基于keras的波士顿房价预测

数据来源kaggle,官网都有介绍。https://www.kaggle.com/c/boston-housing

from keras.datasets import boston_housing

(train_data,train_targets),(test_data,test_targets) = boston_housing.load_data()
#训练集形状：
print(train_data.shape)
#测试集形状:
print(test_data.shape)

D:\ANACONDA\ana3.5.2\lib\site-packages\h5py\__init__.py:36: FutureWarning: Conversion of the second argument of issubdtype from `float` to `np.floating` is deprecated. In future, it will be treated as `np.float64 == np.dtype(float).type`.
  from ._conv import register_converters as _register_converters
Using TensorFlow backend.


(404, 13)
(102, 13)

#房价数据，房价范围在10,000到50,000。
train_targets

array([15.2, 42.3, 50. , 21.1, 17.7, 18.5, 11.3, 15.6, 15.6, 14.4, 12.1,
       17.9, 23.1, 19.9, 15.7,  8.8, 50. , 22.5, 24.1, 27.5, 10.9, 30.8,
       32.9, 24. , 18.5, 13.3, 22.9, 34.7, 16.6, 17.5, 22.3, 16.1, 14.9,
       23.1, 34.9, 25. , 13.9, 13.1, 20.4, 20. , 15.2, 24.7, 22.2, 16.7,
       12.7, 15.6, 18.4, 21. , 30.1, 15.1, 18.7,  9.6, 31.5, 24.8, 19.1,
       22. , 14.5, 11. , 32. , 29.4, 20.3, 24.4, 14.6, 19.5, 14.1, 14.3,
       15.6, 10.5,  6.3, 19.3, 19.3, 13.4, 36.4, 17.8, 13.5, 16.5,  8.3,
       14.3, 16. , 13.4, 28.6, 43.5, 20.2, 22. , 23. , 20.7, 12.5, 48.5,
       14.6, 13.4, 23.7, 50. , 21.7, 39.8, 38.7, 22.2, 34.9, 22.5, 31.1,
       28.7, 46. , 41.7, 21. , 26.6, 15. , 24.4, 13.3, 21.2, 11.7, 21.7,
       19.4, 50. , 22.8, 19.7, 24.7, 36.2, 14.2, 18.9, 18.3, 20.6, 24.6,
       18.2,  8.7, 44. , 10.4, 13.2, 21.2, 37. , 30.7, 22.9, 20. , 19.3,
       31.7, 32. , 23.1, 18.8, 10.9, 50. , 19.6,  5. , 14.4, 19.8, 13.8,
       19.6, 23.9, 24.5, 25. , 19.9, 17.2, 24.6, 13.5, 26.6, 21.4, 11.9,
       22.6, 19.6,  8.5, 23.7, 23.1, 22.4, 20.5, 23.6, 18.4, 35.2, 23.1,
       27.9, 20.6, 23.7, 28. , 13.6, 27.1, 23.6, 20.6, 18.2, 21.7, 17.1,
        8.4, 25.3, 13.8, 22.2, 18.4, 20.7, 31.6, 30.5, 20.3,  8.8, 19.2,
       19.4, 23.1, 23. , 14.8, 48.8, 22.6, 33.4, 21.1, 13.6, 32.2, 13.1,
       23.4, 18.9, 23.9, 11.8, 23.3, 22.8, 19.6, 16.7, 13.4, 22.2, 20.4,
       21.8, 26.4, 14.9, 24.1, 23.8, 12.3, 29.1, 21. , 19.5, 23.3, 23.8,
       17.8, 11.5, 21.7, 19.9, 25. , 33.4, 28.5, 21.4, 24.3, 27.5, 33.1,
       16.2, 23.3, 48.3, 22.9, 22.8, 13.1, 12.7, 22.6, 15. , 15.3, 10.5,
       24. , 18.5, 21.7, 19.5, 33.2, 23.2,  5. , 19.1, 12.7, 22.3, 10.2,
       13.9, 16.3, 17. , 20.1, 29.9, 17.2, 37.3, 45.4, 17.8, 23.2, 29. ,
       22. , 18. , 17.4, 34.6, 20.1, 25. , 15.6, 24.8, 28.2, 21.2, 21.4,
       23.8, 31. , 26.2, 17.4, 37.9, 17.5, 20. ,  8.3, 23.9,  8.4, 13.8,
        7.2, 11.7, 17.1, 21.6, 50. , 16.1, 20.4, 20.6, 21.4, 20.6, 36.5,
        8.5, 24.8, 10.8, 21.9, 17.3, 18.9, 36.2, 14.9, 18.2, 33.3, 21.8,
       19.7, 31.6, 24.8, 19.4, 22.8,  7.5, 44.8, 16.8, 18.7, 50. , 50. ,
       19.5, 20.1, 50. , 17.2, 20.8, 19.3, 41.3, 20.4, 20.5, 13.8, 16.5,
       23.9, 20.6, 31.5, 23.3, 16.8, 14. , 33.8, 36.1, 12.8, 18.3, 18.7,
       19.1, 29. , 30.1, 50. , 50. , 22. , 11.9, 37.6, 50. , 22.7, 20.8,
       23.5, 27.9, 50. , 19.3, 23.9, 22.6, 15.2, 21.7, 19.2, 43.8, 20.3,
       33.2, 19.9, 22.5, 32.7, 22. , 17.1, 19. , 15. , 16.1, 25.1, 23.7,
       28.7, 37.2, 22.6, 16.4, 25. , 29.8, 22.1, 17.4, 18.1, 30.3, 17.5,
       24.7, 12.6, 26.5, 28.7, 13.3, 10.4, 24.4, 23. , 20. , 17.8,  7. ,
       11.8, 24.4, 13.8, 19.4, 25.2, 19.4, 19.4, 29.1])

