Keras自定义RMSLE指标 [英] Keras custom RMSLE metric

问题描述

如何在Keras中实施此指标？我的以下代码给出了错误的结果！
注意，我正在通过exp（x）-1撤消以前的log（x + 1）转换，否定预测也被裁剪为0：

How do I implement this metric in Keras? My code below gives the wrong result! Note that I'm undoing a previous log(x + 1) transformation via exp(x) - 1, also negative predictions are clipped to 0:

def rmsle_cust(y_true, y_pred):
    first_log = K.clip(K.exp(y_pred) - 1.0, 0, None)
    second_log = K.clip(K.exp(y_true) - 1.0, 0, None)
    return K.sqrt(K.mean(K.square(K.log(first_log + 1.) - K.log(second_log + 1.)), axis=-1)

为进行比较，下面是标准的numpy实现：

For comparison, here's the standard numpy implementation:

def rmsle_cust_py(y, y_pred, **kwargs):
    # undo 1 + log
    y = np.exp(y) - 1
    y_pred = np.exp(y_pred) - 1

    y_pred[y_pred < 0] = 0.0
    to_sum = [(math.log(y_pred[i] + 1) - math.log(y[i] + 1)) ** 2.0 for i,pred in enumerate(y_pred)]
    return (sum(to_sum) * (1.0/len(y))) ** 0.5

我做错了吗？

编辑：设置 axis = 0 的值似乎非常接近正确，但是我不确定，因为我似乎所有的代码都使用 axis = -1 。

Setting axis=0 seems to give a value very close to the correct one, but I'm not sure since all the code I've seem uses axis=-1.

推荐答案

我遇到了同样的问题并进行了搜索，这是我发现的问题

I ran into the same problem and searched for it, here is what I found

https://www.kaggle.com/jpopham91/rmlse-vectorized

稍作修改后，这似乎对我有用， rmsle_K 方法通过 Keras 和<$实现c $ c> TensorFlow 。

After modified a bit, this seems to work for me,rmsle_K method implemented with Keras and TensorFlow.

import numpy as np
import math
from keras import backend as K
import tensorflow as tf

def rmsle(y, y0):
    assert len(y) == len(y0)
    return np.sqrt(np.mean(np.power(np.log1p(y)-np.log1p(y0), 2)))

def rmsle_loop(y, y0):
    assert len(y) == len(y0)
    terms_to_sum = [(math.log(y0[i] + 1) - math.log(y[i] + 1)) ** 2.0 for i,pred in enumerate(y0)]
    return (sum(terms_to_sum) * (1.0/len(y))) ** 0.5

def rmsle_K(y, y0):
    return K.sqrt(K.mean(K.square(tf.log1p(y) - tf.log1p(y0))))

r = rmsle(y=[5, 20, 12], y0=[8, 16, 12])
r1 = rmsle_loop(y=[5, 20, 12], y0=[8, 16, 12])
r2 = rmsle_K(y=[5., 20., 12.], y0=[8., 16., 12.])

print(r)

print(r1)

sess = tf.Session()

print(sess.run(r2))

结果：

使用TensorFlow后端

Using TensorFlow backend

0.263978210565

0.263978210565

0.263978

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