Data Analytics Projects - SQL, Excel, Tableau, Python & Power BI Interview Resources
Covering all technical and popular stuff about anything related to Data Science: AI, Big Data, Machine Learning, Statistics, general Math and the applications of former. Ads/ Promo: @love_data
Ko'proq ko'rsatish๐ Telegram kanali Data Analytics Projects - SQL, Excel, Tableau, Python & Power BI Interview Resources analitikasi
Data Analytics Projects - SQL, Excel, Tableau, Python & Power BI Interview Resources (@sqlproject) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 39 485 obunachidan iborat bo'lib, Taสผlim toifasida 4 741-o'rinni va Hindiston mintaqasida 10 461-o'rinni egallagan.
๐ Auditoriya koโrsatkichlari va dinamika
ะฝะตะฒัะดะพะผะพ sanasidan buyon loyiha tez oโsib, 39 485 obunachiga ega boโldi.
06 Iyun, 2026 dagi oxirgi maโlumotlarga koโra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 225 ga, soโnggi 24 soatda esa 12 ga oโzgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya oโrtacha 2.63% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.95% ini tashkil etuvchi reaksiyalarni toโplaydi.
- Post qamrovi: Har bir post oโrtacha 1 037 marta koโriladi; birinchi sutkada odatda 376 ta koโrish yigโiladi.
- Reaksiyalar va oโzaro taโsir: Auditoriya faol: har bir postga oโrtacha 3 ta reaksiya keladi.
- Tematik yoโnalishlar: Kontent analytic, dataset, visualization, sql, learning kabi asosiy mavzularga jamlangan.
๐ Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida taโriflaydi:
โCovering all technical and popular stuff about anything related to Data Science: AI, Big Data, Machine Learning, Statistics, general Math and the applications of former.
Ads/ Promo: @love_dataโ
Yuqori yangilanish chastotasi (oxirgi maโlumot 08 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli boโlib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taสผlim toifasidagi muhim taโsir nuqtasiga aylantirishini koโrsatadi.
COALESCE or IS NULL checks
3๏ธโฃ Wrong JOIN Type
โข INNER instead of LEFT
โข Data silently disappears
โข Always ask: Do you need unmatched rows?
4๏ธโฃ Missing JOIN Conditions
โข Creates cartesian product
โข Rows explode
โข Always join on keys
5๏ธโฃ Filtering After JOIN Instead of Before
โข Processes more rows than needed
โข Slower performance
โข Filter early using WHERE or subqueries
6๏ธโฃ Using WHERE Instead of HAVING
โข WHERE filters rows
โข HAVING filters groups
โข Aggregates fail without HAVING
7๏ธโฃ Not Using Indexes
โข Full table scans
โข Slow dashboards
โข Index columns used in JOIN, WHERE, ORDER BY
8๏ธโฃ Relying on ORDER BY in Subqueries
โข Order not guaranteed
โข Results change
โข Use ORDER BY only in final query
9๏ธโฃ Mixing Data Types
โข Implicit conversions
โข Index not used
โข Match column data types
๐ No Query Validation
โข Results look right but are wrong
โข Always cross-check counts and totals
๐ง Practice Task
โข Rewrite one query
โข Remove SELECT *
โข Add proper JOIN
โข Handle NULLs
โข Compare result count
SQL Resources: https://whatsapp.com/channel/0029VanC5rODzgT6TiTGoa1v
โค๏ธ Double Tap For MoreSELECT MAX(salary)
FROM employees
WHERE salary < (SELECT MAX(salary) FROM employees);
2๏ธโฃ List employees who earn more than the average salary.
SELECT name, salary
FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees);
3๏ธโฃ Show department-wise highest paid employee.
SELECT department, name, salary
FROM (
SELECT *,
RANK() OVER (PARTITION BY department ORDER BY salary DESC) AS rnk
FROM employees
) AS ranked
WHERE rnk = 1;
4๏ธโฃ Display total sales made by each employee in 2023.
SELECT emp_id, SUM(amount) AS total_sales
FROM sales
WHERE YEAR(sale_date) = 2023
GROUP BY emp_id;
5๏ธโฃ Retrieve products with price above average in their category.
SELECT p.name, p.category, p.price
FROM products p
WHERE price > (
SELECT AVG(price)
FROM products
WHERE category = p.category
);
6๏ธโฃ Identify duplicate emails in the users table.
SELECT email, COUNT(*)
FROM users
GROUP BY email
HAVING COUNT(*) > 1;
7๏ธโฃ Rank customers based on total purchase amount.
SELECT customer_id,
SUM(amount) AS total_spent,
RANK() OVER (ORDER BY SUM(amount) DESC) AS rank
FROM orders
GROUP BY customer_id;
๐ฌ Double Tap โค๏ธ For More!
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