tensorflow2.0 - tensorflow 20 - 간단한 기능 최소화

   import tensorflow as tf
    import numpy as np
    x = tf.Variable(2, name='x', trainable=True, dtype=tf.float32)
    with tf.GradientTape() as t:
        t.watch(x)
        log_x = tf.math.log(x)
        y = tf.math.square(log_x)
    opt = tf.optimizers.Adam(0.5)
    # train = opt.minimize(lambda: y, var_list=[x]) # FAILS
    @tf.function
    def f(x):
        log_x = tf.math.log(x)
        y = tf.math.square(log_x)
        return y
    yy = f(x)
    train = opt.minimize(lambda: yy, var_list=[x]) # ALSO FAILS

수율 ValueError :

No gradients provided for any variable: ['x:0'].

이것은 부분적으로 제시 한 예처럼 보입니다. 이것이 버그 또는 2.0 또는 내가 잘못하고있는 버그인지 확실하지 않습니다.

업데이트 :

일부 문제와 흥미로운 메모가 있었기 때문에 아래의 솔루션을 꾸미 셨습니다.

   import numpy as np
    import tensorflow as tf
    x = tf.Variable(3, name='x', trainable=True, dtype=tf.float32)
    with tf.GradientTape(persistent=True) as t:
        # log_x = tf.math.log(x)
        # y = tf.math.square(log_x)
        y = (x - 1) ** 2
    opt = tf.optimizers.Adam(learning_rate=0.001)
    def get_gradient_wrong(x0):
        # this does not work, it does not actually update the value of x
        x.assign(x0)
        return t.gradient(y, [x])
    def get_gradient(x0):
        # this works
        x.assign(x0)
        with tf.GradientTape(persistent=True) as t:
            y = (x - 1) ** 2
        return t.gradient(y, [x])
    #### Option 1
    def a(x0, tol=1e-8, max_iter=10000):
        # does not appear to work properly
        x.assign(x0)
        err = np.Inf # step error (banach), not actual erro
        i = 0
        while err > tol:
            x0 = x.numpy()
            # IMPORTANT: WITHOUT THIS INSIDE THE LOOP THE GRADIENTS DO NOT UPDATE
            with tf.GradientTape(persistent=True) as t:
                y = (x - 1) ** 2
            gradients = t.gradient(y, [x])
            l = opt.apply_gradients(zip(gradients, [x]))
            err = np.abs(x.numpy() - x0)
            print(err, x.numpy(), gradients[0].numpy())
            i += 1
            if i > max_iter:
                print(f'stopping at max_iter={max_iter}')
                return x.numpy()
        print(f'stopping at err={err}<{tol}')
        return x.numpy()
    #### Option 2
    def b(x0, tol=1e-8, max_iter=10000):
        x.assign(x0)
        # To use minimize you have to define your loss computation as a funcction
        def compute_loss():
            log_x = tf.math.log(x)
            y = tf.math.square(log_x)
            return y
        err = np.Inf # step error (banach), not actual erro
        i = 0
        while err > tol:
            x0 = x.numpy()
            train = opt.minimize(compute_loss, var_list=[x])
            err = np.abs(x.numpy() - x0)
            print(err, x.numpy())
            i += 1
            if i > max_iter:
                print(f'stopping at max_iter={max_iter}')
                return x.numpy()
        print(f'stopping at err={err}<{tol}')
        return x.numpy()