Обратное распространение с PyTorch и автодифференциация

Question

У меня есть набор данных - X -, состоящий из 15 переменных и 64 наблюдений, и вектор столбца - Y - из 64 значений, представляющих цель (метку). Я пытаюсь согласовать параметры с квадратичной функцией c, чтобы вернуть наблюдаемые значения (Y) с помощью PyTorch, но я получаю ошибку. Я предоставляю набор данных в конце поста в формате json для воспроизводимости.

Если бы у меня был один пример, мой код мог бы быть:

X = torch.from_numpy(X)
X.requires_grad = True
W = np.random.randn(15,15)
W = np.triu(W, k=0)
W = torch.from_numpy(W)
W.requires_grad = True

# define parameters for gradient descent
max_iter=100
lr_rate = 1e-3

# we will do gradient descent for max_iter iteration 
for i in range(max_iter):

        # compute the loss
        loss = Y - (X@torch.transpose(X, 1,0) * W).sum()
        # use torch.autograd.grad to compute the gradient
        W = W - lr_rate*torch.autograd.grad(out, W)[0]
        print(f"{i}: {out}")

Не могли бы вы привести пример правильной реализации с использованием данных, которые я предоставляю ниже, которые позволят достичь заявленной цели (подгонка параметров к данным) в векторизованном виде?

Данные: X:

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Y

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Answer 1

Я думаю, что ваша функция потери неверна. В настоящее время вы использовали

loss = Y - (X@torch.transpose(X, 1,0) * W).sum()

В идеале, поскольку мы используем SGD и пытаемся минимизировать потери, мы хотим, чтобы потери были минимальными тогда и только тогда, когда Y равно f(x). Тем не менее, в вашем случае потери сводятся к минимуму, когда f(x) максимально велико, так как тогда Y-f(x) будет максимально малым. Обратите внимание, что потеря будет только 0, когда y=f(x), иначе она будет положительной или отрицательной. Быстрое исправление может заключаться в том, чтобы просто возместить потери, чтобы они всегда были положительными и равными 0 только тогда, когда y=f(x).

loss = (Y - (X@torch.transpose(X, 1,0) * W).sum())**2

Также, если Y,X - вектор и В матрицах (более одного образца) можно суммировать потери или вычислять среднее значение

loss = ((Y - (X@torch.transpose(X, 1,0) * W).sum())**2).mean()