python人工智能自定义求导tf_diffs详解

自定义求导:(近似求导数的方法)

让x向左移动eps得到一个点,向右移动eps得到一个点,这两个点形成一条直线,这个点的斜率就是x这个位置的近似导数。

eps足够小,导数就足够真。

def f(x):
    return 3. * x ** 2 + 2. * x - 1
def approximate_derivative(f, x, eps=1e-3):
    return (f(x + eps) - f(x - eps)) / (2. * eps)
print(approximate_derivative(f, 1.))

运行结果:

7.999999999999119

多元函数的求导

def g(x1, x2):
    return (x1 + 5) * (x2 ** 2)
def approximate_gradient(g, x1, x2, eps=1e-3):
    dg_x1 = approximate_derivative(lambda x: g(x, x2), x1, eps)
    dg_x2 = approximate_derivative(lambda x: g(x1, x), x2, eps)
    return dg_x1, dg_x2
print(approximate_gradient(g, 2., 3.))

运行结果:

(8.999999999993236, 41.999999999994486)

在tensorflow中的求导

x1 = tf.Variable(2.0)
x2 = tf.Variable(3.0)
with tf.GradientTape() as tape:
    z = g(x1, x2)
dz_x1 = tape.gradient(z, x1)
print(dz_x1)

运行结果:

tf.Tensor(9.0, shape=(), dtype=float32)

但是tf.GradientTape()只能使用一次,使用一次之后就会被消解

try:
    dz_x2 = tape.gradient(z, x2)
except RuntimeError as ex:
    print(ex)

运行结果:

A non-persistent GradientTape can only be used to compute one set of gradients (or jacobians)

解决办法:设置persistent = True,记住最后要把tape删除掉

x1 = tf.Variable(2.0)
x2 = tf.Variable(3.0)
with tf.GradientTape(persistent = True) as tape:
    z = g(x1, x2)
dz_x1 = tape.gradient(z, x1)
dz_x2 = tape.gradient(z, x2)
print(dz_x1, dz_x2)
del tape

运行结果:

tf.Tensor(9.0, shape=(), dtype=float32) tf.Tensor(42.0, shape=(), dtype=float32)

使用tf.GradientTape()

同时求x1,x2的偏导

x1 = tf.Variable(2.0)
x2 = tf.Variable(3.0)
with tf.GradientTape() as tape:
    z = g(x1, x2)
dz_x1x2 = tape.gradient(z, [x1, x2])
print(dz_x1x2)

运行结果:

[, ]

对常量求偏导

x1 = tf.constant(2.0)
x2 = tf.constant(3.0)
with tf.GradientTape() as tape:
    z = g(x1, x2)
dz_x1x2 = tape.gradient(z, [x1, x2])
print(dz_x1x2)

运行结果:

[None, None]

可以使用watch函数关注常量上的导数

x1 = tf.constant(2.0)
x2 = tf.constant(3.0)
with tf.GradientTape() as tape:
    tape.watch(x1)
    tape.watch(x2)
    z = g(x1, x2)
dz_x1x2 = tape.gradient(z, [x1, x2])
print(dz_x1x2)

运行结果:

[, ]

也可以使用两个目标函数对一个变量求导:

x = tf.Variable(5.0)
with tf.GradientTape() as tape:
    z1 = 3 * x
    z2 = x ** 2
tape.gradient([z1, z2], x)

运行结果:

结果13是z1对x的导数加上z2对于x的导数

求二阶导数的方法

x1 = tf.Variable(2.0)
x2 = tf.Variable(3.0)
with tf.GradientTape(persistent=True) as outer_tape:
    with tf.GradientTape(persistent=True) as inner_tape:
        z = g(x1, x2)
    inner_grads = inner_tape.gradient(z, [x1, x2])
outer_grads = [outer_tape.gradient(inner_grad, [x1, x2])
               for inner_grad in inner_grads]
print(outer_grads)
del inner_tape
del outer_tape

运行结果:

[[None, ], [, ]]

结果是一个2x2的矩阵,左上角是z对x1的二阶导数,右上角是z先对x1求导,在对x2求导

左下角是z先对x2求导,在对x1求导,右下角是z对x2的二阶导数

学会自定义求导就可以模拟梯度下降法了,梯度下降就是求导,再在导数的位置前进一点点 模拟梯度下降法:

learning_rate = 0.1
x = tf.Variable(0.0)
for _ in range(100):
    with tf.GradientTape() as tape:
        z = f(x)
    dz_dx = tape.gradient(z, x)
    x.assign_sub(learning_rate * dz_dx)
print(x)

运行结果:

结合optimizers进行梯度下降法

learning_rate = 0.1
x = tf.Variable(0.0)
optimizer = keras.optimizers.SGD(lr = learning_rate)
for _ in range(100):
    with tf.GradientTape() as tape:
        z = f(x)
    dz_dx = tape.gradient(z, x)
    optimizer.apply_gradients([(dz_dx, x)])
print(x)

运行结果:

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