mean = train_data.mean(axis=0)
train_data -= mean # 减去均值
std = train_data.std(axis=0) # 特征标准差
train_data /= std
test_data -= mean #测试集处理：使用训练集的均值和标准差；不用重新计算
test_data /= std

from keras import models
from keras import layers

def build_model():
    model = models.Sequential()

    model.add(layers.Dense(64, activation='relu',input_shape=(train_data.shape[1],)))
    model.add(layers.Dense(64, activation='relu'))
    model.add(layers.Dense(1))

    model.compile(optimizer='rmsprop', loss='mse', metrics=['mae'])

    return model

K折交叉验证
当调整模型参数时，为了评估模型，我们通常将数据集分成训练集和验证集。但是当数据量过小时，验证集数目也变得很小，导致验证集上的评估结果相互之间差异性很大—与训练集和测试集的划分结果相关。评估结果可信度不高。 
最好的评估方式是采用K折交叉验证–将数据集分成K份(K=4或5)，实例化K个模型，每个模型在K-1份数据上进行训练，在1份数据上进行评估，最后用K次评估分数的平均值做最后的评估结果。

import numpy as np

k = 4
num_val_samples = len(train_data) // k
num_epochs = 100
all_scores = []
for i in range(k):
    print('processing fold #',i)
    val_data = train_data[i*num_val_samples : (i+1)*num_val_samples] # 划分出验证集部分
    val_targets = train_targets[i*num_val_samples : (i+1)*num_val_samples]

    partial_train_data = np.concatenate([train_data[:i*num_val_samples],train_data[(i+1)* num_val_samples:] ],axis=0) # 将训练集拼接到一起
    partial_train_targets = np.concatenate([train_targets[:i*num_val_samples],train_targets[(i+1)* num_val_samples:] ],axis=0)

    model = build_model()
    model.fit(partial_train_data,partial_train_targets,epochs=num_epochs,batch_size=16,verbose=0)#模型训练silent模型
    val_mse, val_mae = model.evaluate(val_data, val_targets, verbose=0) # 验证集上评估
    all_scores.append(val_mae)

processing fold # 0
processing fold # 1
processing fold # 2
processing fold # 3

#模型训练
model = build_model()
model.fit(train_data, train_targets,epochs=80, batch_size=16, verbose=0)

test_mse_score, test_mae_score = model.evaluate(test_data, test_targets)
print(test_mse_score, test_mae_score)

102/102 [==============================] - 0s 538us/step
18.04363639681947 2.575085022870232

小结
回归问题：损失函数通常为MSE均方误差；
模型评估监测指标通常为MAE(mean absolute error);
当数据取值范围不一致时，需要对特征进行预处理；
数据量小时，可以采用Ｋ折验证来衡量模型；
数据量小时，模型复杂度也应该相应的简单，可以避免模型过拟合。