AquilaEdu-Instruct

# Data Description
The dataset comes from 5 different tasks: math word problems, commonsense reasoning, reading comprehension, subject knowledge, and instruction following. It involves 11 datasets, including Math23k, Ape210k, MetaMath, OpenbookQA, CommensenseQA, Arc-e, Arc-c, Race, MCTest. COIG, and Taoli.

## Data Selection
1. We first supplemented the COT process for the data and translated some English data sets into Chinese data sets.
2. Deita: We score the complexity of instrction and the quality of response.

## Data Statistics
The dataset contains 270,000 high-quality SFT education instances, in which the ratio of Chinese and English is 76%:24%. The Figure is as follows:


## Sample
```json
{
{
"instruction": "请依据以下问题，提出答案。",
"output": "假设乙盒中原来的糖果数量为x颗。\n根据题意，先从乙盒中取走了(1/4)x颗糖果，剩下的糖果数量为(3/4)x颗。\n再从乙盒中取走了剩下的(1/2)(3/4)x = (3/8)x颗糖果。\n所以，最后乙盒中剩下的糖果数量为(3/4)x - (3/8)x = (3/8)x颗。\n根据题意，甲、乙两盒的糖果总数为原来两盒糖果总数的50%，即：\n40 + (3/8)x = (1/2)(40 + x)\n将等式两边化简，得：\n40 + (3/8)x = 20 + (1/2)x\n将等式两边同时移项，得：\n(3/8)x - (1/2)x = 20 - 40\n将等式两边化简，得：\n(-1/8)x = -20\n将等式两边同时乘以-8，得：\nx = (-20)(-8) = 160\n所以，原来甲、乙两个盒子中的糖果共有40 + 160 = 200颗。 \n#### 答案是: 200.0",
"answer": "200.0",
"quality": 4.260541501059948,
"complexity": 3.0842046494197084,
"c_q_score": 13.140381906614715,
"input": "甲、乙两个盒子内分别装有糖果，已知甲盒内有40颗，先从乙盒内取走(1/4)，再从乙盒内取走余下的(1/2)，这时甲、乙两盒的糖果总数为原来两盒糖果总数的50%．则原来甲、乙两个盒子中的糖果共有多少颗．"
}
}
```

# 数据集说明
本数据集涵盖5类不同任务：数学应用题、常识推理、阅读理解、学科知识以及指令遵循，共包含11个数据集，分别为Math23k、Ape210k、MetaMath、OpenbookQA、CommensenseQA、Arc-e、Arc-c、Race、MCTest、COIG及Taoli。

## 数据选取
1. 首先为数据集补充思维链（Chain-of-Thought，COT）流程，并将部分英文数据集翻译为中文数据集。
2. Deita评分：我们对指令的复杂度与回复质量进行打分。

## 数据统计
本数据集包含27万个高质量监督微调（Supervised Fine-Tuning，SFT）教育类样本，其中中文与英文样本占比为76%:24%。数据分布如图所示：


## 样本示例
json
{
{
"instruction": "请依据以下问题，提出答案。",
"output": "假设乙盒中原来的糖果数量为x颗。
根据题意，先从乙盒中取走了(1/4)x颗糖果，剩下的糖果数量为(3/4)x颗。
再从乙盒中取走了剩下的(1/2)(3/4)x = (3/8)x颗糖果。
所以，最后乙盒中剩下的糖果数量为(3/4)x - (3/8)x = (3/8)x颗。
根据题意，甲、乙两盒的糖果总数为原来两盒糖果总数的50%，即：
40 + (3/8)x = (1/2)(40 + x)
将等式两边化简，得：
40 + (3/8)x = 20 + (1/2)x
将等式两边同时移项，得：
(3/8)x - (1/2)x = 20 - 40
将等式两边化简，得：
(-1/8)x = -20
将等式两边同时乘以-8，得：
x = (-20)(-8) = 160
所以，原来甲、乙两个盒子中的糖果共有40 + 160 = 200颗。
#### 答案是: 200.0",
"answer": "200.0",
"quality": 4.260541501059948,
"complexity": 3.0842046494197084,
"c_q_score": 13.140381906614715,
"input": "甲、乙两个盒子内分别装有糖果，已知甲盒内有40颗，先从乙盒内取走(1/4)，再从乙盒内取走余下的(1/2)，这时甲、乙两盒的糖果总数为原来两盒糖果总数的50%．则原来甲、乙两个盒子中的糖果共有多少颗．"
}
}