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.
https://t.iss.one/DataScienceQ
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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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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