Data Analytics
Perfect channel to learn Data Analytics Learn SQL, Python, Alteryx, Tableau, Power BI and many more For Promotions: @coderfun @love_data
نمایش بیشتر📈 تحلیل کانال تلگرام Data Analytics
کانال Data Analytics (@sqlspecialist) در بخش زبانی انگلیسی بازیگری فعال است. در حال حاضر جامعه شامل 110 115 مشترک است و جایگاه 1 107 را در دسته فناوری و برنامهها و رتبه 2 303 را در منطقه الهند دارد.
📊 شاخصهای مخاطب و پویایی
از زمان ایجاد در невідомо، پروژه رشد سریعی داشته و 110 115 مشترک جذب کرده است.
بر اساس آخرین دادهها در تاریخ 13 ژوئیه, 2026، کانال فعالیت پایداری دارد. در ۳۰ روز گذشته تغییر اعضا برابر 614 و در ۲۴ ساعت گذشته برابر 7 بوده و همچنان دسترسی گستردهای حفظ شده است.
- وضعیت تأیید: تأیید نشده
- نرخ تعامل (ER): میانگین تعامل مخاطب 3.42% است و در ۲۴ ساعت نخست پس از انتشار، محتوا معمولاً 1.66% واکنش نسبت به کل مشترکان کسب میکند.
- دسترسی پستها: هر پست به طور میانگین 3 766 بازدید دریافت میکند. در اولین روز معمولاً 1 832 بازدید جمعآوری میشود.
- واکنشها و تعامل: مخاطبان بهطور فعال حمایت میکنند؛ میانگین واکنش به هر پست 9 است.
- علایق موضوعی: محتوا بر موضوعات کلیدی مانند row, sql, analytic, analyst, visualization تمرکز دارد.
📝 توضیح و سیاست محتوایی
نویسنده این فضا را محل بیان دیدگاههای شخصی توصیف میکند:
“Perfect channel to learn Data Analytics
Learn SQL, Python, Alteryx, Tableau, Power BI and many more
For Promotions: @coderfun @love_data”
به لطف بهروزرسانیهای پرتکرار (آخرین داده در تاریخ 14 ژوئیه, 2026)، کانال همواره بهروز و دارای دسترسی بالاست. تحلیلها نشان میدهد مخاطبان بهطور فعال با محتوا تعامل دارند و آن را به نقطه اثرگذاری مهم در دسته فناوری و برنامهها تبدیل کردهاند.
SELECT
employee_id,
total_projects
FROM (
SELECT
employee_id,
COUNT(DISTINCT project_id) AS total_projects,
DENSE_RANK() OVER (
ORDER BY COUNT(DISTINCT project_id) DESC
) AS rnk
FROM employee_projects
GROUP BY employee_id
) ranked
WHERE rnk = 1;
💡 Explanation:
This query counts the number of unique projects each employee has worked on and identifies those with the highest count.
• COUNT(DISTINCT project_id) counts unique projects for each employee
• GROUP BY employee_id creates one record per employee
• DENSE_RANK() ranks employees based on the number of projects
• The outer query returns all employees tied for the highest number of projects
This question tests your understanding of:
✅ COUNT(DISTINCT)
✅ GROUP BY
✅ Window Functions DENSE_RANK
✅ Ranking Aggregated Results
🎯 Expected Output Example
Employee ID | Total Projects
101 | 12
205 | 12
Both employees have worked on the highest number of distinct projects.
🚀 Alternative Without Window Functions
SELECT
employee_id,
COUNT(DISTINCT project_id) AS total_projects
FROM employee_projects
GROUP BY employee_id
HAVING COUNT(DISTINCT project_id) = (
SELECT MAX(project_count)
FROM (
SELECT
COUNT(DISTINCT project_id) AS project_count
FROM employee_projects
GROUP BY employee_id
) t
);
This solution uses nested subqueries and MAX() instead of window functions.
🚀 Tip for SQL Job Seekers:
Many interview questions involve ranking aggregated results, such as:
Highest number of projects, Most orders, Maximum sales, Highest attendance, Most logins
Practice combining GROUP BY with window functions like DENSE_RANK() to solve these efficiently.
❤️ React with ❤️ for more interview challenges!WITH salary_changes AS (
SELECT
employee_id,
salary,
effective_date,
salary - LAG(salary) OVER (
PARTITION BY employee_id
ORDER BY effective_date
) AS salary_increment
FROM salary_history
)
SELECT
employee_id,
salary_increment
FROM (
SELECT
employee_id,
salary_increment,
DENSE_RANK() OVER (
ORDER BY salary_increment DESC
) AS rnk
FROM salary_changes
WHERE salary_increment IS NOT NULL
) ranked
WHERE rnk = 1;
💡 Explanation:
This query calculates each employee's salary increment and then finds the highest increment across all employees.
