Numpy set all indexes to a value

Question

I have a numpy ndarray a = np.ndarray((3,3)) and I want all of the indexes to start at the same value, e.g. 5:

array([[5., 5., 5.],
       [5., 5., 5.],
       [5., 5., 5.]])

Note: I'm posting this Q&A style because every time I look this up I always find a bunch of random questions about complex slicing, but not a simple example of casting everything at once. Hopefully this will pop up as a more immediate result next time I search this question. But I also hope that others have good ideas I can adopt.

I sense tomfoolery is afoot. You posted and answered your question within one second... — Reedinationer, Apr 04 '19 at 04:06
It's using the stackoverlow Q&A. I mentioned it in my note at the end of my question. — loganjones16, Apr 04 '19 at 04:09
Oh I see! I didn't know that was a feature...I thought you had like scripted a way to post a question and then immediately answer it. That's a neat feature :D — Reedinationer, Apr 04 '19 at 04:16

score 3 · Answer 1 · answered Apr 04 '19 at 04:05

Here are some solid ways:

# Use the function built for this very purpose
>>> a = np.full((3, 3), 5)
>>> a
array([[5., 5., 5.],
       [5., 5., 5.],
       [5., 5., 5.]])

or

# [:] is shorthand for every index.
>>> a = np.ndarray((3,3))
>>> a[:] = 5
>>> a
array([[5., 5., 5.],
       [5., 5., 5.],
       [5., 5., 5.]])

or

# multiply a single value over every index (currently all 1s)
>>> a = np.ones((3,3)) * 5   
>>> a
array([[5., 5., 5.],
       [5., 5., 5.],
       [5., 5., 5.]])

Check out the documentation on indexing for more details and examples of complex indexing/slicing

score 3 · Accepted Answer · answered Apr 04 '19 at 04:20

There are several alternatives to achieve this, but I think the point will be to analyze which will give you the most optimal results so:

In[1]: %timeit np.ones((3,3)) * 5
6.82 µs ± 374 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)    
In[2]: %%timeit
       np.ndarray((3,3))
       a[:] = 5
1.96 µs ± 29.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In[3]: %timeit np.full((3, 3), 5)
4.13 µs ± 59.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

So, probably the best way to do it is creating the array and assigning the value 5 to all elements, that means to use the second option.

U13-Forward · Answer 3 · 2019-04-04T04:20:38.203

2

I seem to have another way of doing this, just use a plus operator:

>>> import numpy as np
>>> a = np.zeros((3,3))
>>> a
array([[ 0.,  0.,  0.],
       [ 0.,  0.,  0.],
       [ 0.,  0.,  0.]])
>>> a + 5
array([[ 5.,  5.,  5.],
       [ 5.,  5.,  5.],
       [ 5.,  5.,  5.]])
>>>

(P.S using zeros instead of ndarray)

edited Apr 04 '19 at 04:20

answered Apr 04 '19 at 04:10

U13-Forward

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Nice, so it's the same idea as np.ones() * 5 – loganjones16 Apr 04 '19 at 04:11
@loganjones16 Similar, just with plus operator – U13-Forward Apr 04 '19 at 04:11

Numpy set all indexes to a value

3 Answers3