EDA中级-Kaggle学习2-特征工程序章

前言

原文来源：这里！
数据集：根据不同人的数据来判断他有没有可能再次偿还信用借款（Home credit default risk）
看完的感想是：

每个不同的数据集的数据都是不一样的，背后表现的现象也是有差异。
但是还是有处理数据的规律和套路可以值得学习的。
本文是特征工程处理的第一章，提供了一些简单的方法来新建特征数据，之后还有2章讲述新的方法

原文的工作流程：outline of a machine learning project

理解问题，数据背景（understand the problem and the data）
数据清理和格式清理（data cleaning and formatting）
EDA研究数据
使用数据建立最简单的模型（baseline model） ----> 线性回归line regression比如说（得到最低的acc，然后之后都要在这个基础上改进，不能比这个还差）
改进模型（improved model）----> 使用特征工程改进的数据或者使用新的模型
对于模型进行基本解释（model interpretaion） ---> 为啥比baseline好等等的总结

特征工程feature engineering

下面列举本文新建特征数据的几个方法，本文新建的角度有两种：
- 1. 从统计学的角度
- 1. 从应用场景的背景知识来的角度（domain knowledge）
本文还强调了kaggle的关键也就是特征工程，好的特征可以比起HP Tuning和模型选择更能促进gradient boosting，从而获得高acc。
特征工程是：
- 新建特征
- 选择重要特征(删除其他不需要的特征):从而reduce dimentionality的集合

统计学的角度

多项式特征
我们本来有['EXT_SOURCE_1', 'EXT_SOURCE_2', 'EXT_SOURCE_3', 'DAYS_BIRTH']这些特征，然后我们可以将他们两两组合(degree=2)变成下面的特征组合。

['1',
 'EXT_SOURCE_1',
 'EXT_SOURCE_2',
 'EXT_SOURCE_3',
 'DAYS_BIRTH',
 'EXT_SOURCE_1^2',
 'EXT_SOURCE_1 EXT_SOURCE_2',
 'EXT_SOURCE_1 EXT_SOURCE_3',
 'EXT_SOURCE_1 DAYS_BIRTH',
 'EXT_SOURCE_2^2',
 'EXT_SOURCE_2 EXT_SOURCE_3',
 'EXT_SOURCE_2 DAYS_BIRTH',
 'EXT_SOURCE_3^2',
 'EXT_SOURCE_3 DAYS_BIRTH',
 'DAYS_BIRTH^2']

实现方法：

from sklearn.preprocessing import PolynomialFeatures

# Make a new dataframe for polynomial features
poly_features = app_train[['EXT_SOURCE_1', 'EXT_SOURCE_2', 'EXT_SOURCE_3', 'DAYS_BIRTH', 'TARGET']]
                                  
# Create the polynomial object with specified degree
poly_transformer = PolynomialFeatures(degree = 3)
# Train the polynomial features
poly_transformer.fit(poly_features)

# Transform the features
poly_features = poly_transformer.transform(poly_features)

注：对于poly的新特征有必要用scaler来调整一些参数取值范围（原文用了minmaxscaler）

从经济学的角度（domain knowledge）

CREDIT_INCOME_PERCENT: 借贷与收入比 the percentage of the credit amount relative to a client's income
ANNUITY_INCOME_PERCENT: 贷款年金和收入比the percentage of the loan annuity relative to a client's income
CREDIT_TERM: 连续还款月长度 the length of the payment in months (since the annuity is the monthly amount due
DAYS_EMPLOYED_PERCENT: 受雇佣天数与年龄比the percentage of the days employed relative to the client's age
实现方式

app_train_domain['CREDIT_INCOME_PERCENT'] = app_train_domain['AMT_CREDIT'] / app_train_domain['AMT_INCOME_TOTAL']
app_train_domain['ANNUITY_INCOME_PERCENT'] = app_train_domain['AMT_ANNUITY'] / app_train_domain['AMT_INCOME_TOTAL']
app_train_domain['CREDIT_TERM'] = app_train_domain['AMT_ANNUITY'] / app_train_domain['AMT_CREDIT']
app_train_domain['DAYS_EMPLOYED_PERCENT'] = app_train_domain['DAYS_EMPLOYED'] / app_train_domain['DAYS_BIRTH']

其他

除了重点是特征工程以外，本文还提供了一些新的画图/处理数据的方法，这里简单总结一下

1.画图

sns.kdeplot: 内核密度函数kernel density estimation
也就是之前sns.distplot之后hist的拟合曲线？

plt.figure(figsize = (10, 8))

# KDE plot of loans that were repaid on time
sns.kdeplot(app_train.loc[app_train['TARGET'] == 0, 'DAYS_BIRTH'] / 365, label = 'target == 0')

# KDE plot of loans which were not repaid on time
sns.kdeplot(app_train.loc[app_train['TARGET'] == 1, 'DAYS_BIRTH'] / 365, label = 'target == 1')

# Labeling of plot
plt.xlabel('Age (years)'); plt.ylabel('Density'); plt.title('Distribution of Ages');

如何通过kde看出相关性：比如下图EXT_SOURCE_3比起其他2个更能分辨出target=0/1的区别，那就说明EXT_SOURCE_3比起其他2个更好，更有相关性。

