如下所示：
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)]
輸出結果
[[<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])
輸出結果
[<tf.Tensor: id=57075, shape=(), dtype=float32, numpy=136.0>,
<tf.Tensor: id=57076, shape=(), dtype=float32, numpy=30.0>]
總結：如果對一個listz=[z1,z2,z3]求微分，其結果將自動求和，而不是返回z1、z2和z3各自對[w1,w2]的微分。
補充知識：Python/Numpy 矩陣運算符號@
如下所示：
A = np.matrix('3 1; 8 2')
B = np.matrix('6 1; 7 9')
A@B matrix([[25, 12], [62, 26]])
[wooen ] TensorFlow Autodiff自動微分詳解已經有354次圍觀
