is the mse layer really divide by the number of size ? #2246
Description
In the mse_cost in Layers.py. We see the description is
However, when I check the implementation it calls sumofsquarediff, which does not do a normalization of layer size N. I think it is a wrong description of the mse_cost layer ?
@wrap_name_default() @layer_support() def mse_cost(input, label, weight=None, name=None, layer_attr=None): """ mean squared error cost: .. math:: $\frac{1}{N}\sum_{i=1}^N(t _i- y_i)^2$ :param name: layer name. :type name: basestring :param input: Network prediction. :type input: LayerOutput :param label: Data label. :type label: LayerOutput :param weight: The weight affects the cost, namely the scale of cost. It is an optional argument. :type weight: LayerOutput :param layer_attr: layer's extra attribute. :type layer_attr: ExtraLayerAttribute :return: LayerOutput object. :rtype: LayerOutput """ ipts, parents = __cost_input__(input, label, weight) Layer( inputs=ipts, type="square_error", name=name, **ExtraLayerAttribute.to_kwargs(layer_attr)) return LayerOutput(name, LayerType.COST, parents=parents, size=1)