第13名解决方案

大家好！祝贺大家取得了优异的成绩，非常感谢我的队友 @karakasatarik！这又是一场让我们学习了新指标和概念的练习赛。我想简要介绍一下我们的解决方案。

采用的方法：

• 符号回归 (公开方案)
• 多分类神经网络 (该公开方案的修改版)
• XGB/LGB 精选集成 (公开方案)
• CatBoost 嵌套交叉验证融合 (私有方案)

使用的特征

df['mso2'] = df['free sulfur dioxide']/(1+ 10**(df['pH'] -1.81))

df['acidity_ratio'] = df['fixed acidity'] / df['volatile acidity']
df['total_acid'] = df['fixed acidity'] + df['volatile acidity'] + df['citric acid']
df['mean_acid'] = df[['fixed acidity','volatile acidity','citric acid']].mean(axis=1)
df['std_acid'] =  df[['fixed acidity','volatile acidity','citric acid']].std(axis=1)

df['free_sulfur/total_sulfur'] = df['free sulfur dioxide'] / df['total sulfur dioxide']
df['sugar/alcohol'] = df['residual sugar'] / df['alcohol']

df['sugar/citric'] = df['residual sugar'] / df['citric acid']

df['BSO2'] = df['total sulfur dioxide'] - df['free sulfur dioxide']
df['FSO2/alcohol'] = df['free sulfur dioxide'] / df['alcohol']
df['TSO2/alcohol'] = df['total sulfur dioxide'] / df['alcohol']
df['BSO2/alcohol'] = df['BSO2'] / df['alcohol']

df['chlorides/TSO2'] = df['chlorides'] / df['total sulfur dioxide']
df['sulphates/pH'] = df['sulphates'] / df['pH']

df['alcohol/density'] = df['alcohol'] / df['density']
df['alcohol_density'] = df['alcohol']  * df['density']
df['sulphates/chlorides'] = df['sulphates'] / df['chlorides']
df['alcohol/pH'] = df['alcohol'] / df['pH']
df['alcohol/acidity'] = df['alcohol'] / df['total_acid']
df['alkalinity'] = df['pH'] + df['alcohol']
df['mineral'] = df['chlorides'] + df['sulphates'] + df['residual sugar']
df['density/pH'] = df['density'] / df['pH']
df['total_alcohol'] = df['alcohol'] + df['residual sugar']

df['acid/density'] = df['total_acid']  / df['density']
df['sulphate/density'] = df['sulphates']  / df['density']
df['sulphates/acid'] = df['sulphates'] / df['volatile acidity']
df['sulphates*alcohol'] = df['sulphates'] * df['alcohol']

提交结果