Creating a numpy array from an integer

Question

I created a multiplication table from an integer input using numpy. I created a list from the integer, and created an array with the shape of the integer, but it seems I'm ending up doing things the non-numpy way. Is there a more numpionic way of doing this?

TY

def multiplication_table(n):
    import numpy as np
    base = list(range(1, n+1))
    array_shell = np.zeros(shape=(n,n), dtype=int)
    array_shell[0] = base
    for row in range(1,len(base)):
        array_shell[row][0] = base[row]
    for row in range(1, len(base)):
        for idx in range(1, len(base)):
            array_shell[row][idx] = base[idx]*array_shell[row][0]
    return (array_shell)

my_int = 8
print(multiplication_table(my_int))

Divakar · Accepted Answer · 2016-03-01 13:02:40Z

2

Here's a vectorized approach that uses NumPy's powerful broadcasting feature -

def multiplication_table_vectorized(n):
    base = np.arange(n)+1
    return base[:,None]*base

Runtime test -

In [33]: n = 100

In [34]: np.allclose(multiplication_table(n),multiplication_table_vectorized(n))
Out[34]: True

In [35]: %timeit multiplication_table(n)
100 loops, best of 3: 10.1 ms per loop

In [36]: %timeit multiplication_table_vectorized(n)
10000 loops, best of 3: 58.9 µs per loop

Explanation -

Let's take a toy example for explaining things here.

In [72]: n = 4 # Small n for toy example

In [73]: base = np.arange(n)+1 # Same as original: "base = list(range(1, n+1))"

In [74]: base                  # Checkback
Out[74]: array([1, 2, 3, 4])

In [75]: base[:,None]   # Major thing happening as we extend base to a 2D array
                        # with all elements "pushed" as rows (axis=0) and thus
                        # creating a singleton dimension along columns (axis=1)
Out[75]: 
array([[1],
       [2],
       [3],
       [4]])

In [76]: base[:,None]*base    # Broadcasting happens as elementwise multiplications
                              # take place between 2D extended version of 'base' 
                              # and original 'base'. This is our desired output.

                              # To visualize a broadcasting :
                              # |--------->
                              # |
                              # |
                              # |
                              # V

Out[76]: 
array([[ 1,  2,  3,  4],
       [ 2,  4,  6,  8],
       [ 3,  6,  9, 12],
       [ 4,  8, 12, 16]])

For more information and examples on broadcasting, there's nothing better than the official docs. Broadcasting is one of the best vectorization tools around available with NumPy that allows for such automatic expansions.

edited Mar 1, 2016 at 13:02

answered Mar 1, 2016 at 3:38

Divakar

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4 Comments

TASC Solutions Over a year ago

Thanks, that is exactly what I was looking for. Could you please explain how the return makes it work? I understand making the base an ndarray, but I'm confused on how the return/math works to populate.

Divakar Over a year ago

@TASCSolutions See if the added comments help to understand what's going on in there.

TASC Solutions Over a year ago

Yes I see clearly now: I needed to arange my integer so numpy can use it and the base[:,None] replaced my for loop to get the vertical row and then broadcasting and basic math did the rest.

Divakar Over a year ago

@TASCSolutions You got it! Good job on it.

maxymoo · Accepted Answer · 2016-03-01 03:40:04Z

0

One "numpyonic" way to approach this would be as a matrix multiplication (of a column vector with a row vector):

base = np.arange(my_int)
array = base.reshape((my_int,1)) * base

answered Mar 1, 2016 at 3:40

maxymoo

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2 Comments

TASC Solutions Over a year ago

Thanks, it appears I need to take another numpy tutorial, these solutions are much faster and cleaner than my hybrid approach.

maxymoo Over a year ago

don't worry if you don't get it straight away, it took me a very long time for this way of thinking to "click" for me

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