149 lines
8.1 KiB
Python
149 lines
8.1 KiB
Python
import asyncio
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import os
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import uuid
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from sqlmodel import Session, select, delete
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from datetime import datetime
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from database import engine, init_db
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import models
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import storage
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from pdf_parser import parse_pdf_to_questions
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async def seed_pdfs():
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admin_id_str = "b0e17378-864d-439e-8806-708dc48a891b"
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with Session(engine) as session:
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admin = session.exec(select(models.User).limit(1)).first()
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admin_id = admin.id if admin else uuid.UUID(admin_id_str)
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# Delete previously seeded AI DPPs to prevent duplicates
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statement = select(models.Test).where(models.Test.title.like("AI DPP:%"))
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old_tests = session.exec(statement).all()
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for t in old_tests:
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session.delete(t)
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session.commit()
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# Realistic Mock Data
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mock_data = [
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{
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"topic": "Rotational Dynamics",
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"subject": "Physics",
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"difficulty": "Hard",
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"questions": [
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{
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"question_text": r"A solid sphere of mass $M$ and radius $R$ is rolling without slipping on a horizontal surface. Its center of mass is moving with velocity $v$. What is its total kinetic energy?",
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"options": {"A": r"$\frac{1}{2}Mv^2$", "B": r"$\frac{7}{10}Mv^2$", "C": r"$\frac{5}{7}Mv^2$", "D": r"$\frac{3}{4}Mv^2$"},
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"correct_answer": {"value": "B"},
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"solution_text": r"Total KE = Translational KE + Rotational KE. $KE_{tot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. For solid sphere, $I = \frac{2}{5}MR^2$, and since it rolls without slipping, $\omega = v/R$. Thus, $KE_{tot} = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(v/R)^2 = \frac{7}{10}Mv^2$."
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},
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{
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"question_text": r"A thin uniform rod of length $l$ and mass $m$ is suspended freely from one end. It is pulled aside through an angle $\theta$ and released. What is the angular velocity when it passes through the lowest point?",
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"options": {"A": r"$\sqrt{\frac{3g}{l}(1-\cos\theta)}$", "B": r"$\sqrt{\frac{g}{l}(1-\cos\theta)}$", "C": r"$\sqrt{\frac{2g}{l}(1-\cos\theta)}$", "D": r"$\sqrt{\frac{6g}{l}(1-\cos\theta)}$"},
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"correct_answer": {"value": "A"},
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"solution_text": r"By conservation of energy: $mgh_{cm} = \frac{1}{2}I\omega^2$. $h_{cm} = \frac{l}{2}(1-\cos\theta)$. $I = \frac{ml^2}{3}$. Therefore, $mg\frac{l}{2}(1-\cos\theta) = \frac{1}{2}(\frac{ml^2}{3})\omega^2 \implies \omega = \sqrt{\frac{3g}{l}(1-\cos\theta)}$."
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}
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]
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},
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{
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"topic": "Chemical Kinetics",
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"subject": "Chemistry",
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"difficulty": "Medium",
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"questions": [
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{
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"question_text": r"For a first-order reaction $A \rightarrow P$, the rate constant is $k = 2.303 \times 10^{-3} s^{-1}$. What is the half-life ($t_{1/2}$) of this reaction?",
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"options": {"A": r"$300 s$", "B": r"$301 s$", "C": r"$69.3 s$", "D": r"$0.693 s$"},
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"correct_answer": {"value": "B"},
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"solution_text": r"For first-order reaction, $t_{1/2} = \frac{0.693}{k}$. Substituting $k$, $t_{1/2} = \frac{0.693}{2.303 \times 10^{-3}} \approx 301 s$."
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},
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{
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"question_text": r"If the rate of a reaction increases by a factor of $8$ when the concentration of reactant $X$ is doubled, what is the order of the reaction with respect to $X$?",
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"options": {"A": "1", "B": "2", "C": "3", "D": "0"},
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"correct_answer": {"value": "C"},
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"solution_text": r"Rate $R \propto [X]^n$. If $[X]$ is doubled, new rate $R' = 2^n \cdot R$. Given $R' = 8R$, so $2^n = 8 \implies n = 3$."
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}
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]
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},
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{
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"topic": "Genetics",
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"subject": "Biology",
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"difficulty": "Easy",
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"questions": [
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{
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"question_text": r"In a Mendelian dihybrid cross, what is the phenotypic ratio of the F2 generation?",
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"options": {"A": "9:3:3:1", "B": "1:2:1", "C": "3:1", "D": "1:1:1:1"},
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"correct_answer": {"value": "A"},
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"solution_text": r"A standard dihybrid cross between two heterozygotes ($AaBb \times AaBb$) yields offspring with phenotypes in a 9:3:3:1 ratio."
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},
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{
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"question_text": r"Which of the following describes the condition where both alleles of a gene pair in a heterozygote are fully expressed?",
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"options": {"A": "Incomplete Dominance", "B": "Codominance", "C": "Pleiotropy", "D": "Epistasis"},
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"correct_answer": {"value": "B"},
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"solution_text": r"Codominance is the phenomenon where both alleles are simultaneously expressed in the heterozygote, such as in AB blood type."
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}
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]
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},
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{
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"topic": "Calculus",
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"subject": "Mathematics",
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"difficulty": "Hard",
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"questions": [
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{
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"question_text": r"Evaluate the integral $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.",
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"options": {"A": r"$\pi/4$", "B": r"$\pi/2$", "C": r"$\pi$", "D": r"$0$"},
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"correct_answer": {"value": "A"},
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"solution_text": r"Let $I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get $I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$. Adding both, $2I = \int_0^{\pi/2} 1 dx = \pi/2 \implies I = \pi/4$."
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},
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{
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"question_text": r"Find the limit $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$.",
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"options": {"A": "1", "B": r"$1/2$", "C": "0", "D": "Does not exist"},
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"correct_answer": {"value": "B"},
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"solution_text": r"Using L'Hôpital's Rule twice: $\lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$."
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}
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]
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}
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]
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for idx, mock in enumerate(mock_data):
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question_ids = []
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for q_data in mock["questions"]:
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new_q = models.Question(
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subject=mock["subject"],
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topic=mock["topic"],
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subtopic="General",
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difficulty=mock["difficulty"],
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type=models.QuestionTypeEnum.MCQ_SINGLE,
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question_text=q_data["question_text"],
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options=q_data["options"],
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correct_answer=q_data["correct_answer"],
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solution_text=q_data["solution_text"],
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language=models.MediumEnum.ENGLISH,
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exam=models.ExamEnum.JEE_MAIN,
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created_by=admin_id
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)
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session.add(new_q)
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session.flush()
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question_ids.append(str(new_q.id))
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test_obj = models.Test(
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title=f"AI DPP: {mock['topic']}",
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type="DPS",
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exam_pattern=models.ExamEnum.JEE_MAIN,
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sections={
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"Section 1": {
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"questions": question_ids,
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"time_limit": 60
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}
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},
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total_time=60,
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difficulty=mock["difficulty"],
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scheduled_date=datetime.utcnow(),
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is_active=True,
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created_by=admin_id
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)
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session.add(test_obj)
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session.commit()
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print(f"Successfully created DPS Test: {test_obj.title} with {len(question_ids)} questions.")
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if __name__ == "__main__":
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asyncio.run(seed_pdfs())
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