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Data Analysis Useful Resources #dataanalysis #dataanalysisbooks #sqlbooks #pythonbooks #tableau #powerbi #datavisualization For promotions: @coderfun
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El canal Data Analysis Books | Python | SQL | Excel | Artificial Intelligence | Power BI | Tableau | AI Resources (@learndataanalysis) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 53 329 suscriptores, ocupando la posición 3 291 en la categoría Educación y el puesto 6 583 en la región India.
📊 Métricas de audiencia y dinámica
Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 53 329 suscriptores.
Según los últimos datos del 26 octubre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de 73, y en las últimas 24 horas de -2, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 7.02%. Durante las primeras 24 horas tras publicar, el contenido suele obtener N/A% de reacciones respecto al total de suscriptores.
- Alcance de las publicaciones: Cada publicación recibe en promedio 0 visualizaciones. En el primer día suele acumular 0 visualizaciones.
- Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 0.
- Intereses temáticos: El contenido se centra en temas clave como analyst, |--, excel, visualization, analytic.
📝 Descripción y política de contenido
El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“Data Analysis Useful Resources
#dataanalysis
#dataanalysisbooks
#sqlbooks
#pythonbooks
#tableau
#powerbi
#datavisualization
For promotions: @coderfun”
Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 27 octubre, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.
SELECT department, AVG(salary)
FROM employees
WHERE salary > 3000
GROUP BY department
HAVING AVG(salary) > 5000;
2. Write a query to find the second-highest salary.
Solution:
SELECT MAX(salary) AS second_highest_salary
FROM employees
WHERE salary < (SELECT MAX(salary) FROM employees);
3. How do you fetch the first 5 rows of a table?
Solution:
SELECT * FROM employees
LIMIT 5; -- (MySQL/PostgreSQL)
For SQL Server:
SELECT TOP 5 * FROM employees;
4. Write a query to find duplicate records in a table.
Solution:
SELECT column1, column2, COUNT(*)
FROM table_name
GROUP BY column1, column2
HAVING COUNT(*) > 1;
5. How do you find employees who don’t belong to any department?
Solution:
SELECT *
FROM employees
WHERE department_id IS NULL;
6. What is a JOIN, and write a query to fetch data using INNER JOIN.
Solution:
A JOIN combines rows from two or more tables based on a related column.
SELECT e.name, d.department_name
FROM employees e
INNER JOIN departments d ON e.department_id = d.id;
7. Write a query to find the total number of employees in each department.
Solution:
SELECT department_id, COUNT(*) AS total_employees
FROM employees
GROUP BY department_id;
8. How do you fetch the current date in SQL?
Solution:
SELECT CURRENT_DATE; -- MySQL/PostgreSQL
SELECT GETDATE(); -- SQL Server
9. Write a query to delete duplicate rows but keep one.
Solution:
WITH CTE AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY column1, column2 ORDER BY id) AS rn
FROM table_name
)
DELETE FROM CTE WHERE rn > 1;
10. What is a Common Table Expression (CTE), and how do you use it?
Solution:
A CTE is a temporary result set defined within a query.
WITH EmployeeCTE AS (
SELECT department_id, COUNT(*) AS total_employees
FROM employees
GROUP BY department_id
)
SELECT * FROM EmployeeCTE WHERE total_employees > 10;
Hope it helps :)
#sql #dataanalystsSELECT MAX(salary)
FROM employees
WHERE salary < (SELECT MAX(salary) FROM employees);
2️⃣ Count employees in each department:
SELECT department, COUNT(*)
FROM employees
GROUP BY department;
3️⃣ Fetch duplicate emails:
SELECT email, COUNT(*)
FROM users
GROUP BY email
HAVING COUNT(*) > 1;
4️⃣ Join orders with customer names:
SELECT c.name, o.order_date
FROM customers c
JOIN orders o ON c.id = o.customer_id;
5️⃣ Get top 3 highest salaries:
SELECT DISTINCT salary
FROM employees
ORDER BY salary DESC
LIMIT 3;
6️⃣ Retrieve latest 5 logins:
SELECT * FROM logins
ORDER BY login_time DESC
LIMIT 5;
7️⃣ Employees with no manager:
SELECT name
FROM employees
WHERE manager_id IS NULL;
8️⃣ Search names starting with ‘S’:
SELECT * FROM employees
WHERE name LIKE 'S%';
9️⃣ Total sales per month:
SELECT MONTH(order_date) AS month, SUM(amount)
FROM sales
GROUP BY MONTH(order_date);
🔟 Delete inactive users:
DELETE FROM users
WHERE last_active < '2023-01-01';
✅ Tip: Master subqueries, joins, groupings & filters – they show up in nearly every interview!
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