draw_function_on_2x2_grid(x):
fig, ((ax1, ax2), (ax3, ax4)) = plt.subplots(2, 2)
plt.subplots_adjust(wspace=.5)
plt.subplots_adjust(hspace=.5)
ax1.plot(x, normalized_sigmoid_fkt( .5, 18, x))
ax1.set_title('1')
ax2.plot(x, normalized_sigmoid_fkt(0.518, 10.549, x))
ax2.set_title('2')
ax3.plot(x, normalized_sigmoid_fkt( .7, 11, x))
ax3.set_title('3')
ax4.plot(x, normalized_sigmoid_fkt( .2, 14, x))
ax4.set_title('4')
plt.suptitle('Different normalized (sigmoid) function',size=10 )
return fig
Finally:
最后：
x = np.linspace(0,1,100)
Travel_function = draw_function_on_2x2_grid(x)
回答by Enrique Pérez Herrero
Tensorflow includes also a sigmoidfunction:
https://www.tensorflow.org/versions/r1.2/api_docs/python/tf/sigmoid
Tensorflow 还包括一个sigmoid函数：https:
//www.tensorflow.org/versions/r1.2/api_docs/python/tf/sigmoid
import tensorflow as tf
sess = tf.InteractiveSession()
x = 0.458
y = tf.sigmoid(x)
u = y.eval()
print(u)
# 0.6125396
回答by Yash Khare
A numerically stable version of the logistic sigmoid function.
逻辑 sigmoid 函数的数值稳定版本。
def sigmoid(x):
pos_mask = (x >= 0)
neg_mask = (x < 0)
z = np.zeros_like(x,dtype=float)
z[pos_mask] = np.exp(-x[pos_mask])
z[neg_mask] = np.exp(x[neg_mask])
top = np.ones_like(x,dtype=float)
top[neg_mask] = z[neg_mask]
return top / (1 + z)
回答by Diatche
Use the numpy package to allow your sigmoid function to parse vectors.
使用 numpy 包允许您的 sigmoid 函数解析向量。
In conformity with Deeplearning, I use the following code:
根据 Deeplearning，我使用以下代码：
import numpy as np
def sigmoid(x):
s = 1/(1+np.exp(-x))
return s
