作为度量函数的平衡错误率 [英] Balanced Error Rate as metric function

问题描述

我正在尝试使用 Keras 的序列模型解决二元分类问题
并且必须满足给定的平衡错误率 (BER)

所以我认为使用 BER 而不是准确性作为衡量标准是个好主意.

我的 BER 自定义指标实现如下所示:

defbalanced_error_rate(y_true, y_pred):标签 = theano.shared(np.asmatrix([[0, 1]], dtype='int8'))label_matrix = K.repeat_elements(labels, K.shape(y_true)[0],axis=1)true_matrix = K.repeat_elements(y_true, K.shape(labels)[0], axis=1)pred_matrix = K.repeat_elements(K.round(y_pred), K.shape(labels)[0],axis=1)class_lens = K.sum(K.equal(label_matrix, true_matrix),axis=1)return K.sum(K.sum(class_lens - K.sum(K.equal(label_matrix, K.not_equal(true_matrix,pred_matrix)),axis=1),axis=0)/class_lens,axis=0)/2

这个想法是从可用标签创建一个矩阵，并将其与输入数据进行比较(然后求和)以获得此标签的元素数....

我的问题是:

<代码>>K.shape(y_true)形状.0>类型信息:>类型(y_true)<class 'theano.tensor.var.TensorVariable'>>类型(K.shape(y_true))<class 'theano.tensor.var.TensorVariable'>

...我不知道为什么.

<小时>

我正在寻找:

获取数组维度的方法/解释为什么 shape 表现得像它/为什么 y_true 似乎具有 0 维度

或

一种通过重复给定的行/列向量来创建具有给定宽度/高度的张量矩阵的方法.

或

使用张量函数计算 BER 的更智能解决方案.

解决方案

获取数组维度的方法/解释为什么形状像它一样/为什么 y_true 似乎有 0 个维度

print 和像 Theano 这样的抽象库的问题是，你通常不会得到值，而是值的表示.所以如果你这样做

print(foo.shape)

您不会得到实际的形状，而是在运行时完成的操作的表示.由于这都是在外部设备上计算的，因此计算不会立即运行，而是在创建具有适当输入的函数(或调用 foo.shape.eval())后才运行.

打印值的另一种方法是在使用值时使用theano.printing.Print，例如:

shape = theano.printing.Print('foo的形状')(foo.shape)# 使用形状(不是 foo.shape！)

<块引用>

一种通过重复给定的行/列向量来创建具有给定宽度/高度的张量矩阵的方法.

参见theano.tensor.重复.numpy 中的示例(用法非常相似):

<预><代码>>>>X数组([[1, 2, 3]])>>>x.repeat(3,axis=0)数组([[1, 2, 3],[1, 2, 3],[1, 2, 3]])

I am trying to solve a binary classification problem with the sequential model from Keras
and have to meet a given Balanced Error Rate (BER)

So I thought it would be a good idea to use the BER instead of accuracy as a metric.

My custom metric implementation for BER looks like this:

def balanced_error_rate(y_true, y_pred):
    labels = theano.shared(np.asmatrix([[0, 1]], dtype='int8'))
    label_matrix = K.repeat_elements(labels, K.shape(y_true)[0], axis=1)
    true_matrix = K.repeat_elements(y_true, K.shape(labels)[0], axis=1)
    pred_matrix = K.repeat_elements(K.round(y_pred), K.shape(labels)[0], axis=1)

    class_lens = K.sum(K.equal(label_matrix, true_matrix), axis=1)
    return K.sum(K.sum(class_lens - K.sum(K.equal(label_matrix, K.not_equal(true_matrix,pred_matrix)), axis=1), axis=0)/class_lens, axis=0)/2

The idea is to create a matrix from the available labels and compare it to the input data (then sum the ones) to get the number of elements of this label....

My problem is that:

> K.shape(y_true) 
Shape.0    


> Typeinfo:

> type(y_true)
<class 'theano.tensor.var.TensorVariable'>

> type(K.shape(y_true))
<class 'theano.tensor.var.TensorVariable'>

...and I can't find out why.

---

I am now looking for:

A way to get the array dimensions / an explanation why shape acts like it does / the reason why y_true seems to have 0 dimensions

or

A method to create a tensor matrix with a given with/height by repeating a given row/column vector.

or

A smarter solution to calculate the BER using tensor functions.

解决方案

A way to get the array dimensions / an explanation why shape acts like it does / the reason why y_true seems to have 0 dimensions

The deal with print and abstraction libraries like Theano is that you usually do not get the values but a represenation of the value. So if you do

print(foo.shape)

You won't get the actual shape but a representation of the operation that is done at runtime. Since this is all computed on an external device the computation is not run immediately but only after creating a function with appropriate inputs (or calling foo.shape.eval()).

Another way to print the value is to use theano.printing.Print when using the value, e.g.:

shape = theano.printing.Print('shape of foo')(foo.shape)
# use shape (not foo.shape!)

A method to create a tensor matrix with a given with/height by repeating a given row/column vector.

See theano.tensor.repeat for that. Example in numpy (usage is quite similar):

>>> x
array([[1, 2, 3]])
>>> x.repeat(3, axis=0)
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

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