Calculating cumulative minimum with numpy arrays

For any 2-argument NumPy universal function, its accumulate method is the cumulative version of that function. Thus, numpy.minimum.accumulate is what you're looking for:

>>> numpy.minimum.accumulate([5,4,6,10,3])
array([5, 4, 4, 4, 3])

Create a matrix which lower triangle (np.tril) is filled with values of your array nums and your upper triangle (np.triu, with second parameter 1, so the diagonal stays free) is filled with the maximum of the array. (EDIT: instead of the maximum, the first element of the array is the better way. -> comments)

nums = np.array([5.,3.,4.,2.,1.,1.,2.,0.])
oneSquare = np.ones((nums.size, nums.size))

A = nums * np.tril(oneSquare)
B = np.triu(oneSquare, 1) * nums[0]
A, B

Out:

(array([[ 5.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 5.,  3.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 5.,  3.,  4.,  0.,  0.,  0.,  0.,  0.],
       [ 5.,  3.,  4.,  2.,  0.,  0.,  0.,  0.],
       [ 5.,  3.,  4.,  2.,  1.,  0.,  0.,  0.],
       [ 5.,  3.,  4.,  2.,  1.,  1.,  0.,  0.],
       [ 5.,  3.,  4.,  2.,  1.,  1.,  2.,  0.],
       [ 5.,  3.,  4.,  2.,  1.,  1.,  2.,  0.]]),
 array([[ 0.,  5.,  5.,  5.,  5.,  5.,  5.,  5.],
       [ 0.,  0.,  5.,  5.,  5.,  5.,  5.,  5.],
       [ 0.,  0.,  0.,  5.,  5.,  5.,  5.,  5.],
       [ 0.,  0.,  0.,  0.,  5.,  5.,  5.,  5.],
       [ 0.,  0.,  0.,  0.,  0.,  5.,  5.,  5.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  5.,  5.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  5.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]]))

Now take the minimum of each row:

(A+B).min(axis=1)

Out:

array([ 5.,  3.,  3.,  2.,  1.,  1.,  1.,  0.])