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Discussion on: TIL: Python Comprehensions Make Matrices Work

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rhymes profile image
rhymes • Edited

Hi Ry! I laughed reading this so it's a good start :-)

What you're experiencing is a combination of how Python syntax works and the fact that lists are mutable objects.

In the first example:

sleeve = [0] * 5
shirt = [sleeve] * 5

you're telling Python: "make sleeve a list of five numbers and then make shirt a list of five sleeves". So Python is going to copy the object pointed by the variable sleeve inside shirt five time, but it's the same object. You can easily see it like this:

>>> sleeve = [0] * 5
>>> shirt = [sleeve] * 5
>>> shirt
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> [print(id(l)) for l in shirt]
4384048520
4384048520
4384048520
4384048520
4384048520

id() is a function that returns the memory identity (address) of the object. As you can see, it's all the same object

The second example is basically the same, you're telling Python to multiply by five times the same list:

>>> shirt = [[0] * 5] * 5
>>> [print(id(l)) for l in shirt]
4387549256
4387549256
4387549256
4387549256
4387549256

In the third example you're actually building a new list for each iteration in the list comprehension, so it works:

>>> shirt = [[0 for i in range(5)] for j in range(5)]
>>> [print(id(l)) for l in shirt]
4387549896
4387550472
4387550344
4387549704
4387205896

:-)

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rsdesoto profile image
Ry

! This makes significantly more sense -- thank you!

Why does the first [0]*5 not run into the same issues? Is it because this is updating a single list?

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rhymes profile image
rhymes

Why does the first [0]*5 not run into the same issues? Is it because this is updating a single list?

That's because Python cheats a little bit. I don't know the exact list of optimizations but you're creating a list of integers so I think it reuses the same object in memory. An integer is an immutable object, it detects you're creating a list of 5 integers, so it just fills up a list of 5 zeroes. You can see what's going on:

>>> s
[0, 0, 0, 0, 0]
>>> [id(x) for x in s]
[4501366592, 4501366592, 4501366592, 4501366592, 4501366592]

they all are the same object in memory but since you can't change it, it's fine.

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rsdesoto profile image
Ry

Ooooohh, that makes sense -- thank you!!