Python Projects & Free Books
Python Interview Projects & Free Courses Admin: @Coderfun
Show moreπ Analytical overview of Telegram channel Python Projects & Free Books
Channel Python Projects & Free Books (@pythonfreebootcamp) in the English language segment is an active participant. Currently, the community unites 40 940 subscribers, ranking 3 313 in the Technologies & Applications category and 9 602 in the India region.
π Audience metrics and dynamics
Since its creation on Π½Π΅Π²ΡΠ΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 40 940 subscribers.
According to the latest data from 23 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 142 over the last 30 days and by -13 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 3.59%. Within the first 24 hours after publication, content typically collects 0.93% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 470 views. Within the first day, a publication typically gains 380 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 4.
- Thematic interests: Content is focused on key topics such as learning, analyst, framework, link:-, structure.
π Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
βPython Interview Projects & Free Courses
Admin: @Coderfunβ
Thanks to the high frequency of updates (latest data received on 24 June, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Technologies & Applications category.
items = [1, 2, 2, 3, 4]
for item in items:
if item == 2:
items.remove(item)
print(items)
# Output: [1, 2, 3, 4]
It seems that all 2s should disappear, but one remains. β Why?
After removing an element, the list shifts, but the loop moves on β as a result, some values are simply skipped. ππ«
How to do it correctly β iterate over a copy: β
for item in items[:]:
if item == 2:
items.remove(item)
print(items)
# Output: [1, 3, 4]
Even better β use list comprehension: π
items = [x for x in items if x != 2]
Conclusion: π do not modify a collection during iteration. This can lead to skipped elements, duplication, or even errors during execution. π οΈπ§
#Python #Coding #Programming #Debugging #TechTips #PythonTipsif obj == None, use if obj is None
In Python, when you write:
obj == None
you're not directly checking if obj is the value None. Instead, you're asking if the object is equal to None.
Yes, in many cases, the result will be the same as for the code:
obj is None
But the behavior of these two variants is different, and this difference is important.
When you use:
obj == None
Python calls the __eq__ method on the object. That is, the object itself decides what it means to be "equal to None". And this method can be overridden.
If obj is an instance of a class in which __eq__ is implemented so that when compared with None, it returns True (even if the object is not actually None), then obj == None may mistakenly give True.
Example:
class Weird:
def __eq__(self, other):
return True # Always asserts that it's equal
obj = Weird()
print(obj == None) # True
print(obj is None) # False
Here, it can be seen that obj == None returns True due to the custom behaeqf the __eq__ operator in the class.
Therefore, when using obj == None, the result is not always predictable.
On the other hand, when you write:
obj is None
you're using the is operator, which cannot be overridden. This means that the result will always be the same and predictable.
The is operator checks the identity of objects, that is, whether two references point to the same object. Since None is a singleton (the only instance), obj is None is the correct and most efficient way to perform such a check.
β€οΈ Therefore, it is always recommended, and this is best practice, to use obj is None instead of obj == None for predictability and efficiency.
π https://t.me/DataScienceQdef first_duplicate(lst):
seen = set()
for x in lst:
if x in seen:
return x
seen.add(x)
return None
print(first_duplicate([3, 1, 3, 4, 2])) # Output: 3
2οΈβ£ Q: Check whether a number is a palindrome.
def is_pal_num(n):
return str(n) == str(n)[::-1]
print(is_pal_num(121)) # True
print(is_pal_num(123)) # False
3οΈβ£ Q: Sort a dictionary by values.
def sort_by_value(d):
return dict(sorted(d.items(), key=lambda x: x[1]))
print(sort_by_value({'a': 3, 'b': 1, 'c': 2}))
Output: {'b': 1, 'c': 2, 'a': 3}
4οΈβ£ Q: Return all prime numbers in a given range.
def primes_upto(n):
primes = []
for num in range(2, n + 1):
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
break
else:
primes.append(num)
return primes
print(primes_upto(10)) # [2, 3, 5, 7]
5οΈβ£ Q: Convert a list of numbers into a string.
def list_to_string(lst):
return "".join(map(str, lst))
print(list_to_string([1, 2, 3])) # Output: 123
π¬ Double Tap β€οΈ for Part-12
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