What does it mean that a QuerySet in Django is "lazy"?
Answer:
The actual database access happens only when the results are really needed: when iterating over the QuerySet, calling list(), count(), first(), exists(), and other methods that require data.
This approach helps avoid unnecessary database hits and improves performance — queries are executed only at the moment of real necessity.
tags: #interview
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What is the difference between calling
start() and run() on threading.Thread?Answer:
If you call run() directly, it will execute in the current thread like a normal function — without creating a new thread and without parallelism.
This is the key difference: start() launches a separate execution thread, while run() just runs the code in the same thread.
tags: #interview
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What does
nonlocal do and where can it be used?Answer:
This is often used in closures to maintain and update state between calls to the nested function.
tags: #interview
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🐍 Tricky Python Interview Question
> What will this code output and why?
❓Question: Why are list1 and list3 the same?
🔍 Explanation:
Default arguments in Python are evaluated once — at function definition, not at each call.
So lst=[] is created once and preserved between calls if you don't explicitly pass your own list.
🧠 What happens:
- extend_list(10) → uses the shared list [], now it is [10]
- extend_list(123, []) → creates a new list [123]
- extend_list('a') → again uses the shared list → [10, 'a']
👉 Result:
✅ How to fix:
If you want a new list created by default on each call, do this:
This is a classic Python interview trap — mutable default arguments.
It tests if you understand how default values and memory scope work.
https://t.iss.one/DataScienceQ⭐️
> What will this code output and why?
def extend_list(val, lst=[]):
lst.append(val)
return lst
list1 = extend_list(10)
list2 = extend_list(123, [])
list3 = extend_list('a')
print(list1, list2, list3)
❓Question: Why are list1 and list3 the same?
🔍 Explanation:
Default arguments in Python are evaluated once — at function definition, not at each call.
So lst=[] is created once and preserved between calls if you don't explicitly pass your own list.
🧠 What happens:
- extend_list(10) → uses the shared list [], now it is [10]
- extend_list(123, []) → creates a new list [123]
- extend_list('a') → again uses the shared list → [10, 'a']
👉 Result:
[10, 'a'] [123] [10, 'a']✅ How to fix:
If you want a new list created by default on each call, do this:
def extend_list(val, lst=None):
if lst is None:
lst = []
lst.append(val)
return lst
This is a classic Python interview trap — mutable default arguments.
It tests if you understand how default values and memory scope work.
https://t.iss.one/DataScienceQ
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