Numpy: Transform list into square array

Question

I have a list t with a square number of elements, e.g. 16, 25, 400. Now want to create a numpy array a full of zeros with the same size but in a square shape, e.g. 4x4, 5x5, 20x20.

I found a solution:

a = np.zeros(2 * (int(np.sqrt(len(t))),))

or

a = np.zeros(len(t)).reshape(int(np.sqrt(len(t))), int(np.sqrt(len(t))))

It is working, but it is very ugly and I am sure there must be a better way to do this. Something like a = np.zeros(t, 2). Any idea for that?

Thank you! :)

Your first suggestion is the shortest you can make it. If it's unreadable, extract a square_shape(length) function, and use np.zeros(square_shape(len(l))) — Eric
– Eric, Commented Feb 22, 2019 at 18:03

sacuL · Accepted Answer · 2019-02-22 16:40:38Z

1

You can try:

shp = int(np.sqrt(len(l))
a = np.zeros((shp, shp))

answered Feb 22, 2019 at 16:40

sacuL

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Comments

liamhawkins · Accepted Answer · 2019-02-22 16:42:08Z

1

You can clean it up like so:

size = len(l)
sqrt = int(np.sqrt(size))
a = np.zeros((sqrt, sqrt))

Any time you are writing the same fragment of code multiple times, it is good to replace with a variable, function etc.

answered Feb 22, 2019 at 16:42

liamhawkins

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1 Comment

Max16hr Over a year ago

This is what i tried in the first attempt. But it is so bad readable. And so many variables for such a small task, I also do not like^^

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