creating a 2D numpy array from the min of the i-th and j-th elements of a 1D array

Question

Given a 1D array, I would like to create a 2D numpy array where the element at arr2d[i,j] is equal to min(arr1d[i], arr1d[j]).

I can achieve this with the code below:

arr1d = np.array([5, 4, 3, 4, 2, 3])
arr2d = np.zeros((arr1d.shape[0], arr1d.shape[0]))
for i in range(arr1d.shape[0]):
    for j in range(arr1d.shape[0]):
        arr2d[i, j] = min(arr1d[i], arr1d[j])

...but I feel like there should be a more idiomatic and/or efficient way to achieve this with numpy. Can anyone suggest a cleaner solution?

Answer 1

Imagine you repeated arr1d to make a square array. Then, arr2d is the elementwise minimum of that square array and its transpose (since that will compare i with j).

a1 = np.repeat(arr1d[None,:], arr1d.shape[0], axis=0)
a2 = a1.T
arr2d = np.minimum(a1, a2)  # elementwise minimum

Here is what a1 looks like:

array([[5, 4, 3, 4, 2, 3],
       [5, 4, 3, 4, 2, 3],
       [5, 4, 3, 4, 2, 3],
       [5, 4, 3, 4, 2, 3],
       [5, 4, 3, 4, 2, 3],
       [5, 4, 3, 4, 2, 3]])

And here's arr2d:

array([[5, 4, 3, 4, 2, 3],
       [4, 4, 3, 4, 2, 3],
       [3, 3, 3, 3, 2, 3],
       [4, 4, 3, 4, 2, 3],
       [2, 2, 2, 2, 2, 2],
       [3, 3, 3, 3, 2, 3]])

Or, without the intermediate arrays (as suggested in the comments):

arr2d = np.minimum(arr1d, arr1d[:,None])