From 004fea8295417aaf8a17590b24ef8695fe76acf5 Mon Sep 17 00:00:00 2001
From: benjas <909336740@qq.com>
Date: Mon, 30 Aug 2021 11:50:57 +0800
Subject: [PATCH] Add. Combining / Transforming / Interaction

---
 ...re Engineering Techniques-checkpoint.ipynb | 47 +++++++++++++++++--
 .../Feature Engineering Techniques.ipynb      | 47 +++++++++++++++++--
 2 files changed, 86 insertions(+), 8 deletions(-)

diff --git a/竞赛优胜技巧/.ipynb_checkpoints/Feature Engineering Techniques-checkpoint.ipynb b/竞赛优胜技巧/.ipynb_checkpoints/Feature Engineering Techniques-checkpoint.ipynb
index c48a018..e502a2a 100644
--- a/竞赛优胜技巧/.ipynb_checkpoints/Feature Engineering Techniques-checkpoint.ipynb	
+++ b/竞赛优胜技巧/.ipynb_checkpoints/Feature Engineering Techniques-checkpoint.ipynb	
@@ -243,7 +243,7 @@
   },
   {
    "cell_type": "markdown",
-   "id": "f81f2de6",
+   "id": "6091bf47",
    "metadata": {},
    "source": [
     "## 分类特征\n",
@@ -253,7 +253,7 @@
   {
    "cell_type": "code",
    "execution_count": 10,
-   "id": "bf7c9e8a",
+   "id": "6061ae00",
    "metadata": {},
    "outputs": [
     {
@@ -280,7 +280,7 @@
   },
   {
    "cell_type": "markdown",
-   "id": "3412c82b",
+   "id": "94a95b95",
    "metadata": {},
    "source": [
     "## Splitting\n",
@@ -289,10 +289,49 @@
     "例如，id_30诸如\"Mac OS X 10_9_5\"之类的字符串列可以拆分为操作系统\"Mac OS X\"和版本\"10_9_5\"。或者例如数字\"1230.45\"可以拆分为元\" 1230\"和分\"45\"。LGBM 无法单独看到这些片段，需要将它们拆分。"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "87e8b887",
+   "metadata": {},
+   "source": [
+    "## 组合/转化/交互\n",
+    "两个（字符串或数字）列可以合并为一列。例如card1，card2可以成为一个新列"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "92515211",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "df['uid'] = df[‘card1’].astype(str)+’_’+df[‘card2’].astype(str)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "b9c66a13",
+   "metadata": {},
+   "source": [
+    "这有助于LGBM将card1和card2一起去与目标关联，并不会在树节点分裂他们。\n",
+    "\n",
+    "但这种uid = card1_card2可能与目标相关，现在LGBM会将其拆分。数字列可以与加法、减法、乘法等组合使用。一个数字示例是"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "d50f2c15",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "df['x1_x2'] = df['x1'] * df['x2']"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,
-   "id": "e78b77c4",
+   "id": "3ce3cb5f",
    "metadata": {},
    "outputs": [],
    "source": []
diff --git a/竞赛优胜技巧/Feature Engineering Techniques.ipynb b/竞赛优胜技巧/Feature Engineering Techniques.ipynb
index c48a018..e502a2a 100644
--- a/竞赛优胜技巧/Feature Engineering Techniques.ipynb	
+++ b/竞赛优胜技巧/Feature Engineering Techniques.ipynb	
@@ -243,7 +243,7 @@
   },
   {
    "cell_type": "markdown",
-   "id": "f81f2de6",
+   "id": "6091bf47",
    "metadata": {},
    "source": [
     "## 分类特征\n",
@@ -253,7 +253,7 @@
   {
    "cell_type": "code",
    "execution_count": 10,
-   "id": "bf7c9e8a",
+   "id": "6061ae00",
    "metadata": {},
    "outputs": [
     {
@@ -280,7 +280,7 @@
   },
   {
    "cell_type": "markdown",
-   "id": "3412c82b",
+   "id": "94a95b95",
    "metadata": {},
    "source": [
     "## Splitting\n",
@@ -289,10 +289,49 @@
     "例如，id_30诸如\"Mac OS X 10_9_5\"之类的字符串列可以拆分为操作系统\"Mac OS X\"和版本\"10_9_5\"。或者例如数字\"1230.45\"可以拆分为元\" 1230\"和分\"45\"。LGBM 无法单独看到这些片段，需要将它们拆分。"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "87e8b887",
+   "metadata": {},
+   "source": [
+    "## 组合/转化/交互\n",
+    "两个（字符串或数字）列可以合并为一列。例如card1，card2可以成为一个新列"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "92515211",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "df['uid'] = df[‘card1’].astype(str)+’_’+df[‘card2’].astype(str)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "b9c66a13",
+   "metadata": {},
+   "source": [
+    "这有助于LGBM将card1和card2一起去与目标关联，并不会在树节点分裂他们。\n",
+    "\n",
+    "但这种uid = card1_card2可能与目标相关，现在LGBM会将其拆分。数字列可以与加法、减法、乘法等组合使用。一个数字示例是"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "d50f2c15",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "df['x1_x2'] = df['x1'] * df['x2']"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,
-   "id": "e78b77c4",
+   "id": "3ce3cb5f",
    "metadata": {},
    "outputs": [],
    "source": []