台大李宏毅教授机器学习HW2 作业二 赢家还是输家

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台大李宏毅教授机器学习HW2 作业二 赢家还是输家

首先数据文件

好像在网上只能搜到这一个文件，是一个4000*58的表格，其中最后一列表示标签，如果是0的话表示工资小于50k，如果是1表工资大于50k，前57列是一个人的不同属性，我们需要用这57个属性来预测一个人的工资是否会大于50k

由于只有一个文件，所以将文件的80%用作训练，其余部分用作测试。

import numpy as np
import pandas as pddef read2train():path = 'F:\\python_book\\machine_learning\\spam_train.csv't = pd.read_csv(path)t = t.iloc[:, 1:]data = np.array(t, float)index = int(data.shape[0]*0.8)train_x = data[:index, :-1]train_y = data[:index, -1]# 由于文件中的标签只有一列，但是这是一个二分类问题，所以我在这里进行了处理# 已知1表示大于50k 0表示小于50k，那么1-train_y的值若是1就表示小于50k的概率为1# 若是0就表示小于50k的概率为0，这样做是为了与sigmoid的结果求误差a = 1 - train_ytrain_y = np.vstack((train_y, a)).reshape(train_x.shape[0], -1)test_x = data[index:, :-1]test_y = data[index:, -1].reshape(test_x.shape[0], -1)a = 1 - test_ytest_y = np.vstack((test_y, a)).reshape(test_x.shape[0], -1)return train_x, train_y, test_x, test_y

import numpy as npdef softmax(x):x = x - np.max(x)  # 溢出对策return np.exp(x) / np.sum(np.exp(x))'''
由于one-hot表示中t为0的元素的交叉熵也为0，因此针对这些元素的计算可以忽略。
只需要获得网络中在正确解标签处的输出即可计算交叉熵误差，1e-7 是为了防止出现负无穷大的结果
'''def cross_entry_error(y, t):if y.ndim == 1:t = t.reshape(1, t.size)y = y.reshape(1, y.size)# 监督数据是one-hot-vector的情况下，转换为正确解标签的索引if t.size == y.size:t = t.argmax(axis=1)batch_size = int(y.shape[0])return -np.sum(np.log(y[np.arange(batch_size), t] + 1e-7)) / batch_size# 动量法更新参数class Momentum:def __init__(self, lr=0.01, momentum=0.9):self.lr = lrself.momentum = momentumself.v = Nonedef update(self, params, grads):if self.v is None:self.v = {}for key, val in params.items():self.v[key] = np.zeros_like(val)for key in params.keys():self.v[key] = self.momentum * self.v[key] - self.lr * grads[key]params[key] += self.v[key]class AdaGrad:def __init__(self, lr=0.01):self.lr = lrself.h = Nonedef update(self, params, grads):if self.h is None:self.h = {}for key, val in params.items():self.h[key] = np.zeros_like(val)for key in params.keys():self.h[key] += grads[key]**2params[key] -= self.lr * grads[key] / (np.sqrt(self.h[key]) + 1e-7)# 乘法层的实现，用来进行w*x+b的前向和后向传播class Affine:def __init__(self, w, b):self.w = wself.b = bself.x = Noneself.dw = Noneself.db = Nonedef forward(self, x):self.x = xreturn np.dot(x, self.w) + self.bdef backward(self, dout):self.dw = np.dot(self.x.T, dout)dx = np.dot(dout, self.w.T)self.db = np.sum(dout, axis=0)return dx# Sigmoid用来实现Sigmoid函数的前向和后向传播class Sigmoid:def __init__(self):self.out = Nonedef forward(self, x):self.out = 1 / (1 + np.exp(-x))return self.outdef backward(self, dout):dx = dout * (1 - self.out) * self.outreturn dx# 对进过进行softmax计算出概率之后，求出交叉熵误差,这层反向传播的结果是y-t
# 最后面除以batch_size，则传给前面层的是单个数据的误差
# 我试了一下不写这句，好像收敛的幅度更大，但这种写法是我在参考书上看的，所以就照做了class SoftMaxWithLoss:def __init__(self):self.loss = None  # 交叉熵误差self.y = None  # softmax的输出self.t = None  # 监督数据def forward(self, x, t):self.t = tself.y = softmax(x)self.loss = cross_entry_error(self.y, self.t)return self.lossdef backward(self):batch_size = self.y.shape[0]dx = (self.y - self.t) / batch_sizereturn dx

from util import *
from collections import OrderedDictclass Net:def __init__(self, input_size, output_size):self.params = {}self.params['w1'] = np.random.randn(input_size, output_size)self.params['b1'] = np.zeros(output_size)self.layers = OrderedDict()self.layers['Affine1'] = Affine(self.params['w1'], self.params['b1'])self.layers['Sigmoid1'] = Sigmoid()self.lastLayer = SoftMaxWithLoss()def predict(self, x):for layer in self.layers.values():x = layer.forward(x)return xdef loss(self, x, t):y = self.predict(x)return self.lastLayer.forward(y, t)def gradient(self, x, t):loss = self.loss(x, t)dout = self.lastLayer.backward()layers = list(self.layers.values())layers.reverse()for layer in layers:dout = layer.backward(dout)grads = {}grads['w1'] = self.layers['Affine1'].dwgrads['b1'] = self.layers['Affine1'].dbreturn gradsdef accuracy(self, x, t):y = self.predict(x)accuracy = np.sum(y == t) / float(x.shape[0])return accuracy

from read_Data import *
from logistic_regression import *train_x, train_y, test_x, test_y = read2train()
network = Net(train_x.shape[1], 2)
Ada = AdaGrad(lr=0.001)
Mom = Momentum()
for i in range(100000):grads = network.gradient(train_x, train_y)# Mom.update(network.params, grads)Ada.update(network.params, grads)if i % 1000 == 0:acc = network.accuracy(test_x, test_y)print(i, acc)

最后准确率能达到90以上

下面这些图片来帮助大家理解相关公式的来由

补充：由于本人是新手，且网上关于李宏毅教授的作业二的资料较少，我就写了一个发出来，不管是代码的深度还是规范程度上来讲都得往后稍一稍，但我还是希望能帮助到在入门的各位同行们！冲就完事儿了

欢迎大家在下面评论讨论共同进步！