• LAG(salary) retrieves the employee's previous salary
• The difference between the current and previous salary gives the increment
• DENSE_RANK() ranks increments from highest to lowest
• The outer query returns all employees tied for the highest salary increment
This question tests your understanding of:
✅ LAG() Window Function
✅ Common Table Expressions (CTEs)
✅ DENSE_RANK()
✅ Time-Series Data Analysis
🎯 Expected Output Example
Employee ID | Salary Increment
101 | 20,000
205 | 20,000
Both employees received the largest salary increase.
🚀 Why Interviewers Ask This?
This is a classic window function interview question. It evaluates your ability to compare a row with its previous row—a common requirement in payroll, finance, and audit systems.
🚀 Tip for SQL Job Seekers:
Master these analytical window functions:
LAG() / LEAD() / FIRST_VALUE() / LAST_VALUE() / NTILE()
These functions are frequently tested in product-based companies and data-focused interviews because they simplify complex row-by-row comparisons.
❤️ React with ❤️ for more interview challenges!SELECT
customer_id
FROM orders
WHERE YEAR(order_date) = 2025
GROUP BY customer_id
HAVING COUNT(DISTINCT MONTH(order_date)) = 12;
💡 Explanation:
This query identifies customers who placed at least one order in every month of 2025.
• WHERE YEAR(order_date) = 2025 filters orders from the year 2025
• GROUP BY customer_id groups all orders by customer
• COUNT(DISTINCT MONTH(order_date)) counts the unique months in which each customer placed an order
• HAVING ... = 12 ensures the customer has orders in all 12 months
This question tests your understanding of:
✅ Date Functions (YEAR, MONTH)
✅ GROUP BY
✅ HAVING
✅ COUNT(DISTINCT)
🎯 Expected Output Example
| Customer ID |
|-------------|
| 101 |
| 205 |
These customers placed at least one order in every month of 2025.
🚀 Alternative (Database-Agnostic SQL)
SELECT
customer_id
FROM orders
WHERE EXTRACT(YEAR FROM order_date) = 2025
GROUP BY customer_id
HAVING COUNT(DISTINCT EXTRACT(MONTH FROM order_date)) = 12;
This version works with databases like PostgreSQL and Oracle that support the EXTRACT() function.
🚀 Tip for SQL Job Seekers:
Whenever you see interview questions containing phrases like:
"Every month" / "Every quarter" / "Every year" / "Every category"
Think of COUNT(DISTINCT ...) combined with GROUP BY and HAVING. This is a very common SQL interview pattern.
❤️ React with ❤️ for more interview challenges!The GigaChat team has released GigaChat 3.5 Ultra as open source—a new 432B model under the MIT license. This is the first open-source hybrid of GatedDeltaNet and MLA scaled to hundreds of billions of parameters, featuring a proprietary training recipe we refined through more than 1,500 experiments. The model has grown in terms of code, mathematics, agent scenarios, and application domains—yet it’s 40% smaller than GigaChat 3.1 Ultra.What’s inside: 🔘A proprietary hybrid MLA + Gated DeltaNet architecture with a dedicated stabilization framework, without which this hybrid setup would not train reliably at this scale; 🔘 Gated Attention: the model can locally down-weight overly strong signals from the attention layer; 🔘GatedNorm: normalization with an explicit gate that controls signal magnitude across features; 🔘Approximately 4x lower KV cache per token: with the same memory budget, the model can support 2.14x longer context and deliver a 20% throughput increase under load; 🔘Two MTP heads, enabling up to 2.2x faster generation; 🔘FP8 across all training stages with no quality degradation compared with bf16, enabled by custom Triton and CUDA kernels; 🔘A new online RL stage after SFT and DPO. Results: 🔘 GigaChat-3.5-Ultra-Base outperforms DeepSeek V3.2 Exp Base and DeepSeek V4 Flash Base on average across a set of general, math, and code benchmarks: 🔘 GigaChat-3.5-Ultra-Instruct is comparable to DeepSeek V3.2 in terms of average score, despite having half the size; 🔘 According to the MiniMax-M2.7 LLM judge, the average win rate against GigaChat 3.1 Ultra is 75.9%, and against GPT-5 is 68.7%.
The entire stack — data (our own LLM-filtered Common Crawl, 600+ programming languages in the code), architecture, training methodology, and infrastructure — was built end-to-end by GigaChat team.➡️ HuggingFace
