这是我对深度学习类(class)中 Andrew NG 的神经网络之一的自定义扩展，我正在尝试为二进制分类生成 0 或 1，而不是生成 0 或 1
对多个示例进行分类。

输入和输出都是一种热编码。

在没有太多训练的情况下，我的准确度为 'train accuracy: 67.51658067499625 %'
如何对单个训练示例进行分类而不是对所有训练示例进行分类？

我认为我的实现中存在一个错误，因为该网络的一个问题是训练示例 (train_set_x) 和输出值 (train_set_y) 都需要具有相同的维度，否则会收到与矩阵维度相关的错误。
例如使用:

train_set_x = np.array([
    [1,1,1,1],[0,1,1,1],[0,0,1,1]
])

train_set_y = np.array([
    [1,1,1],[1,1,0],[1,1,1]
])

ValueError                                Traceback (most recent call last)
<ipython-input-11-0d356e8d66f3> in <module>()
     27 print(A)
     28 
---> 29 np.multiply(train_set_y,A)
     30 
     31 def initialize_with_zeros(numberOfTrainingExamples):

import numpy as np
import matplotlib.pyplot as plt
import h5py
import scipy
from scipy import ndimage
import pandas as pd
%matplotlib inline

train_set_x = np.array([
    [1,1,1,1],[0,1,1,1],[0,0,1,1]
])

train_set_y = np.array([
    [1,1,1,0],[1,1,0,0],[1,1,1,1]
])

numberOfFeatures = 4
numberOfTrainingExamples = 3

def sigmoid(z):
    s = 1 / (1 + np.exp(-z))  
    return s

w = np.zeros((numberOfTrainingExamples , 1))
b = 0
A = sigmoid(np.dot(w.T , train_set_x))    
print(A)

np.multiply(train_set_y,A)

def initialize_with_zeros(numberOfTrainingExamples):
    w = np.zeros((numberOfTrainingExamples , 1))
    b = 0
    return w, b

def propagate(w, b, X, Y):

    m = X.shape[1]

    A = sigmoid(np.dot(w.T , X) + b)    

    cost = -(1/m)*np.sum(np.multiply(Y,np.log(A)) + np.multiply((1-Y),np.log(1-A)), axis=1)

    dw =  ( 1 / m ) *   np.dot( X, ( A - Y ).T )    # consumes ( A - Y )
    db =  ( 1 / m ) *   np.sum( A - Y )    # consumes ( A - Y ) again

#     cost = np.squeeze(cost)

    grads = {"dw": dw,
             "db": db}

    return grads, cost

def optimize(w, b, X, Y, num_iterations, learning_rate, print_cost = True):

    costs = []

    for i in range(num_iterations):

        grads, cost = propagate(w, b, X, Y)

        dw = grads["dw"]
        db = grads["db"]

        w = w - (learning_rate * dw)
        b = b - (learning_rate * db)

        if i % 100 == 0:
            costs.append(cost)

        if print_cost and i % 10000 == 0:
            print(cost)

    params = {"w": w,
              "b": b}

    grads = {"dw": dw,
             "db": db}

    return params, grads, costs

def model(X_train, Y_train, num_iterations, learning_rate = 0.5, print_cost = False):

    w, b = initialize_with_zeros(numberOfTrainingExamples)

    parameters, grads, costs = optimize(w, b, X_train, Y_train, num_iterations, learning_rate, print_cost = True)
    w = parameters["w"]
    b = parameters["b"]
    Y_prediction_train = sigmoid(np.dot(w.T , X_train) + b) 

    print("train accuracy: {} %".format(100 - np.mean(np.abs(Y_prediction_train - Y_train)) * 100))

model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.0001, print_cost = True)

train_set_x = np.array([
    [1,1,1,1,1],[0,1,1,1,1],[0,0,1,1,0],[0,0,1,0,1]
])

train_set_y = np.array([
    [1,0,0],[0,0,1],[0,1,0],[1,0,1]
])

grads = model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.001, print_cost = True)

# To classify single training example : 

print(sigmoid(dw @ [0,0,1,1,0] + db))

