with tf.GradientTape(persistent=True) as tape:
z1 = f(w1, w2 + 2.)
z2 = f(w1, w2 + 5.)
z3 = f(w1, w2 + 7.)
z = [z1,z3,z3]
[tape.gradient(z, [w1, w2]) for z in (z1, z2, z3)]
输出结果
[[<tf.Tensor: id=56906, shape=(), dtype=float32, numpy=40.0>,
<tf.Tensor: id=56898, shape=(), dtype=float32, numpy=10.0>],
[<tf.Tensor: id=56919, shape=(), dtype=float32, numpy=46.0>,
<tf.Tensor: id=56911, shape=(), dtype=float32, numpy=10.0>],
[<tf.Tensor: id=56932, shape=(), dtype=float32, numpy=50.0>,
<tf.Tensor: id=56924, shape=(), dtype=float32, numpy=10.0>]]
with tf.GradientTape(persistent=True) as tape:
z1 = f(w1, w2 + 2.)
z2 = f(w1, w2 + 5.)
z3 = f(w1, w2 + 7.)
z = [z1,z2,z3]
tape.gradient(z, [w1, w2])
输出结果
[<tf.Tensor: id=57075, shape=(), dtype=float32, numpy=136.0>,
<tf.Tensor: id=57076, shape=(), dtype=float32, numpy=30.0>]
总结:如果对一个listz=[z1,z2,z3]
求微分,其结果将自动求和,而不是返回z1
、z2
和z3
各自对[w1,w2]
的微分。