PyData Careers
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Python Data Science jobs, interview tips, and career insights for aspiring professionals.
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❔ Interview Question

What does nonlocal do and where can it be used?

Answer: nonlocal allows you to modify a variable from the nearest enclosing function without creating a new local one. It only works inside a nested function when you need to change a variable declared in the outer, but not global, scope.

This is often used in closures to maintain and update state between calls to the nested function.


tags: #interview

➑ @DataScienceQ
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🐍 Tricky Python Interview Question

> 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?

πŸ” Explanation:

Default arguments in Python are evaluated once β€” 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) β†’ uses the shared list [], now it is [10]

- extend_list(123, []) β†’ creates a new list [123]

- extend_list('a') β†’ again uses the shared list β†’ [10, 'a']

πŸ‘‰ Result:

[10, 'a'] [123] [10, 'a']

βœ… 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 β€” mutable default arguments.

It tests if you understand how default values and memory scope work.

https://t.iss.one/DataScienceQ ⭐️
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⛓️ Tricky Python Interview Question

> What will this code output and why?

funcs = []
for i in range(3):
funcs.append(lambda: print(i))

for f in funcs:
f()


❓ Question: Why does this code print 2, 2, 2 instead of 0, 1, 2?

πŸ” Explanation:

This is a classic example of late binding in Python closures. The lambda functions don't capture the value of i at each step of the loop. Instead, they all hold a reference to the same variable i.

🧠 What happens:

β€’ The for loop completes. By the end of the loop, the variable i holds the value 2.
β€’ The funcs list contains three functions, but each one is defined to print the value of whatever i is at the time of execution.
β€’ When you call each function f(), it looks up the current value of i in its enclosing scope, which is 2.

πŸ‘‰ Result:

2
2
2

βœ… How to fix:

You can force the lambda to capture the value of i at the time of definition by using a default argument.

funcs = []
for i in range(3):
# The default argument captures i's value at definition time
funcs.append(lambda num=i: print(num))

for f in funcs:
f()

This trick works because default arguments are evaluated when the function is defined, not when it's called. This will correctly print:

0
1
2

This is a common interview question that tests your understanding of closures, scope, and late binding in Python.

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By: @DataScienceQ ✨
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❔ Interview question

How does Django handle an HTTP request?

Answer: When Django receives an HTTP request, it goes through several stages. First, the request enters the routing system (URLconf), where the appropriate view is selected by the address. Then a function or class-based view is called, which executes the logic: it may access models, prepare data, and select the appropriate template.

After that, the template forms an HTML response based on the provided data, and Django sends it back to the client.

This is how Django organizes work following the MVT pattern: URL β†’ view β†’ logic and data β†’ template β†’ HTTP response.


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➑ @DataScienceQ
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❔ Interview Question

How does the module import mechanism work in Python and what is sys.path?

Answer: When importing a module, Python searches for it in the directories listed in sys.path. This list includes the current directory, standard Python installation paths, and manually added paths. If the module is not found, a ModuleNotFoundError is raised.

tags: #interview

➑️ @DataScienceQ
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PyData Careers
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Interview question

What will this code output and why?


a = [1, 2, 3]
b = a

a += [4, 5]

print(a)
print(b)

a = (1, 2, 3)
b = a

a += (4, 5)

print(a)
print(b)


Try to answer yourself first, then check the explanation.


Part 1. Lists
a = [1, 2, 3]
b = a

a += [4, 5]

print(a)
print(b)


Step-by-step:

a = [1, 2, 3]


Variable a refers to the list [1, 2, 3].

b = a
Now b refers to the same list as a.
Not a copy, but the very same object in memory.

a += [4, 5]
Important: for lists, += works as an in-place modification:
this is roughly the same as:

a.extend([4, 5])


That is, elements 4 and 5 are added to the existing list that both a and b refer to.

Therefore:

print(a)  # [1, 2, 3, 4, 5]
print(b)  # [1, 2, 3, 4, 5]


Both point to the same modified list.

Part 2. Tuples
a = (1, 2, 3)
b = a

a += (4, 5)

print(a)
print(b)


Tuples are immutable, and here the magic begins.

a = (1, 2, 3)
a refers to the tuple (1, 2, 3).

b = a
b refers to the same tuple (1, 2, 3).

a += (4, 5)
For tuples, += cannot modify the object in place (they are immutable).
So Python does this:

a = a + (4, 5)


That is, a new tuple (1, 2, 3, 4, 5) is created and variable a is reassigned to it.

b still points to the old tuple (1, 2, 3).

Therefore:

print(a)  # (1, 2, 3, 4, 5)
print(b)  # (1, 2, 3)

The trick

For the list, a += [4, 5] mutates the object in place, visible through all variables referencing it.

For the tuple, a += (4, 5) creates a new object and reassigns only a.

In summary:


# List part:
[1, 2, 3, 4, 5]
[1, 2, 3, 4, 5]

# Tuple part:
(1, 2, 3, 4, 5)
(1, 2, 3)


If you want, I can explain another tricky question β€” about mutable default arguments in functions or about loops and closures.

https://t.iss.one/DataScienceQ
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❔ Interview Question

What are optimistic and pessimistic locking in the context of databases?

Answer: These are two approaches to managing concurrent data access.

Pessimistic locking assumes conflicts are likely. Therefore, data is locked immediately upon reading or writing and remains locked until the end of the transaction. This prevents concurrent modifications but reduces scalability and can lead to deadlocks.

Optimistic locking assumes conflicts are rare. Data is read without locking, and before committing changes, a version check is performed to see if someone else has modified the data. If so, the transaction is rolled back and retried. This approach offers better performance under low contention.


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➑ @DataScienceQ
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❔ Interview question

What is ORM and what is SQLAlchemy used for?

Answer: ORM (Object-Relational Mapping) allows interacting with the database through Python classes and objects instead of writing SQL queries manually.

With SQLAlchemy, you can describe tables as classes, rows as objects, and perform SELECT, INSERT, UPDATE, DELETE operations through Python methods.

This simplifies working with databases, makes the code more readable, reduces the risk of SQL injections, and facilitates maintenance and migrations.


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➑ @DataScienceQ
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❔ Interview Question

What is the difference between pass, continue, and break?

Answer:

▢️ pass β€” this is a no-op, which does nothing. It is used as a placeholder when syntax requires the presence of code (for example, inside a function, class, or condition), but the logic is not yet implemented

▢️ continue β€” terminates the current iteration of the loop and moves to the next, skipping the remaining code in the loop body

▢️ break β€” completely terminates the execution of the loop, exiting it prematurely, regardless of the condition

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➑ @DataScienceQ
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Python Data Science jobs, interview tips, and career insights for aspiring professionals.
https://t.iss.one/DataScienceQ

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Admin: @HusseinSheikho
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❔ Interview Question

How does the map() function work?

Answer:map() takes a function and an iterable object, and returns an iterator that sequentially yields the result of applying this function to each element. The map() itself does not create a list β€” it lazily forms values on demand. If a list is needed, the result can be wrapped in list().

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➑ @DataScienceQ
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