Question

我正在尝试训练这个神经网络来预测某些数据。我在一个小数据集（大约100条记录）上尝试过它，它就像一个魅力。然后我插入新的数据集，我发现NN收敛到0输出，误差大致收敛到正例数和例子总数之间的比率。

我的数据集由是/否功能（1.0 / 0.0）组成，基本事实也是/否。

我的假设：
1）输出为0的局部最小值（但我尝试了许多学习率和初始权重值，似乎总是在那里收敛）
2）我的体重更新错了（但对我来说很好看）
3）它只是一个输出缩放问题。我试图缩放输出（即输出/最大（输出）和输出/平均值（输出））但结果并不好，您可以在下面提供的代码中看到。我应该以不同的方式扩展它吗？ SOFTMAX？

这是代码：

import pandas as pd
import numpy as np
import pickle
import random
from collections import defaultdict

alpha = 0.1
N_LAYERS = 10
N_ITER = 10
#N_FEATURES = 8
INIT_SCALE = 1.0

train = pd.read_csv("./data/prediction.csv")

y = train['y_true'].as_matrix()
y = np.vstack(y).astype(float)
ytest = y[18000:]
y = y[:18000]

X = train.drop(['y_true'], axis = 1).as_matrix()
Xtest = X[18000:].astype(float)
X = X[:18000]

def tanh(x,deriv=False):
    if(deriv==True):
        return (1 - np.tanh(x)**2) * alpha
    else:
        return np.tanh(x)

def sigmoid(x,deriv=False):
    if(deriv==True):
        return x*(1-x)
    else:
        return 1/(1+np.exp(-x))

def relu(x,deriv=False):
    if(deriv==True):
        return 0.01 + 0.99*(x>0)
    else:
        return 0.01*x + 0.99*x*(x>0)

np.random.seed()

syn = defaultdict(np.array)

for i in range(N_LAYERS-1):
    syn[i] = INIT_SCALE * np.random.random((len(X[0]),len(X[0]))) - INIT_SCALE/2
syn[N_LAYERS-1] = INIT_SCALE * np.random.random((len(X[0]),1)) - INIT_SCALE/2

l = defaultdict(np.array)
delta = defaultdict(np.array)

for j in xrange(N_ITER):
    l[0] = X
    for i in range(1,N_LAYERS+1):
        l[i] = relu(np.dot(l[i-1],syn[i-1]))

    error = (y - l[N_LAYERS])

    e = np.mean(np.abs(error))
    if (j% 1) == 0:
        print "\nIteration " + str(j) + " of " + str(N_ITER)
        print "Error: " + str(e)

    delta[N_LAYERS] = error*relu(l[N_LAYERS],deriv=True) * alpha
    for i in range(N_LAYERS-1,0,-1):
        error = delta[i+1].dot(syn[i].T)
        delta[i] = error*relu(l[i],deriv=True) * alpha

    for i in range(N_LAYERS):
        syn[i] += l[i].T.dot(delta[i+1])



pickle.dump(syn, open('neural_weights.pkl', 'wb'))

# TESTING with f1-measure
# RECALL = TRUE POSITIVES / ( TRUE POSITIVES + FALSE NEGATIVES)
# PRECISION = TRUE POSITIVES / (TRUE POSITIVES + FALSE POSITIVES)

l[0] = Xtest
for i in range(1,N_LAYERS+1):
    l[i] = relu(np.dot(l[i-1],syn[i-1]))

out = l[N_LAYERS]/max(l[N_LAYERS])

tp = float(0)
fp = float(0)
fn = float(0)
tn = float(0)

for i in l[N_LAYERS][:50]:
    print i

for i in range(len(ytest)):
    if out[i] > 0.5 and ytest[i] == 1:
        tp += 1
    if out[i] <= 0.5 and ytest[i] == 1:
        fn += 1
    if out[i] > 0.5 and ytest[i] == 0:
        fp += 1
    if out[i] <= 0.5 and ytest[i] == 0:
        tn += 1

print "tp: " + str(tp)
print "fp: " + str(fp)
print "tn: " + str(tn)
print "fn: " + str(fn)

print "\nprecision: " + str(tp/(tp + fp))
print "recall: " + str(tp/(tp + fn))

f1 = 2 * tp /(2 * tp + fn + fp)
print "\nf1-measure:" + str(f1)

这是输出：

Iteration 0 of 10
Error: 0.222500767998

Iteration 1 of 10
Error: 0.222500771157

Iteration 2 of 10
Error: 0.222500774321

Iteration 3 of 10
Error: 0.22250077749

Iteration 4 of 10
Error: 0.222500780663

Iteration 5 of 10
Error: 0.222500783841

Iteration 6 of 10
Error: 0.222500787024

Iteration 7 of 10
Error: 0.222500790212

Iteration 8 of 10
Error: 0.222500793405

Iteration 9 of 10
Error: 0.222500796602


[ 0.]
[ 0.]
[  5.58610895e-06]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[  4.62182626e-06]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[  5.58610895e-06]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[  4.62182626e-06]
[ 0.]
[ 0.]
[  5.04501079e-10]
[  5.58610895e-06]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[ 0.]
[  5.04501079e-10]
[ 0.]
[ 0.]
[  4.62182626e-06]
[ 0.]
[  5.58610895e-06]
[ 0.]
[ 0.]
[ 0.]
[  5.58610895e-06]
[ 0.]
[ 0.]
[ 0.]
[  5.58610895e-06]
[ 0.]
[  1.31432294e-05]

tp: 28.0
fp: 119.0
tn: 5537.0
fn: 1550.0

precision: 0.190476190476
recall: 0.0177439797212

f1-measure:0.0324637681159

Answer 1

根据您的模型，您的网络不太可能需要10层才能收敛。

尝试使用更多隐藏节点的3层网络。对于大多数前馈问题，您只需要1个隐藏层即可有效收敛。

Deep NN的训练要比浅的训练困难得多。

像其他人一样，你说学习率应该小得多[.01，.3]是一个不错的范围，另外迭代次数需要更大。

10层太多了。

神经网络收敛到零输出

1 个答案: