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天池_二手车价格预测_Task_3_特征工程

Excalibur
2020-04-02 / 0 评论 / 0 点赞 / 32 阅读 / 24,301 字
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特征工程🃏

  特征工程,是指用一系列工程化的方式从原始数据中筛选出更好的数据特征,以提升模型的训练效果。业内有一句广为流传的话是:数据和特征决定了机器学习的上限,而模型和算法是在逼近这个上限而已。由此可见,好的数据和特征是模型和算法发挥更大的作用的前提。特征工程通常包括数据预处理、特征选择、降维等环节。如下图所示:
在这里插入图片描述

0️⃣ 前言

  我们经常在处理数据时,会面临以下问题:

  • 收集的数据格式不对(如 SQL 数据库、JSONCSV 等)
  • 缺失值和异常值
  • 标准化
  • 减少数据集中存在的固有噪声(部分存储数据可能已损坏)
  • 数据集中的某些功能可能无法收集任何信息以供分析

而减少统计分析期间要使用的特征的数量可能会带来一些好处,例如:

  • 提高精度
  • 降低过拟合风险
  • 加快训练速度
  • 改进数据可视化
  • 增加我们模型的可解释性

事实上,统计上证明,当执行机器学习任务时,存在针对每个特定任务应该使用的最佳数量的特征(图 1)。如果添加的特征比必要的特征,那么我们的模型性能将下降(因为添加了噪声)。真正的挑战是找出哪些特征是最佳的使用特征(这实际上取决于我们提供的数据量和我们正在努力实现的任务的复杂性)。这就是特征选择技术能够帮到我们的地方!

在这里插入图片描述

0️⃣.1️⃣ 赛题重述

  这是一道来自于天池的新手练习题目,用数据分析机器学习等手段进行 二手车售卖价格预测 的回归问题。赛题本身的思路清晰明了,即对给定的数据集进行分析探讨,然后设计模型运用数据进行训练,测试模型,最终给出选手的预测结果。前面我们已经进行过EDA分析在这里天池_二手车价格预测_Task1-2_赛题理解与数据分析

0️⃣.2️⃣ 数据集概述

  赛题官方给出了来自Ebay Kleinanzeigen的二手车交易记录,总数据量超过40w,包含31列变量信息,其中15列为匿名变量,即v0v15。并从中抽取15万条作为训练集,5万条作为测试集A,5万条作为测试集B,同时对namemodelbrandregionCode等信息进行脱敏。具体的数据表如下图:

Field Description
SaleID 交易ID,唯一编码
name 汽车交易名称,已脱敏
regDate 汽车注册日期,例如20160101,2016年01月01日
model 车型编码,已脱敏
brand 汽车品牌,已脱敏
bodyType 车身类型:豪华轿车:0,微型车:1,厢型车:2,大巴车:3,敞篷车:4,双门汽车:5,商务车:6,搅拌车:7
fuelType 燃油类型:汽油:0,柴油:1,液化石油气:2,天然气:3,混合动力:4,其他:5,电动:6
gearbox 变速箱:手动:0,自动:1
power 发动机功率:范围 [ 0, 600 ]
kilometer 汽车已行驶公里,单位万km
notRepairedDamage 汽车有尚未修复的损坏:是:0,否:1
regionCode 地区编码,已脱敏
seller 销售方:个体:0,非个体:1
offerType 报价类型:提供:0,请求:1
creatDate 汽车上线时间,即开始售卖时间
price 二手车交易价格(预测目标)
v系列特征 匿名特征,包含v0-14在内15个匿名特征

1️⃣ 异常缺失值删除

1️⃣.1️⃣ 导入库与数据

import pandas as pd
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
import seaborn as sns
from operator import itemgetter

%matplotlib inline
train = pd.read_csv('used_car_train_20200313.csv', sep=' ')
test = pd.read_csv('used_car_testA_20200313.csv', sep=' ')
print(train.shape)
print(test.shape)
(150000, 31)
(50000, 30)
train.head()
SaleID name regDate model brand bodyType fuelType gearbox power kilometer ... v_5 v_6 v_7 v_8 v_9 v_10 v_11 v_12 v_13 v_14
0 0 736 20040402 30.0 6 1.0 0.0 0.0 60 12.5 ... 0.235676 0.101988 0.129549 0.022816 0.097462 -2.881803 2.804097 -2.420821 0.795292 0.914762
1 1 2262 20030301 40.0 1 2.0 0.0 0.0 0 15.0 ... 0.264777 0.121004 0.135731 0.026597 0.020582 -4.900482 2.096338 -1.030483 -1.722674 0.245522
2 2 14874 20040403 115.0 15 1.0 0.0 0.0 163 12.5 ... 0.251410 0.114912 0.165147 0.062173 0.027075 -4.846749 1.803559 1.565330 -0.832687 -0.229963
3 3 71865 19960908 109.0 10 0.0 0.0 1.0 193 15.0 ... 0.274293 0.110300 0.121964 0.033395 0.000000 -4.509599 1.285940 -0.501868 -2.438353 -0.478699
4 4 111080 20120103 110.0 5 1.0 0.0 0.0 68 5.0 ... 0.228036 0.073205 0.091880 0.078819 0.121534 -1.896240 0.910783 0.931110 2.834518 1.923482

5 rows × 31 columns

train.columns
Index(['SaleID', 'name', 'regDate', 'model', 'brand', 'bodyType', 'fuelType',
       'gearbox', 'power', 'kilometer', 'notRepairedDamage', 'regionCode',
       'seller', 'offerType', 'creatDate', 'price', 'v_0', 'v_1', 'v_2', 'v_3',
       'v_4', 'v_5', 'v_6', 'v_7', 'v_8', 'v_9', 'v_10', 'v_11', 'v_12',
       'v_13', 'v_14'],
      dtype='object')

1️⃣.2️⃣ 异常值删除

  这里可以将箱型图中的超过上下限的那些值作为异常值删除。如下图所示,箱型图中间是一个箱体,也就是粉红色部分,箱体左边,中间,右边分别有一条线,左边是下分位数($Q1$),右边是上四分位数($Q3$),中间是中位数($Median$),上下四分位数之差是四分位距$IQR(Interquartile Range$,用$Q1-1.5IQR$得到下边缘(最小值),$Q3+1.5IQR$得到上边缘(最大值)。在上边缘之外的数据就是极大异常值,在下边缘之外的数据就是极小异常值。
在这里插入图片描述

搞清楚原理那我们就构造一个实现上述功能的函数吧!

def drop_outliers(data, col_name, scale = 1.5):
    """
    用于清洗异常值,默认用 box_plot(scale=1.5)进行清洗
    :param data: 接收 pandas 数据格式
    :param col_name: pandas 列名
    :param scale: 尺度
    :return:
    """
    data_n = data.copy()
    data_series = data_n[col_name]
    
    IQR = scale * (data_series.quantile(0.75) - data_series.quantile(0.25)) # quantile是pd内置的求四分位的函数
    val_low = data_series.quantile(0.25) - IQR # 下边缘
    val_up = data_series.quantile(0.75) + IQR # 上边缘
    rule_low = (data_series < val_low) # 下边缘的极小异常值的下标列表
    rule_up = (data_series > val_up)   # 上边缘的极大异常值的下标列表
    
    index = np.arange(data_series.shape[0])[rule_low | rule_up] # | 运算就是说只要rule_low和rule_up中只要有一个值为True,就把这个下标取出来
    print(index)
    print("Delete number is: {}".format(len(index)))
    
    data_n = data_n.drop(index) # 删除index对应下标的元素
    data_n.reset_index(drop=True, inplace=True) #下文有介绍
    print("Now column number is: {}".format(data_n.shape[0]))
    
    index_low = np.arange(data_series.shape[0])[rule_low] # 下边缘的异常数据的描述统计量
    outliers = data_series.iloc[index_low]
    print("Description of data less than the lower bound is:")
    print(pd.Series(outliers).describe())
    
    index_up = np.arange(data_series.shape[0])[rule_up] # 上边缘的异常数据的描述统计量
    outliers = data_series.iloc[index_up]
    print("Description of data larger than the upper bound is:")
    print(pd.Series(outliers).describe())
    
    fig, ax = plt.subplots(1, 2, figsize = (10, 7))
    sns.boxplot(y = data[col_name], data = data, palette = "Set1", ax = ax[0])
    sns.boxplot(y = data_n[col_name], data = data_n, palette = "Set1", ax = ax[1])
    return data_n

这里reset_index可以还原索引,重新变为默认的整型索引

DataFrame.reset_index(level=None, drop=False, inplace=False, col_level=0, col_fill=”)

  • levelintstrtuplelist,默认无,仅从索引中删除给定级别。默认情况下移除所有级别。控制了具体要还原的那个等级的索引
  • dropdropFalse则索引列会被还原为普通列,否则会丢失
  • inplace:默认为False,适当修改DataFrame(不要创建新对象)
  • col_levelintstr,默认值为0,如果列有多个级别,则确定将标签插入到哪个级别。默认情况下,它将插入到第一级。
  • col_fill:对象,默认‘’,如果列有多个级别,则确定其他级别的命名方式。如果没有,则重复索引名
drop_outliers(train, 'power', scale=1.5)
[    33     77    104 ... 149967 149981 149984]
Delete number is: 4878
Now column number is: 145122
Description of data less than the lower bound is:
count    0.0
mean     NaN
std      NaN
min      NaN
25%      NaN
50%      NaN
75%      NaN
max      NaN
Name: power, dtype: float64
Description of data larger than the upper bound is:
count     4878.000000
mean       410.132021
std        884.219933
min        264.000000
25%        286.000000
50%        306.000000
75%        349.000000
max      19312.000000
Name: power, dtype: float64
SaleID name regDate model brand bodyType fuelType gearbox power kilometer ... v_5 v_6 v_7 v_8 v_9 v_10 v_11 v_12 v_13 v_14
0 0 736 20040402 30.0 6 1.0 0.0 0.0 60 12.5 ... 0.235676 0.101988 0.129549 0.022816 0.097462 -2.881803 2.804097 -2.420821 0.795292 0.914762
1 1 2262 20030301 40.0 1 2.0 0.0 0.0 0 15.0 ... 0.264777 0.121004 0.135731 0.026597 0.020582 -4.900482 2.096338 -1.030483 -1.722674 0.245522
2 2 14874 20040403 115.0 15 1.0 0.0 0.0 163 12.5 ... 0.251410 0.114912 0.165147 0.062173 0.027075 -4.846749 1.803559 1.565330 -0.832687 -0.229963
3 3 71865 19960908 109.0 10 0.0 0.0 1.0 193 15.0 ... 0.274293 0.110300 0.121964 0.033395 0.000000 -4.509599 1.285940 -0.501868 -2.438353 -0.478699
4 4 111080 20120103 110.0 5 1.0 0.0 0.0 68 5.0 ... 0.228036 0.073205 0.091880 0.078819 0.121534 -1.896240 0.910783 0.931110 2.834518 1.923482
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
145117 149995 163978 20000607 121.0 10 4.0 0.0 1.0 163 15.0 ... 0.280264 0.000310 0.048441 0.071158 0.019174 1.988114 -2.983973 0.589167 -1.304370 -0.302592
145118 149996 184535 20091102 116.0 11 0.0 0.0 0.0 125 10.0 ... 0.253217 0.000777 0.084079 0.099681 0.079371 1.839166 -2.774615 2.553994 0.924196 -0.272160
145119 149997 147587 20101003 60.0 11 1.0 1.0 0.0 90 6.0 ... 0.233353 0.000705 0.118872 0.100118 0.097914 2.439812 -1.630677 2.290197 1.891922 0.414931
145120 149998 45907 20060312 34.0 10 3.0 1.0 0.0 156 15.0 ... 0.256369 0.000252 0.081479 0.083558 0.081498 2.075380 -2.633719 1.414937 0.431981 -1.659014
145121 149999 177672 19990204 19.0 28 6.0 0.0 1.0 193 12.5 ... 0.284475 0.000000 0.040072 0.062543 0.025819 1.978453 -3.179913 0.031724 -1.483350 -0.342674

145122 rows × 31 columns

在这里插入图片描述

从这张删除异常值前后的箱型图对比可以看出,剔除异常值后,数据的分布就很均匀了。

下面我们就批量对所有的特征进行一次异常数据删除:

def Bach_drop_outliers(data,scale=1.5):
    dataNew = data.copy()

    for fea in data.columns:
        try:
            IQR = scale * (dataNew[fea].quantile(0.75) - dataNew[fea].quantile(0.25)) # quantile是pd内置的求四分位的函数
        except:
            continue
        val_low = dataNew[fea].quantile(0.25) - IQR # 下边缘
        val_up = dataNew[fea].quantile(0.75) + IQR # 上边缘
        rule_low = (dataNew[fea] < val_low) # 下边缘的极小异常值的下标列表
        rule_up = (dataNew[fea] > val_up)   # 上边缘的极大异常值的下标列表
        
        index = np.arange(dataNew[fea].shape[0])[rule_low | rule_up] # | 运算就是说只要rule_low和rule_up中只要有一个值为True,就把这个下标取出来
        print("feature %s deleted number is %d"%(fea, len(index)))
        
        dataNew = dataNew.drop(index)# 删除index对应下标的元素
        dataNew.reset_index(drop=True, inplace=True)
    fig, ax = plt.subplots(5, 6, figsize = (20, 15))
    x = 0
    y = 0
    for fea in dataNew.columns:
        try:
            sns.boxplot(y = dataNew[fea], data =dataNew, palette = "Set2", ax = ax[x][y])
            y+=1
            if y == 6:
                y = 0
                x += 1
        except:
            print(fea)
            y+=1
            if y == 6:
                y = 0
                x += 1
            continue
    return dataNew

train = Bach_drop_outliers(train)
feature SaleID deleted number is 0
feature name deleted number is 0
feature regDate deleted number is 0
feature model deleted number is 9720
feature brand deleted number is 4032
feature bodyType deleted number is 5458
feature fuelType deleted number is 333
feature gearbox deleted number is 26829
feature power deleted number is 1506
feature kilometer deleted number is 15306
feature regionCode deleted number is 4
feature seller deleted number is 1
feature offerType deleted number is 0
feature creatDate deleted number is 13989
feature price deleted number is 4527
feature v_0 deleted number is 2558
feature v_1 deleted number is 0
feature v_2 deleted number is 487
feature v_3 deleted number is 173
feature v_4 deleted number is 61
feature v_5 deleted number is 0
feature v_6 deleted number is 0
feature v_7 deleted number is 64
feature v_8 deleted number is 0
feature v_9 deleted number is 24
feature v_10 deleted number is 0
feature v_11 deleted number is 0
feature v_12 deleted number is 4
feature v_13 deleted number is 0
feature v_14 deleted number is 1944
notRepairedDamage
v_14

在这里插入图片描述

可以看出,经过箱型图异常值删除后,新数据的箱型图的数据几乎没有异常值了,甚至有些箱型图的数据是一条直线,当然那是因为数据本身就是种类非0即1。

2️⃣ 树模型的特征构造

  训练集和测试集放在一起,方便构造特征

train['train'] = 1
test['train'] = 0
data = pd.concat([train, test], ignore_index=True, sort=False)
data
SaleID name regDate model brand bodyType fuelType gearbox power kilometer ... v_6 v_7 v_8 v_9 v_10 v_11 v_12 v_13 v_14 train
0 1 2262 20030301 40.0 1 2.0 0.0 0.0 0 15.0 ... 0.121004 0.135731 0.026597 0.020582 -4.900482 2.096338 -1.030483 -1.722674 0.245522 1
1 5 137642 20090602 24.0 10 0.0 1.0 0.0 109 10.0 ... 0.000518 0.119838 0.090922 0.048769 1.885526 -2.721943 2.457660 -0.286973 0.206573 1
2 7 165346 19990706 26.0 14 1.0 0.0 0.0 101 15.0 ... 0.000000 0.122943 0.039839 0.082413 3.693829 -0.245014 -2.192810 0.236728 0.195567 1
3 10 18961 20050811 19.0 9 3.0 1.0 0.0 101 15.0 ... 0.105385 0.077271 0.042445 0.060794 -4.206000 1.060391 -0.647515 -0.191194 0.349187 1
4 13 8129 20041110 65.0 1 0.0 0.0 0.0 150 15.0 ... 0.106950 0.134945 0.050364 0.051359 -4.614692 0.821889 0.753490 -0.886425 -0.341562 1
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
112975 199995 20903 19960503 4.0 4 4.0 0.0 0.0 116 15.0 ... 0.130044 0.049833 0.028807 0.004616 -5.978511 1.303174 -1.207191 -1.981240 -0.357695 0
112976 199996 708 19991011 0.0 0 0.0 0.0 0.0 75 15.0 ... 0.108095 0.066039 0.025468 0.025971 -3.913825 1.759524 -2.075658 -1.154847 0.169073 0
112977 199997 6693 20040412 49.0 1 0.0 1.0 1.0 224 15.0 ... 0.105724 0.117652 0.057479 0.015669 -4.639065 0.654713 1.137756 -1.390531 0.254420 0
112978 199998 96900 20020008 27.0 1 0.0 0.0 1.0 334 15.0 ... 0.000490 0.137366 0.086216 0.051383 1.833504 -2.828687 2.465630 -0.911682 -2.057353 0
112979 199999 193384 20041109 166.0 6 1.0 NaN 1.0 68 9.0 ... 0.000300 0.103534 0.080625 0.124264 2.914571 -1.135270 0.547628 2.094057 -1.552150 0

112980 rows × 32 columns

2️⃣.1️⃣ 时间特征构造

  • 使用时间:data['creatDate'] - data['regDate'],反应汽车使用时间,一般来说价格与使用时间成反比
  • 不过要注意,数据里有时间出错的格式,所以我们需要 errors='coerce'
data['used_time'] = (pd.to_datetime(data['creatDate'], format='%Y%m%d', errors='coerce') - 
                            pd.to_datetime(data['regDate'], format='%Y%m%d', errors='coerce')).dt.days
data['used_time']
0         4757.0
1         2482.0
2         6108.0
3         3874.0
4         4154.0
           ...  
112975    7261.0
112976    6014.0
112977    4345.0
112978       NaN
112979    4151.0
Name: used_time, Length: 112980, dtype: float64
  • 看一下空数据,有 7.6k 个样本的时间是有问题的,我们可以选择删除,也可以选择放着。
  • 但是这里不建议删除,因为删除缺失数据占总样本量过大,3.8%
  • 我们可以先放着,因为如果我们 XGBoost 之类的决策树,其本身就能处理缺失值,所以可以不用管;
data['used_time'].isnull().sum()
8591
data.isnull().sum().sum()
70585

2️⃣.2️⃣ 城市信息特征提取

  • 从邮编中提取城市信息,因为是德国的数据,所以参考德国的邮编,相当于加入了先验知识
data['city'] = data['regionCode'].apply(lambda x : str(x)[:-3])
data['city']
0         4
1         3
2         4
3         1
4         3
         ..
112975    3
112976    1
112977    3
112978    1
112979    3
Name: city, Length: 112980, dtype: object

2️⃣.3️⃣ 品牌特征提取

  计算某品牌的销售统计量,这里要以 train 的数据计算统计量。

train_gb = train.groupby("brand")
all_info = {}
for kind, kind_data in train_gb:
    info = {}
    kind_data = kind_data[kind_data['price'] > 0] # kind_data['price'] > 0 返回的是下标再取一次列表就得到了数据
    info['brand_amount'] = len(kind_data)
    info['brand_price_max'] = kind_data.price.max()
    info['brand_price_median'] = kind_data.price.median()
    info['brand_price_min'] = kind_data.price.min()
    info['brand_price_sum'] = kind_data.price.sum()
    info['brand_price_std'] = kind_data.price.std()
    info['brand_price_average'] = round(kind_data.price.sum() / (len(kind_data) + 1), 2)
    all_info[kind] = info
brand_fe = pd.DataFrame(all_info).T.reset_index().rename(columns={"index": "brand"})
data = data.merge(brand_fe, how='left', on='brand')
data
SaleID name regDate model brand bodyType fuelType gearbox power kilometer ... train used_time city brand_amount brand_price_max brand_price_median brand_price_min brand_price_sum brand_price_std brand_price_average
0 1 2262 20030301 40.0 1 2.0 0.0 0.0 0 15.0 ... 1 4757.0 4 4940.0 9500.0 2999.0 149.0 17934852.0 2537.956443 3629.80
1 5 137642 20090602 24.0 10 0.0 1.0 0.0 109 10.0 ... 1 2482.0 3 3557.0 9500.0 2490.0 200.0 10936962.0 2180.881827 3073.91
2 7 165346 19990706 26.0 14 1.0 0.0 0.0 101 15.0 ... 1 6108.0 4 8784.0 9500.0 1350.0 13.0 17445064.0 1797.704405 1985.78
3 10 18961 20050811 19.0 9 3.0 1.0 0.0 101 15.0 ... 1 3874.0 1 4487.0 9500.0 1250.0 55.0 7867901.0 1556.621159 1753.10
4 13 8129 20041110 65.0 1 0.0 0.0 0.0 150 15.0 ... 1 4154.0 3 4940.0 9500.0 2999.0 149.0 17934852.0 2537.956443 3629.80
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
112975 199995 20903 19960503 4.0 4 4.0 0.0 0.0 116 15.0 ... 0 7261.0 3 6368.0 9500.0 3000.0 150.0 24046576.0 2558.650243 3775.57
112976 199996 708 19991011 0.0 0 0.0 0.0 0.0 75 15.0 ... 0 6014.0 1 16371.0 9500.0 2150.0 50.0 46735356.0 2276.755156 2854.59
112977 199997 6693 20040412 49.0 1 0.0 1.0 1.0 224 15.0 ... 0 4345.0 3 4940.0 9500.0 2999.0 149.0 17934852.0 2537.956443 3629.80
112978 199998 96900 20020008 27.0 1 0.0 0.0 1.0 334 15.0 ... 0 NaN 1 4940.0 9500.0 2999.0 149.0 17934852.0 2537.956443 3629.80
112979 199999 193384 20041109 166.0 6 1.0 NaN 1.0 68 9.0 ... 0 4151.0 3 5778.0 9500.0 1400.0 50.0 11955982.0 1871.933447 2068.87

112980 rows × 41 columns

brand_fe
brand brand_amount brand_price_max brand_price_median brand_price_min brand_price_sum brand_price_std brand_price_average
0 0 16371.0 9500.0 2150.0 50.0 46735356.0 2276.755156 2854.59
1 1 4940.0 9500.0 2999.0 149.0 17934852.0 2537.956443 3629.80
2 3 665.0 9500.0 2800.0 99.0 2158773.0 2058.532395 3241.40
3 4 6368.0 9500.0 3000.0 150.0 24046576.0 2558.650243 3775.57
4 5 2842.0 9500.0 1850.0 75.0 6562224.0 1738.415572 2308.20
5 6 5778.0 9500.0 1400.0 50.0 11955982.0 1871.933447 2068.87
6 7 1035.0 9500.0 1500.0 100.0 2372550.0 2071.320262 2290.11
7 8 705.0 9500.0 1100.0 125.0 1077211.0 1318.748474 1525.79
8 9 4487.0 9500.0 1250.0 55.0 7867901.0 1556.621159 1753.10
9 10 3557.0 9500.0 2490.0 200.0 10936962.0 2180.881827 3073.91
10 11 1390.0 9500.0 1750.0 50.0 3513591.0 2151.572044 2525.95
11 12 549.0 9500.0 1850.0 100.0 1413264.0 2091.218447 2569.57
12 13 1689.0 8950.0 1250.0 25.0 2832005.0 1363.018568 1675.74
13 14 8784.0 9500.0 1350.0 13.0 17445064.0 1797.704405 1985.78
14 15 389.0 9500.0 5700.0 1800.0 2247357.0 1795.404288 5762.45
15 16 291.0 8900.0 1950.0 300.0 636703.0 1223.490908 2180.49
16 17 542.0 9500.0 1970.0 150.0 1444129.0 2136.402905 2659.54
17 18 66.0 8990.0 1650.0 150.0 167360.0 2514.210817 2497.91
18 19 341.0 9100.0 1200.0 130.0 540335.0 1337.203100 1579.93
19 20 514.0 8150.0 1200.0 100.0 818973.0 1276.623577 1590.24
20 21 527.0 8900.0 1890.0 99.0 1285258.0 1832.524896 2434.20
21 22 222.0 9300.0 1925.0 190.0 592296.0 2118.280894 2656.04
22 23 68.0 9500.0 1194.5 100.0 110253.0 1754.883573 1597.87
23 24 4.0 8600.0 7550.0 5999.0 29699.0 1072.435041 5939.80
24 25 735.0 9500.0 1500.0 100.0 1725999.0 2152.726491 2345.11
25 26 121.0 9500.0 2699.0 300.0 417260.0 2563.586943 3420.16

3️⃣ 树模型的数据分桶

  数据分箱(也称为离散分箱或分段)是一种数据预处理技术,用于减少次要观察误差的影响,是一种将多个连续值分组为较少数量的“分箱”的方法。例如我们有各个年龄的数据的统计值,可以分成某个段的年龄的值。

  1. 离散后稀疏向量内积乘法运算速度更快,计算结果也方便存储,容易扩展;
  2. 离散后的特征对异常值更具鲁棒性,如 age>30 为 1 否则为 0,对于年龄为 200 的也不会对模型造成很大的干扰;
  3. LR 属于广义线性模型,表达能力有限,经过离散化后,每个变量有单独的权重,这相当于引入了非线性,能够提升模型的表达能力,加大拟合;
  4. 离散后特征可以进行特征交叉,提升表达能力,由 M+N 个变量变成 M*N 个变量,进一步引入非线形,提升了表达能力;
  5. 特征离散后模型更稳定,如用户年龄区间,不会因为用户年龄长了一岁就变化

下面以power为例子,做一次数据分桶

bin = [i*10 for i in range(31)]
data['power_bin'] = pd.cut(data['power'], bin, labels=False)
data[['power_bin', 'power']]
power_bin power
0 NaN 0
1 10.0 109
2 10.0 101
3 10.0 101
4 14.0 150
... ... ...
112975 11.0 116
112976 7.0 75
112977 22.0 224
112978 NaN 334
112979 6.0 68

112980 rows × 2 columns

可以看出这个分箱的作用就是将同一个区间段的功率值设为同样的值,比如101~109都设置为10.0。
然后就可以删除掉原数据了:

data = data.drop(['creatDate', 'regDate', 'regionCode'], axis=1)
print(data.shape)
data.columns
(112980, 39)





Index(['SaleID', 'name', 'model', 'brand', 'bodyType', 'fuelType', 'gearbox',
       'power', 'kilometer', 'notRepairedDamage', 'seller', 'offerType',
       'price', 'v_0', 'v_1', 'v_2', 'v_3', 'v_4', 'v_5', 'v_6', 'v_7', 'v_8',
       'v_9', 'v_10', 'v_11', 'v_12', 'v_13', 'v_14', 'train', 'used_time',
       'city', 'brand_amount', 'brand_price_max', 'brand_price_median',
       'brand_price_min', 'brand_price_sum', 'brand_price_std',
       'brand_price_average', 'power_bin'],
      dtype='object')

至此,可以导出给树模型用的数据

data.to_csv('data_for_tree.csv', index=0)

4️⃣ LR与NN模型的特征构造

  上面的步骤就是一次比较完备的特征构造,我们还可以为其他模型构造特征,主要是由于不用模型需要的数据输入是不同的。

4️⃣.1️⃣ $log$与归一化

观察一下数据分布

data['power'].plot.hist()
<matplotlib.axes._subplots.AxesSubplot at 0x2108b6377b8>

在这里插入图片描述

再看看train数据集的分布:

train['power'].plot.hist()
<matplotlib.axes._subplots.AxesSubplot at 0x2108b4ed588>

在这里插入图片描述

我们对其取 log,再做归一化

data['power'] = np.log(data['power'] + 1) 
data['power'] = ((data['power'] - np.min(data['power'])) / (np.max(data['power']) - np.min(data['power'])))
data['power'].plot.hist()
<matplotlib.axes._subplots.AxesSubplot at 0x2108abc1438>

在这里插入图片描述

看看行驶里程的情况,应该是原始数据已经分好了桶

data['kilometer'].plot.hist()
<matplotlib.axes._subplots.AxesSubplot at 0x2108abc1390>

在这里插入图片描述

归一化

data['kilometer'] = ((data['kilometer'] - np.min(data['kilometer'])) / 
                        (np.max(data['kilometer']) - np.min(data['kilometer'])))
data['kilometer'].plot.hist()
<matplotlib.axes._subplots.AxesSubplot at 0x2108aca0898>

在这里插入图片描述

对刚刚构造的统计量进行归一化

def max_min(x):
    return (x - np.min(x)) / (np.max(x) - np.min(x))

data.columns[-10:]
Index(['used_time', 'city', 'brand_amount', 'brand_price_max',
       'brand_price_median', 'brand_price_min', 'brand_price_sum',
       'brand_price_std', 'brand_price_average', 'power_bin'],
      dtype='object')
for i in data.columns[-10:]:
    if np.min(data[i]) != '': # 存在空值的情况
        data[i] = max_min(data[i])

