Data Analyst Interview Resources
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Data Analyst Interview Resources (@dataanalystinterview) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 52 623 obunachidan iborat bo'lib, Taʼlim toifasida 3 247-o'rinni va Hindiston mintaqasida 6 676-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 52 623 obunachiga ega bo‘ldi.
29 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 15 ga, so‘nggi 24 soatda esa 4 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 1.89% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.83% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 995 marta ko‘riladi; birinchi sutkada odatda 438 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent sql, row, |--, dataset, visualization kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“Join our telegram channel to learn how data analysis can reveal fascinating patterns, trends, and stories hidden within the numbers! 📊
For ads & suggestions: @love_data”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 30 Avgust, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
def is_palindrome(s):
return s == s[::-1]
print(is_palindrome("madam")) # True
print(is_palindrome("hello")) # False
2. How to find the factorial of a number using recursion?
def factorial(n):
if n == 0 or n == 1:
return 1
return n * factorial(n - 1)
print(factorial(5)) # 120
3. How to merge two dictionaries in Python?
dict1 = {'a': 1, 'b': 2}
dict2 = {'c': 3, 'd': 4}
# Method 1 (Python 3.5+)
merged_dict = {**dict1, **dict2}
# Method 2 (Python 3.9+)
merged_dict = dict1 | dict2
print(merged_dict)
4. How to find the intersection of two lists?
list1 = [1, 2, 3, 4]
list2 = [3, 4, 5, 6]
intersection = list(set(list1) & set(list2))
print(intersection) # [3, 4]
5. How to generate a list of even numbers from 1 to 100?
even_numbers = [i for i in range(1, 101) if i % 2 == 0]
print(even_numbers)
6. How to find the longest word in a sentence?
def longest_word(sentence):
words = sentence.split()
return max(words, key=len)
print(longest_word("Python is a powerful language")) # "powerful"
7. How to count the frequency of elements in a list?
from collections import Counter
my_list = [1, 2, 2, 3, 3, 3, 4]
frequency = Counter(my_list)
print(frequency) # Counter({3: 3, 2: 2, 1: 1, 4: 1})
8. How to remove duplicates from a list while maintaining the order?
def remove_duplicates(lst):
return list(dict.fromkeys(lst))
my_list = [1, 2, 2, 3, 4, 4, 5]
print(remove_duplicates(my_list)) # [1, 2, 3, 4, 5]
9. How to reverse a linked list in Python?
class Node:
def __init__(self, data):
self.data = data
self.next = None
def reverse_linked_list(head):
prev = None
current = head
while current:
next_node = current.next
current.next = prev
prev = current
current = next_node
return prev
# Create linked list: 1 -> 2 -> 3
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
# Reverse and print the list
reversed_head = reverse_linked_list(head)
while reversed_head:
print(reversed_head.data, end=" -> ")
reversed_head = reversed_head.next
10. How to implement a simple binary search algorithm?
def binary_search(arr, target):
low, high = 0, len(arr) - 1
while low <= high:
mid = (low + high) // 2
if arr[mid] == target:
return mid
elif arr[mid] < target:
low = mid + 1
else:
high = mid - 1
return -1
print(binary_search([1, 2, 3, 4, 5, 6, 7], 4)) # 3
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