2.发现样本目标值不均衡怎么办？

在训练器里面加权目标值:weight classes（XGBOOST）
其他解决方法看这里:这里
下面有class_weight = 'balanced'参数

# Create the model
        model = lgb.LGBMClassifier(n_estimators=10000, objective = 'binary', 
                                   class_weight = 'balanced', learning_rate = 0.05, 
                                   reg_alpha = 0.1, reg_lambda = 0.1, 
                                   subsample = 0.8, n_jobs = -1, random_state = 50)

3.查看数据集里面所有categorical里面的Unique数量

# Number of unique classes in each object column
app_train.select_dtypes('object').apply(pd.Series.nunique, axis = 0)

4.对于categorical的encoding

原文给出的是
小于2个的categorical ---> label encoding

from sklearn.preprocessing import LabelEncoder
le = LabelEncoder()
le.fit(app_train[col])
app_train[col] = le.transform(app_train[col])

大于2个的 ---> one-hot encoding

pd.get_dummies(app_train[col])

5.如何对齐align训练集和测试集

最后训练集要比测试集多一列：target
align的理由是在one-hot encoding之后：训练集可能比测试集多出几列（训练集里面存在的categorical值，在测试集里面不存在）
下面是对齐的实现，注意要事先把target拿出来，防止被align的时候删除。

train_labels = app_train['TARGET']

# Align the training and testing data, keep only columns present in both dataframes
app_train, app_test = app_train.align(app_test, join = 'inner', axis = 1)

# Add the target back in
app_train['TARGET'] = train_labels

print('Training Features shape: ', app_train.shape)
print('Testing Features shape: ', app_test.shape)

得到

Training Features shape:  (307511, 240)
Testing Features shape:  (48744, 239)

6.对于相关性的评估

.00-.19 “very weak”
.20-.39 “weak”
.40-.59 “moderate”
.60-.79 “strong”
.80-1.0 “very strong”

7.简单replace一个值

可以用切片或者df.replace

# Replace the anomalous values with nan
app_train['DAYS_EMPLOYED'].replace({365243: np.nan}, inplace = True)

8.处理outlier/anomalies

将其变成nan，然后之后会在处理nan(插补imputation)的时候一同处理。

9.神奇的anomalies

原文发现有些人的年龄超过了1000岁（异常值）而且这些人欠款违约（default loan）的概率比一般人还小。

anom = app_train[app_train['DAYS_EMPLOYED'] == 365243]
non_anom = app_train[app_train['DAYS_EMPLOYED'] != 365243]
print('The non-anomalies default on %0.2f%% of loans' % (100 * non_anom['TARGET'].mean()))
print('The anomalies default on %0.2f%% of loans' % (100 * anom['TARGET'].mean()))
print('There are %d anomalous days of employment' % len(anom))

得到只有5.4%的人违约欠款了。

The non-anomalies default on 8.66% of loans
The anomalies default on 5.40% of loans
There are 55374 anomalous days of employment

10.插补Imputer填充nan

之前都是直接df.fillna(df.mean())，这里用了Imputer

from sklearn.preprocessing import MinMaxScaler, Imputer
# Median imputation of missing values
imputer = Imputer(strategy = 'median')
# Fit on the training data
imputer.fit(train)
# Transform both training and testing data
train = imputer.transform(train)

11.训练完后展示重要的特征feature_importance

sklearn的基本每个classifer都有feature_importances_这个参数，下面的例子是随机树

# Extract feature importances
feature_importance_values_domain = random_forest_domain.feature_importances_
feature_importances_domain = pd.DataFrame({'feature': domain_features_names, 'importance': feature_importance_values_domain})

获取到之后就可以用barplot来显示

def plot_feature_importances(df):
    """
    Plot importances returned by a model. This can work with any measure of
    feature importance provided that higher importance is better. 
    
    Args:
        df (dataframe): feature importances. Must have the features in a column
        called `features` and the importances in a column called `importance
        
    Returns:
        shows a plot of the 15 most importance features
        
        df (dataframe): feature importances sorted by importance (highest to lowest) 
        with a column for normalized importance
        """
    
    # Sort features according to importance
    df = df.sort_values('importance', ascending = False).reset_index()
    
    # Normalize the feature importances to add up to one
    df['importance_normalized'] = df['importance'] / df['importance'].sum()

    # Make a horizontal bar chart of feature importances
    plt.figure(figsize = (10, 6))
    ax = plt.subplot()
    
    # Need to reverse the index to plot most important on top
    ax.barh(list(reversed(list(df.index[:15]))), 
            df['importance_normalized'].head(15), 
            align = 'center', edgecolor = 'k')
    
    # Set the yticks and labels
    ax.set_yticks(list(reversed(list(df.index[:15]))))
    ax.set_yticklabels(df['feature'].head(15))
    
    # Plot labeling
    plt.xlabel('Normalized Importance'); plt.title('Feature Importances')
    plt.show()
    
    return df
# Show the feature importances for the default features
feature_importances_sorted = plot_feature_importances(feature_importances)

发表于 2021-01-08 18:56 niemand-01 阅读(163) 评论(0) 编辑收藏举报

前言