The term broadcasting describes how NumPy treats arrays with different shapes during arithmetic operations. Subject to certain constraints, the smaller array is “broadcast” across the larger array so that they have compatible shapes.

Broadcasting provides a means of vectorizing array operations so that looping occurs in C instead of Python. It does this without making needless copies of data and usually leads to efficient algorithm implementations. There are, however, cases where broadcasting is a bad idea because it leads to inefficient use of memory that slows computation.

In the following multiplication operation between a (3,1) array and ascalar, numpy broadcasts copies of the scalar b across the required dimension so that the operation can be done as a pair of arrays on an element-by-element basis.

import numpy as np

a = np.array([1.0, 2.0, 3.0])
b = 2.0
a * b
array([2., 4., 6.])

The result of the above is the equivalent of

a = np.array([1.0, 2.0, 3.0])
b = np.array([2.0, 2.0, 2.0])
a * b
>>> array([ 2.,  4.,  6.])

The new elements in b, as shown in Figure below, are simply copies of the original scalar.In fact NumPy uses the original scalar value without actually making copies so that broadcasting operations are as memory and computationally efficient as possible.

Image source:

The code in the first example is more efficient than that in the first because broadcasting moves less memory around during the multiplication (b is a scalar rather than an array).

General Broadcasting Rules

  1. When operating on two arrays, NumPy compares their shapes element-wise. It starts with the trailing (i.e. rightmost) dimensions and works its way left. Two dimensions are compatible when
    • they are equal, or
    • one of them is 1
      If these conditions are not met, a ValueError: operands could not be broadcast together exception is thrown, indicating that the arrays have incompatible shapes. The size of the resulting array is the size that is not 1 along each axis of the inputs.
  2. Arrays do not need to have the same number of dimensions. For example, if you have a 256x256x3 array of RGB values, and you want to scale each color in the image by a different value, you can multiply the image by a one-dimensional array with 3 values. Lining up the sizes of the trailing axes of these arrays according to the broadcast rules, shows that they are compatible:
    Image  (3d array): 256 x 256 x 3
    Scale  (1d array):             3
    Result (3d array): 256 x 256 x 3

    When either of the dimensions compared is one, the other is used. In other words, dimensions with size 1 are stretched or “copied” to match the other.

  3. A set of arrays is called “broadcastable” to the same shape if the above rules produce a valid result.
    For example, if a.shape is (5,1), b.shape is (1,6), c.shape is (6,) and d.shape is () so that d is a scalar, then a, b, c, and d are all broadcastable to dimension (5,6); and
    • a acts like a (5,6) array where a[:,0] is broadcast to the other columns,
    • b acts like a (5,6) array where b[0,:] is broadcast to the other rows,
    • c acts like a (1,6) array and therefore like a (5,6) array where c[:] is broadcast to every row, and finally,
    • d acts like a (5,6) array where the single value is repeated.


Image source:

Application: Euclidean Distance Matrix

To calculate the euclidean distance matrix in a dataset (m observations, n variables), we would have to iterate over the the rows/observations of the data matrix so that a value is calculated all possible combinations of observations based on the following equation:

\[{\displaystyle {\begin{aligned}d(\mathbf {p} ,\mathbf {q}) &={\sqrt {\sum _{i=1}^{n}(q_{i}-p_{i})^{2}}}.\end{aligned}}}\]

We will compare for-loop and a mattrix multiplication operation in terms of speed.

from sklearn.datasets import load_iris

data, _ = load_iris(return_X_y=True)

In our data matrix A, as a first step of the euclideal matrix calculation we would like to

  • Subtract each $m_{i}$ row with every other $m_{j}$ row (data matrix A in figure below)
  • Calculate the square root of the cube each pair subtraction pair
  • Calculate the sum across the dimension of variables.

For-loop calculation

def eucl_loop(A):
    result = np.zeros((data.shape[0],data.shape[0]), dtype = int)
    for i,_ in enumerate(data):
        for j,_ in enumerate(data): 
            result[i][j] = np.sqrt(np.sum((data[i,:]-data[j,:])**2))
    return result

Matrix multiplication calculation

To do the above as a matrix operation with the help of broadcasting we are going to

  1. Generate a $D = (m_{i} \times (m_{j} \times n))$ matrix where $(m_{j} b. times n)$ repsresents a slice “that belongs” to datapoint $m_{i}$ and includes all other datapoints. Subtracting $m_i$ from “their” slice will result in a slice consisting of all subtraction operations of $m_i$ with every other $m_j$ datapoint across all variables. To do the above we will
    • add an additional dummie dimension to $A$ data matrix and rotate it so that $m \times 1$ view is exposed and store it as $B$.
    • rotate $B$ so that $1 \times m$ view is exposed.
    • Given we have a broadcastable pairs matrices because $(1 \times m)$$(m \times 1)$ the subtruct operation is going to strech across $m$ generating the desired $D$ matrix. Screenshot 2022-01-13 at 14.39.46.png
  2. Calculate the cube of all elements of $D$.
  3. Sum across the $n$ dimension of the matrix and calculate the square root. That is the Euclidean distance matrix $E$. Screenshot 2022-01-13 at 14.39.57.png

The above will result the matrix $E$, as seen in the figurbelow. The distance matrix is going to be a symmetric quadratic matrix of size $(m \times m)$ which contains all zeroes along the main diagonal (the distance of each object and a replica of itself is zero).

image source:

Performing A to B calculation above:

 # Adding dummie dimension
B = data[:,:, None]

B = np.rot90(B, 1, (1,2))
(150, 4, 1)
(150, 1, 4)

Performing the B to C calculation from diagram above:

C = np.rot90(B,1,(0,1))
(1, 150, 4)

The dimensions of our matrices appear to be broadcastable. We can proceed with the subtraction operation. See how the first row of the first slice of D matrix is zeroes, because the first datapoint is subtracted to itself. Same applies for the second row of the second slice and so forth.

D = B - C
[[ 0.   0.   0.   0. ]
 [ 0.2  0.5  0.   0. ]
 [ 0.4  0.3  0.1  0. ]
 [ 0.5  0.4 -0.1  0. ]
 [ 0.1 -0.1  0.   0. ]]
[[-0.2 -0.5  0.   0. ]
 [ 0.   0.   0.   0. ]
 [ 0.2 -0.2  0.1  0. ]
 [ 0.3 -0.1 -0.1  0. ]
 [-0.1 -0.6  0.   0. ]]

Calculate finally the Euclidean distance perfor ing final operations.

Euc_mat = np.sqrt(np.sum(D**2, axis=2))
(150, 150)

The above can be summarised in the following function:

def eucl_matmul(A):
    # generate distance matrix:
    B = np.rot90(A[:,:,None],1,(1,2)) #[:,:,None] is needed to add a dimension
    C = np.rot90(B,1,(0,1))
    return np.sqrt(np.sum((B-C)**2, axis=2))

Why though?

Let’s compare runtimes between the for-loop and matmul methodologies.

%timeit A = eucl_matmul(data) 
%timeit B = eucl_loop(data)
1000 loops, best of 5: 1 ms per loop
1 loop, best of 5: 221 ms per loop