main
HuangHai 2 weeks ago
parent 77abf8fc41
commit 60a1fdd11c

@ -15,6 +15,10 @@ EMBED_BASE_URL = "https://api.siliconflow.cn/v1"
EMBED_DIM = 1024
EMBED_MAX_TOKEN_SIZE = 8192
# 视觉模型
VISION_API_KEY = "sk-pbqibyjwhrgmnlsmdygplahextfaclgnedetybccknxojlyl"
VISION_BASE_URL = "https://api.siliconflow.cn/v1/chat/completions"
VISION_MODEL_NAME = "GLM-4.1V-9B-Thinking"
NEO4J_URI = "bolt://localhost:7687"
NEO4J_USERNAME = "neo4j"

@ -1,61 +1,49 @@
### 题目1
**题目序号**: 1
**题目内容**: 已知集合 $A=\left\{x \mid -5 < x^{3} < 5\right\}, B=\left\{-3,-1,0,2,3\right\}$,则 $A \cap B=$
**选项**:
A. $\{-1,0\}$
B. $\{2,3\}$
C. $\{-3,-1,0\}$
D. $\{-1,0,2\}$
**答案**: A
**解析**: $A \cap B=\{-1,0\}$,选 A。
---
### 题目2
**题目序号**: 2
**题目内容**: 若 $\frac{2}{z-1}=1+i$,则 $z=$
**选项**:
A. $-1-i$
B. $-1+i$
C. $1-i$
D. $1+i$
**答案**: C
---
### 题目3
**题目序号**: 3
**题目内容**: 已知向量 $\vec{a}=(0,1)$$\vec{b}=(2,x)$,若 $\vec{b} \perp (\vec{b}-4\vec{a})$,则 $x=$
**选项**:
A. $-2$
B. $-1$
C. $1$
D. $2$
**答案**: D
**解析**: $\vec{b}-4\vec{a}=(2,x-4)$$\vec{b} \perp (\vec{b}-4\vec{a})$$\therefore \vec{b}(\vec{b}-4\vec{a})=0$$\therefore 4+x(x-4)=0$$\therefore x=2$,选 D。
---
### 题目4
**题目序号**: 4
**题目内容**: 已知 $\cos(\alpha+\beta)=m$$\tan \alpha \tan \beta=2$,则 $\cos(\alpha-\beta)=$
**选项**:
A. $-3m$
B. $-\frac{m}{3}$
C. $\frac{m}{3}$
D. $3m$
**答案**: A
**解析**: $\left\{\begin{array}{l}\cos \alpha \cos \beta-\sin \alpha \sin \beta=m \\\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}=2\end{array}\right.$$\therefore \left\{\begin{array}{l}\sin \alpha \sin \beta=-2m \\\cos \alpha \cos \beta=-m\end{array}\right.$ $\cos(\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta=-m-2m=-3m$,选 A。
---
### 题目5
**题目序号**: 5
**题目内容**: 已知圆柱和圆锥的底面半径相等,侧面积相等,且它们的高均为 $\sqrt{3}$,则圆锥的体积为
**选项**:
A. $2\sqrt{3}\pi$
B. $3\sqrt{3}\pi$
C. $6\sqrt{3}\pi$
D. $9\sqrt{3}\pi$
**答案**: B
**解析**: 设它们底面半径为 $r$,圆锥母线 $l$$\therefore 2\pi r\sqrt{3}=\pi rl$$\therefore l=\sqrt{3}$,则圆锥的体积为 $\frac{1}{3}\pi r^{2}h$。
1. 题目序号1
2. 题目内容:已知集合 $A=\left\{x \mid -5 < x^{3} < 5\right\}, B=\left\{-3,-1,0,2,3\right\}$,则 $A \cap B=$
3. 选项:
A. $\{-1,0\}$
B. $\{2,3\}$
C. $\{-3,-1,0\}$
D. $\{-1,0,2\}$
4. 答案A
5. 解析:$A \cap B=\{-1,0\}$,选 A。
1. 题目序号2
2. 题目内容:若 $\frac{2}{z-1}=1+i$,则 $z=$
3. 选项:
A. $-1-i$
B. $-1+i$
C. $1-i$
D. $1+i$
4. 答案C
5. 解析:
1. 题目序号3
2. 题目内容:已知向量 $\vec{a}=(0,1)$$\vec{b}=(2,x)$,若 $\vec{b} \perp (\vec{b}-4\vec{a})$,则 $x=$
3. 选项:
A. $-2$
B. $-1$
C. $1$
D. $2$
4. 答案D
5. 解析:$\vec{b}-4\vec{a}=(2,x-4)$$\vec{b} \perp (\vec{b}-4\vec{a})$$\therefore \vec{b}(\vec{b}-4\vec{a})=0$$\therefore 4+x(x-4)=0$$\therefore x=2$,选 D。
1. 题目序号4
2. 题目内容:已知 $\cos(\alpha+\beta)=m$$\tan \alpha \tan \beta=2$,则 $\cos(\alpha-\beta)=$
3. 选项:
A. $-3m$
B. $-\frac{m}{3}$
C. $\frac{m}{3}$
D. $3m$
4. 答案A
5. 解析:$\left\{\begin{array}{l}\cos \alpha \cos \beta-\sin \alpha \sin \beta=m \\\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}=2\end{array}\right.$$\therefore \left\{\begin{array}{l}\sin \alpha \sin \beta=-2m \\\cos \alpha \cos \beta=-m\end{array}\right.$$\cos(\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta=-m-2m=-3m$,选 A。
1. 题目序号5
2. 题目内容:已知圆柱和圆锥的底面半径相等,侧面积相等,且它们的高均为 $\sqrt{3}$,则圆锥的体积为
3. 选项:
A. $2\sqrt{3}\pi$
B. $3\sqrt{3}\pi$
C. $6\sqrt{3}\pi$
D. $9\sqrt{3}\pi$
4. 答案B
5. 解析:设它们底面半径为 $r$,圆锥母线 $l$$\therefore 2\pi r\sqrt{3}=\pi rl$$\therefore l=\sqrt{3}$,则圆锥的体积为 $\frac{1}{3}\pi r^{2}h$。