4️⃣.2️⃣ $OneEncoder$编码

  对类别特征进行$OneEncoder$

在此之前先介绍一下$OneEncoder$编码:$one-hot$的基本思想,将离散型特征的每一种取值都看成一种状态,若你的这一特征中有$N$个不相同的取值,那么我们就可以将该特征抽象成$N$种不同的状态,$one-hot$编码保证了每一个取值只会使得一种状态处于“激活态”,也就是说这N种状态中只有一个状态位值为1,其他状态位都是0。举个例子,假设我们以学历为例,我们想要研究的类别为小学、中学、大学、硕士、博士五种类别,我们使用$one-hot$对其编码就会得到:

在这里插入图片描述

  • $dummy \quad encoding$

哑变量编码直观的解释就是任意的将一个状态位去除。还是拿上面的例子来说,我们用4个状态位就足够反应上述5个类别的信息,也就是我们仅仅使用前四个状态位 [0,0,0,0] 就可以表达博士了。只是因为对于一个我们研究的样本,他已不是小学生、也不是中学生、也不是大学生、又不是研究生,那么我们就可以默认他是博士,是不是。所以,我们用哑变量编码可以将上述5类表示成:

在这里插入图片描述

data = pd.get_dummies(data, columns=['model', 'brand', 'bodyType', 'fuelType',
                                     'gearbox', 'notRepairedDamage', 'power_bin'])
data
SaleID name power kilometer seller offerType price v_0 v_1 v_2 ... power_bin_0.6896551724137931 power_bin_0.7241379310344828 power_bin_0.7586206896551724 power_bin_0.7931034482758621 power_bin_0.8275862068965517 power_bin_0.8620689655172413 power_bin_0.896551724137931 power_bin_0.9310344827586207 power_bin_0.9655172413793104 power_bin_1.0
0 1 2262 0.000000 1.000000 0 0 3600.0 45.305273 5.236112 0.137925 ... 0 0 0 0 0 0 0 0 0 0
1 5 137642 0.474626 0.655172 0 0 8000.0 46.323165 -3.229285 0.156615 ... 0 0 0 0 0 0 0 0 0 0
2 7 165346 0.467002 1.000000 0 0 1000.0 42.255586 -3.167771 -0.676693 ... 0 0 0 0 0 0 0 0 0 0
3 10 18961 0.467002 1.000000 0 0 3100.0 45.401241 4.195311 -0.370513 ... 0 0 0 0 0 0 0 0 0 0
4 13 8129 0.506615 1.000000 0 0 3100.0 46.844574 4.175332 0.490609 ... 0 0 0 0 0 0 0 0 0 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
112975 199995 20903 0.480856 1.000000 0 0 NaN 45.621391 5.958453 -0.918571 ... 0 0 0 0 0 0 0 0 0 0
112976 199996 708 0.437292 1.000000 0 0 NaN 43.935162 4.476841 -0.841710 ... 0 0 0 0 0 0 0 0 0 0
112977 199997 6693 0.546885 1.000000 0 0 NaN 46.537137 4.170806 0.388595 ... 0 0 1 0 0 0 0 0 0 0
112978 199998 96900 0.587076 1.000000 0 0 NaN 46.771359 -3.296814 0.243566 ... 0 0 0 0 0 0 0 0 0 0
112979 199999 193384 0.427535 0.586207 0 0 NaN 43.731010 -3.121867 0.027348 ... 0 0 0 0 0 0 0 0 0 0

112980 rows × 369 columns

  • 将这份数据输出给LR模型使用
data.to_csv('data_for_lr.csv', index=0)

5️⃣ 特征选择

5️⃣.1️⃣ 过滤式(filter)

相关性分析

print(data['power'].corr(data['price'], method='spearman'))
print(data['kilometer'].corr(data['price'], method='spearman'))
print(data['brand_amount'].corr(data['price'], method='spearman'))
print(data['brand_price_average'].corr(data['price'], method='spearman'))
print(data['brand_price_max'].corr(data['price'], method='spearman'))
print(data['brand_price_median'].corr(data['price'], method='spearman'))
0.4698539569820024
-0.19974282513118508
0.04085800320025127
0.3135239590412946
0.07894119089254827
0.3138873049004745

可以看出powerbrand_price_averagebrand_price_medianprice相关性比较高

data_numeric = data[['power', 'kilometer', 'brand_amount', 'brand_price_average', 
                     'brand_price_max', 'brand_price_median']]
correlation = data_numeric.corr()

f , ax = plt.subplots(figsize = (7, 7))
plt.title('Correlation of Numeric Features with Price',y=1,size=30)
sns.heatmap(correlation, square = True, cmap = 'PuBuGn', vmax=0.8)
<matplotlib.axes._subplots.AxesSubplot at 0x21096f60198>

在这里插入图片描述

看不出来啥。😛

5️⃣.2️⃣ 包裹式(wrapper)

