我正在熟悉 Theano 和机器学习。为了达到这个目的,我想计算一个线性回归。我的代码灵感来自于 Theano 介绍中的逻辑回归示例。
我编写了以下代码:
import numpy import theano import theano.tensor as T class LinearRegression(object): """ 计算线性回归 """ def __init__(self, input): """ 初始化逻辑回归的参数 参数: ----------- :type input: theano.tensor.TensorType :param input: 描述架构输入的符号变量(一个小批量) """ self.W = theano.shared( value=numpy.zeros(1, dtype=theano.config.floatX), name='W', borrow=True ) self.b = theano.shared( value=numpy.zeros(1, dtype=theano.config.floatX), name='b', borrow=True ) self.y_pred = T.dot(input, self.W) + self.b def errors(self, y): """ 平方距离 参数: ---------- :y input: array_like: :param input: 样本数据 """ errors = y- self.y_pred return T.sum(T.pow(errors, 2)) def sgd_optimization(learning_rate=0.0013, n_epochs=100): """ 演示线性模型的随机梯度下降优化 参数: ----- :type learning_rate: float :param learning_rate: 学习率(随机梯度的因子) :type n_epochs: int :param n_epochs: 运行优化器的最大轮数 """ x_train = numpy.random.uniform(low=-2, high = 2, size=(50,1)) epsilon = numpy.random.normal(scale=0.01, size=50) y_train = numpy.squeeze(2*x_train) + epsilon costs = [] eta0, x, y = T.scalar('eta0'), T.matrix(name='x'), T.vector(name='y') classifier = LinearRegression(input = x) cost = classifier.errors(y) g_W = T.grad(cost=cost, wrt=classifier.W) g_b = T.grad(cost=cost, wrt=classifier.b) update = [(classifier.W, classifier.W - eta0 * g_W), (classifier.b, classifier.b - eta0 * g_b)] train = theano.function(inputs = [eta0], outputs = cost, updates = update, givens = {x: x_train, y: y_train}) for _ in range(n_epochs): costs.append(train(learning_rate)) return costs, w SSE, regressor = sgd_optimization()
遗憾的是,当我运行这段代码时,Python 返回了以下错误信息:
ValueError: Input dimension mis-match. (input[0].shape[0] = 1, input[1].shape[0] = 50)Apply node that caused the error: Elemwise{Composite{((-i0) + i1)}}[(0, 1)](b, CGemv{no_inplace}.0)Inputs types: [TensorType(float64, vector), TensorType(float64, vector)]Inputs shapes: [(1,), (50,)]Inputs strides: [(8,), (8,)]Inputs values: [array([ 0.]), 'not shown']HINT: Re-running with most Theano optimization disabled could give you a back-trace of when this node was created. This can be done with by setting the Theano flag 'optimizer=fast_compile'. If that does not work, Theano optimizations can be disabled with 'optimizer=None'.HINT: Use the Theano flag 'exception_verbosity=high' for a debugprint and storage map footprint of this apply node.
我怀疑错误与样本数据的维度为(50,1)而回归器的维度仅为(1,1)有关。尽管如此,我尝试了一段时间也没能纠正代码中的错误。有人能提供一些如何纠正这个错误的提示吗?我对任何帮助都表示感谢!
回答:
你需要对 b
进行广播:
self.y_pred = T.dot(input, self.W) + self.b[:, None]
我原本期望 Theano 会自动完成这个操作,但似乎并非如此。
要定位问题,请按照错误消息的建议,运行 Theano 并设置高异常详细度
$ THEANO_FLAGS='exception_verbosity=high' python path/to/script.py
这会产生大量输出,包括有问题的节点及其操作数
Debugprint of the apply node:Elemwise{Composite{((-i0) + i1)}}[(0, 1)] [@A] <TensorType(float64, vector)> '' |b [@B] <TensorType(float64, vector)> |CGemv{no_inplace} [@C] <TensorType(float64, vector)> '' |<TensorType(float64, vector)> [@D] <TensorType(float64, vector)> |TensorConstant{-1.0} [@E] <TensorType(float64, scalar)> |<TensorType(float64, matrix)> [@F] <TensorType(float64, matrix)> |W [@G] <TensorType(float64, vector)> |TensorConstant{1.0} [@H] <TensorType(float64, scalar)>
该节点对应于从临时节点 CGemv{no_inplace}
中减去 b
。涉及 b
的唯一代码行是
self.y_pred = T.dot(input, self.W) + self.b