-2.09657359028
-3.94918577439
[[ 0.74043089  0.32851512  0.14776077  0.77970162]
 [ 0.04810012  0.08033521  0.72846174  0.1063849 ]
 [ 0.25956911  0.67148488  0.22029838  0.85223923]]
[[1 0 0 1]
 [0 0 1 0]
 [0 1 0 1]]
train accuracy: 79.84462279013312 %
[[ 0.51309252  0.48853845  0.50945862]
 [ 0.5110232   0.48646923  0.50738869]
 [ 0.51354109  0.48898712  0.50990734]]

print(sigmoid(dw @ np.array([[0,0,1,1,0]]).T + db))

array([[ 0.51309252],
       [ 0.48646923],
       [ 0.50990734]])

最佳答案

您的困难来自不匹配的维度，所以让我们解决问题并尝试解决问题。

您的网络有许多输入和功能，让我们称其为 N_in ( numberOfFeatures 在您的代码中)。它有许多对应于不同类别的输出，我们称其为 N_out .输入和输出通过权重连接 w .

现在问题来了。连接是多对多的，所以我们需要为 N_out x N_in 中的每一个添加一个权重。成对的输出和输入。因此在您的代码中 w 的形状必须改为 (N_out, N_in) .您可能还需要一个偏移量 b对于每个输出，所以 b 应该是大小为 (N_out,) 的向量或者更确切地说 (N_out, 1)所以它适用于 2d 术语。

我已经在下面的修改后的代码中修复了这个问题，并试图让它变得非常明确。我还提出了一个模拟数据创建者的讨价还价。

关于 one-hot 编码的分类输出，我不是神经网络方面的专家，但我认为，大多数人都理解它，因此类是互斥的，因此模拟输出中的每个样本都应该有一个 1，其余的都是 0。

边注:

有一次，一个竞争性的答案建议您摆脱 1-...成本函数中的术语。虽然这对我来说似乎是一个有趣的想法，但我的直觉( 编辑 现在使用无梯度最小化器确认；在下面的代码中使用 activation="hybrid"。求解器将简单地最大化所有至少在一个训练示例。)它不会像那样工作，因为成本将无法惩罚误报(详细说明见下文)。为了使它工作，你必须添加某种正则化。一种似乎有效的方法是使用 softmax而不是 sigmoid . softmax是为了一热么sigmoid是二进制。它确保输出是“模糊一热”。

因此我的建议是:

-y log (y') - (1-y) log (1-y')

-log (y')   if   y = 1
-log(1-y')  if   y = 0

import numpy as np
from scipy import optimize as opt
from collections import namedtuple

# First, a few structures to keep ourselves organized

Problem_Size = namedtuple('Problem_Size', 'Out In Samples')
Data = namedtuple('Data', 'Out In')
Network = namedtuple('Network', 'w b activation cost gradient most_likely')

def get_dims(Out, In, transpose=False):
    """extract dimensions and ensure everything is 2d
    return Data, Dims"""
    # gracefully acccept lists etc.
    Out, In = np.asanyarray(Out), np.asanyarray(In)
    if transpose:
        Out, In = Out.T, In.T
    # if it's a single sample make sure it's n x 1
    Out = Out[:, None] if len(Out.shape) == 1 else Out
    In = In[:, None] if len(In.shape) == 1 else In
    Dims = Problem_Size(Out.shape[0], *In.shape)
    if Dims.Samples != Out.shape[1]:
        raise ValueError("number of samples must be the same for Out and In")
    return Data(Out, In), Dims


def sigmoid(z):
    s = 1 / (1 + np.exp(-z))  
    return s

def sig_cost(Net, data):
    A = process(data.In, Net)
    logA = np.log(A)
    return -(data.Out * logA + (1-data.Out) * (1-logA)).sum(axis=0).mean()

def sig_grad (Net, Dims, data):
    A = process(data.In, Net)
    return dict(dw =  (A - data.Out) @ data.In.T / Dims.Samples,
                db =  (A - data.Out).mean(axis=1, keepdims=True))

def sig_ml(z):
    return np.round(z).astype(int)

def sof_ml(z):
    hot = np.argmax(z, axis=0)
    z = np.zeros(z.shape, dtype=int)
    z[hot, np.arange(len(hot))] = 1
    return z

def softmax(z):
    z = z - z.max(axis=0, keepdims=True)
    z = np.exp(z)
    return z / z.sum(axis=0, keepdims=True)

