(自定义)百分比MSE损失函数



我有一个Keras模型,它有输入x_1,。。。,x_ n和d维输出f(x_1(,。。。,f(x_n(。我正在研究一个d维目标y_1,…的回归问题,。。。,y_ n。

我想尽量减少损失函数:对于0和1之间的固定元参数a,返回|f(x_i)-y_i|^2的第a^个(经验(分位数。

以下是我迄今为止编码的内容:

def keras_custom_loss(y_true,y_predicted):
SEs = K.square(y_true-y_predicted)
out_custom = tfp.stats.percentile(SEs, 50.0, interpolation='midpoint')
return out_custom

一个问题是,我希望避免使用tensorflow_propability,并且我希望在Keras上完成整个实现。

然而,我不知道该怎么做。

要将"所有元素"都取到该百分位数以上,您需要一个不同的答案:

import keras.backend as K
from keras.layers import *
from keras.models import Model
import numpy as np
import tensorflow as tf
def above_percentile(x, p): #assuming the input is flattened: (n,)
samples = K.cast(K.shape(x)[0], K.floatx()) #batch size
p =  (100. - p)/100.  #100% will return 0 elements, 0% will return all elements
#samples to get:
samples = K.cast(tf.math.floor(p * samples), 'int32')
#you can choose tf.math.ceil above, it depends on whether you want to
#include or exclude one element. Suppose you you want 33% top,
#but it's only possible to get exactly 30% or 40% top:
#floor will get 30% top and ceil will get 40% top.
#(exact matches included in both cases)
#selected samples
values, indices = tf.math.top_k(x, samples)
return values
def custom_loss(p):
def loss(y_true, y_predicted):
ses = K.square(y_true-y_predicted)
above = above_percentile(K.flatten(ses), p)
return K.mean(above)
return loss

测试:

dataX = np.array([2,3,1,4,7,10,8,5,6]).reshape((-1,1))
dataY = np.ones((9,1))

ins = Input((1,))
outs = Lambda(lambda x: x)(ins)
model = Model(ins, outs)
model.compile(optimizer='adam', loss = custom_loss(70.))
model.fit(dataX, dataY)

损失为65,即130/2(平均值(。而130 = (10-1)² + (8-1)²,即108是输入中的两个顶部k。

对于您的特定用例,您可以使用以下函数,它是tfp.stats.percentile的简化版本(它们使用Apache License 2.0(:

import tensorflow as tf
def percentile(x, p):
with tf.name_scope('percentile'):
y = tf.transpose(x)  # take percentile over batch dimension
sorted_y = tf.sort(y)
frac_idx = tf.cast(p, tf.float64) / 100. * (tf.cast(tf.shape(y)[-1], tf.float64) - 1.)
return 0.5 * (  # using midpoint rule
tf.gather(sorted_y, tf.math.ceil(frac_idx), axis=-1)
+ tf.gather(sorted_y, tf.math.floor(frac_idx), axis=-1))

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