Keras自定义RMSLE指标

Question

如何在Keras中实施此指标？我的代码下面给出了错误的结果!请注意,我正在通过exp(x) - 1撤消先前的log(x + 1)变换,负面预测也会被剪切为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)

为了比较,这是标准的numpy实现:

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

我做错了什么？谢谢!

编辑:设置axis=0似乎给出了一个非常接近正确值的值,但我不确定,因为我似乎使用了所有代码axis=-1.

Answer 1

我遇到了同样的问题并搜索了它,这是我发现的

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

经过修改后,这似乎对我有用,rmsle_K用Keras和实现的方法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

0.263978210565

0.263978210565

0.263978