def sof_cost(Net, data):
    A = process(data.In, Net)
    logA = np.log(A)
    return -(data.Out * logA).sum(axis=0).mean()

sof_grad = sig_grad

def get_net(Dims, activation='softmax'):
    activation, cost, gradient, ml = {
        'sigmoid': (sigmoid, sig_cost, sig_grad, sig_ml),
        'softmax': (softmax, sof_cost, sof_grad, sof_ml),
        'hybrid': (sigmoid, sof_cost, None, sig_ml)}[activation]
    return Network(w=np.zeros((Dims.Out, Dims.In)),
                   b=np.zeros((Dims.Out, 1)),
                   activation=activation, cost=cost, gradient=gradient,
                   most_likely=ml)

def process(In, Net):
    return Net.activation(Net.w @ In + Net.b)

def propagate(data, Dims, Net):
    return Net.gradient(Net, Dims, data), Net.cost(Net, data)

def optimize_no_grad(Net, Dims, data):
    def f(x):
        Net.w[...] = x[:Net.w.size].reshape(Net.w.shape)
        Net.b[...] = x[Net.w.size:].reshape(Net.b.shape)
        return Net.cost(Net, data)
    x = np.r_[Net.w.ravel(), Net.b.ravel()]
    res = opt.minimize(f, x, options=dict(maxiter=10000)).x
    Net.w[...] = res[:Net.w.size].reshape(Net.w.shape)
    Net.b[...] = res[Net.w.size:].reshape(Net.b.shape)

def optimize(Net, Dims, data, num_iterations, learning_rate, print_cost = True):

    w, b = Net.w, Net.b
    costs = []

    for i in range(num_iterations):

        grads, cost = propagate(data, Dims, Net)

        dw = grads["dw"]
        db = grads["db"]

        w -= learning_rate * dw
        b -= learning_rate * db

        if i % 100 == 0:
            costs.append(cost)

        if print_cost and i % 10000 == 0:
            print(cost)

    return grads, costs

def model(X_train, Y_train, num_iterations, learning_rate = 0.5, print_cost = False, activation='sigmoid'):

    data, Dims = get_dims(Y_train, X_train, transpose=True)
    Net = get_net(Dims, activation)

    if Net.gradient is None:
        optimize_no_grad(Net, Dims, data)
    else:
        grads, costs = optimize(Net, Dims, data, num_iterations, learning_rate, print_cost = True)

    Y_prediction_train = process(data.In, Net)

    print(Y_prediction_train)
    print(data.Out)
    print(Y_prediction_train.sum(axis=0))
    print("train accuracy: {} %".format(100 - np.mean(np.abs(Y_prediction_train - data.Out)) * 100))
    return Net

def predict(In, Net, probability=False):
    In = np.asanyarray(In)
    is1d = In.ndim == 1
    if is1d:
        In = In.reshape(-1, 1)
    Out = process(In, Net)
    if not probability:
        Out = Net.most_likely(Out)
    if is1d:
        Out = Out.reshape(-1)
    return Out

def create_data(Dims):
    Out = np.zeros((Dims.Out, Dims.Samples), dtype=int)
    Out[np.random.randint(0, Dims.Out, (Dims.Samples,)), np.arange(Dims.Samples)] = 1
    In = np.random.randint(0, 2, (Dims.In, Dims.Samples))
    return Data(Out, In)

train_set_x = np.array([
    [1,1,1,1,1],[0,1,1,1,1],[0,0,1,1,0],[0,0,1,0,1]
])

train_set_y = np.array([
    [1,0,0],[1,0,0],[0,0,1],[0,0,1]
])

Net1 = model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.001, print_cost = True, activation='sigmoid')

Net2 = model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.001, print_cost = True, activation='softmax')

Net3 = model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.001, print_cost = True, activation='hybrid')

Dims = Problem_Size(8, 100, 50)
data = create_data(Dims)
model(data.In.T, data.Out.T, num_iterations = 40000, learning_rate = 0.001, print_cost = True, activation='softmax') 
model(data.In.T, data.Out.T, num_iterations = 40000, learning_rate = 0.001, print_cost = True, activation='sigmoid') 

关于python - 修改神经网络对单个示例进行分类，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47308181/