βοΈ Tricky Python Interview Question
> What will this code output and why?
β Question: Why does this code print 2, 2, 2 instead of 0, 1, 2?
π Explanation:
This is a classic example of late binding in Python closures. The lambda functions don't capture the value of
π§ What happens:
β’ The
β’ The
β’ When you call each function
π Result:
2
2
2
β How to fix:
You can force the lambda to capture the value of
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 β¨
> What will this code output and why?
funcs = []
for i in range(3):
funcs.append(lambda: print(i))
for f in funcs:
f()
β Question: Why does this code print 2, 2, 2 instead of 0, 1, 2?
π 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.π§ 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.π Result:
2
2
2
β 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 β¨
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Answer:
After that, the template forms an HTML response based on the provided data, and Django sends it back to the client.
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Interview question
What will this code output and why?
Try to answer yourself first, then check the explanation.
Step-by-step:
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:
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:
If you want, I can explain another tricky question β about mutable default arguments in functions or about loops and closures.
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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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Answer:
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What is ORM and what is SQLAlchemy used for?
Answer:
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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βοΈ DON'T CREATE NESTED LISTS WITH THE
Because of this, you create a list containing multiple references to the very same inner list.
It looks like you're making a grid, but modifying one row will surprisingly change all of them. This is because the outer list just holds copies of the reference, not copies of the list itself.
Correct β use a list comprehension to ensure each inner list is a new, independent object.
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By: @DataScienceQ β¨
* OPERATOR.Because of this, you create a list containing multiple references to the very same inner list.
It looks like you're making a grid, but modifying one row will surprisingly change all of them. This is because the outer list just holds copies of the reference, not copies of the list itself.
Correct β use a list comprehension to ensure each inner list is a new, independent object.
Subscribe for more Python secrets!
# hidden error β all inner lists are the same object
matrix = [[]] * 3 # seems to create a 3x0 matrix
# append to the first row
matrix[0].append(99)
# all rows were modified!
print(matrix) # [[99], [99], [99]]
# β correct version β use a list comprehension
matrix_fixed = [[] for _ in range(3)]
# append to the first row
matrix_fixed[0].append(99)
# only the first row is modified, as expected
print(matrix_fixed) # [[99], [], []]
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Checking Memory Usage in Python
psutil is imported, then through psutil.virtual_memory() memory data is obtained.
The function convert_bytes converts bytes to gigabytes.
Then the code calculates:
- total RAM
- available RAM
- used RAM
- percentage usage
And outputs this to the console.
Or simply press CTRL + ALT + DELETE and open Task Manager. It has worked since the days of Windows 95.
The RAM usage percentage loses its meaning if you have Chrome open β it will consume everything on its own π
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psutil is imported, then through psutil.virtual_memory() memory data is obtained.
The function convert_bytes converts bytes to gigabytes.
Then the code calculates:
- total RAM
- available RAM
- used RAM
- percentage usage
And outputs this to the console.
import psutil
memory = psutil.virtual_memory()
def convert_bytes(size):
# Convert bytes to GB
gb = size / (1024 ** 3)
return gb
total_gb = convert_bytes(memory.total)
available_gb = convert_bytes(memory.available)
used_gb = convert_bytes(memory.used)
print(f"Total RAM: {total_gb:.3f} GB")
print(f"Available RAM: {available_gb:.3f} GB")
print(f"Used RAM: {used_gb:.3f} GB")
print(f"RAM Usage: {memory.percent}%")
The RAM usage percentage loses its meaning if you have Chrome open β it will consume everything on its own
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