@ -3,7 +3,7 @@ import os
import shutil
from raganything import RAGAnything, RAGAnythingConfig
from Util.RagUtil import create_llm_model_func, create_vision_model_func, create_embedding_func
from Util.LightRagUtil import create_llm_model_func, create_embedding_func, create_vision_model_func
import logging
@ -24,10 +24,10 @@ logger.addHandler(handler)
async def main():
# 要处理的文件路径
file_path = "Docx/《动能定理》巩固练习.docx"
WORKING_DIR = "../../Topic/MathOcr"
WORKING_DIR = "../Topic/MathOcr"
# 删除output目录下的所有文件
output_dir = "../../output"
output_dir = "../output"
shutil.rmtree(output_dir, ignore_errors=True)
os.makedirs(output_dir, exist_ok=True)

@ -139,4 +139,43 @@ def create_embedding_func():
api_key=EMBED_API_KEY,
base_url=EMBED_BASE_URL,
),
)
)
def create_vision_model_func(llm_model_func):
def vision_model_func(
prompt, system_prompt=None, history_messages=[], image_data=None, **kwargs
):
if image_data:
return openai_complete_if_cache(
VISION_MODEL_NAME,
"",
system_prompt=None,
history_messages=[],
messages=[
{"role": "system", "content": system_prompt}
if system_prompt
else None,
{
"role": "user",
"content": [
{"type": "text", "text": prompt},
{
"type": "image_url",
"image_url": {
"url": f"data:image/jpeg;base64,{image_data}"
},
},
],
}
if image_data
else {"role": "user", "content": prompt},
],
api_key=VISION_API_KEY,
base_url=VISION_BASE_URL,
**kwargs,
)
else:
return llm_model_func(prompt, system_prompt, history_messages, **kwargs)
return vision_model_func
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