!pip install mlxtend
Collecting mlxtend
  Downloading https://files.pythonhosted.org/packages/64/e2/1610a86284029abcad0ac9bc86cb19f9787fe6448ede467188b2a5121bb4/mlxtend-0.17.2-py2.py3-none-any.whl (1.3MB)
Requirement already satisfied: setuptools in d:\software\anaconda\lib\site-packages (from mlxtend) (40.8.0)
Requirement already satisfied: pandas>=0.24.2 in d:\software\anaconda\lib\site-packages (from mlxtend) (0.25.1)
Requirement already satisfied: scipy>=1.2.1 in d:\software\anaconda\lib\site-packages (from mlxtend) (1.2.1)
Requirement already satisfied: matplotlib>=3.0.0 in d:\software\anaconda\lib\site-packages (from mlxtend) (3.0.3)
Requirement already satisfied: numpy>=1.16.2 in d:\software\anaconda\lib\site-packages (from mlxtend) (1.16.2)
Collecting joblib>=0.13.2 (from mlxtend)
  Downloading https://files.pythonhosted.org/packages/28/5c/cf6a2b65a321c4a209efcdf64c2689efae2cb62661f8f6f4bb28547cf1bf/joblib-0.14.1-py2.py3-none-any.whl (294kB)
Requirement already satisfied: scikit-learn>=0.20.3 in d:\software\anaconda\lib\site-packages (from mlxtend) (0.20.3)
Requirement already satisfied: pytz>=2017.2 in d:\software\anaconda\lib\site-packages (from pandas>=0.24.2->mlxtend) (2018.9)
Requirement already satisfied: python-dateutil>=2.6.1 in d:\software\anaconda\lib\site-packages (from pandas>=0.24.2->mlxtend) (2.8.0)
Requirement already satisfied: cycler>=0.10 in d:\software\anaconda\lib\site-packages (from matplotlib>=3.0.0->mlxtend) (0.10.0)
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Requirement already satisfied: pyparsing!=2.0.4,!=2.1.2,!=2.1.6,>=2.0.1 in d:\software\anaconda\lib\site-packages (from matplotlib>=3.0.0->mlxtend) (2.3.1)
Requirement already satisfied: six>=1.5 in d:\software\anaconda\lib\site-packages (from python-dateutil>=2.6.1->pandas>=0.24.2->mlxtend) (1.12.0)
Installing collected packages: joblib, mlxtend
Successfully installed joblib-0.14.1 mlxtend-0.17.2
x
SaleID name power kilometer seller offerType v_0 v_1 v_2 v_3 ... power_bin_0.6896551724137931 power_bin_0.7241379310344828 power_bin_0.7586206896551724 power_bin_0.7931034482758621 power_bin_0.8275862068965517 power_bin_0.8620689655172413 power_bin_0.896551724137931 power_bin_0.9310344827586207 power_bin_0.9655172413793104 power_bin_1.0
0 1 2262 0.000000 1.000000 0 0 45.305273 5.236112 0.137925 1.380657 ... 0 0 0 0 0 0 0 0 0 0
1 5 137642 0.474626 0.655172 0 0 46.323165 -3.229285 0.156615 -1.727217 ... 0 0 0 0 0 0 0 0 0 0
2 7 165346 0.467002 1.000000 0 0 42.255586 -3.167771 -0.676693 1.942673 ... 0 0 0 0 0 0 0 0 0 0
3 10 18961 0.467002 1.000000 0 0 45.401241 4.195311 -0.370513 0.444251 ... 0 0 0 0 0 0 0 0 0 0
4 13 8129 0.506615 1.000000 0 0 46.844574 4.175332 0.490609 0.085718 ... 0 0 0 0 0 0 0 0 0 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
112975 199995 20903 0.480856 1.000000 0 0 45.621391 5.958453 -0.918571 0.774826 ... 0 0 0 0 0 0 0 0 0 0
112976 199996 708 0.437292 1.000000 0 0 43.935162 4.476841 -0.841710 1.328253 ... 0 0 0 0 0 0 0 0 0 0
112977 199997 6693 0.546885 1.000000 0 0 46.537137 4.170806 0.388595 -0.704689 ... 0 0 1 0 0 0 0 0 0 0
112978 199998 96900 0.587076 1.000000 0 0 46.771359 -3.296814 0.243566 -1.277411 ... 0 0 0 0 0 0 0 0 0 0
112979 199999 193384 0.427535 0.586207 0 0 43.731010 -3.121867 0.027348 -0.808914 ... 0 0 0 0 0 0 0 0 0 0

112980 rows × 368 columns

from mlxtend.feature_selection import SequentialFeatureSelector as SFS
from sklearn.linear_model import LinearRegression
sfs = SFS(LinearRegression(),
           k_features=10,
           forward=True,
           floating=False,
           scoring = 'r2',
           cv = 0)
x = data.drop(['price'], axis=1)
x = x.fillna(0)
y = data['price']
x.dropna(axis=0, how='any', inplace=True)
y.dropna(axis=0, how='any', inplace=True)
sfs.fit(x, y)
sfs.k_feature_names_ 

画出来,可以看到边际效益

from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs
import matplotlib.pyplot as plt
fig1 = plot_sfs(sfs.get_metric_dict(), kind='std_dev')
plt.grid()
plt.show()

在这里插入图片描述

5️⃣.3️⃣ 嵌入式(embedding)

Lasso 回归和决策树可以完成嵌入式特征选择,大部分情况下都是用嵌入式做特征筛选。

下一步就是建模了。🤔

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