Max value per diagonal in 2d array

Use ndarray.diagonal

v = [max(c.diagonal(-i)) for i in range(b.shape[0])]print(v) # [0, 13, 3, 6, -4]

Not sure exactly how efficient this is considering the advanced indexing involved, but this is one way to do that:

import numpy as npa = np.array([8, 18, 5, 15, 12])b = a[:, None] - a# Fill lower triangle with largest negativeb[np.tril_indices(len(a))] = np.iinfo(b.dtype).min  # np.finfo for float# Put diagonals as rowss = b.strides[1]diags = np.ndarray((len(a) - 1, len(a) - 1), b.dtype, b, offset=s, strides=(s, (len(a) + 1) * s))# Get maximum from each row and add initial zeroc = np.r_[0, diags.max(1)]print(c)# [ 0 13  3  6 -4]

EDIT:

Another alternative, which may not be what you were looking for though, is just using Numba, for example like this:

import numpy as npimport numba as nbdef max_window_diffs_jdehesa(a):    a = np.asarray(a)    dtinf = np.iinfo(b.dtype) if np.issubdtype(b.dtype, np.integer) else np.finfo(b.dtype)    out = np.full_like(a, dtinf.min)    _pwise_diffs(a, out)    return out@nb.njit(parallel=True)def _pwise_diffs(a, out):    out[0] = 0    for w in nb.prange(1, len(a)):        for i in range(len(a) - w):            out[w] = max(a[i] - a[i + w], out[w])a = np.array([8, 18, 5, 15, 12])print(max_window_diffs(a))# [ 0 13  3  6 -4]

Comparing these methods to the original:

import numpy as npimport numba as nbdef max_window_diffs_orig(a):    a = np.asarray(a)    b = a - a[:, None]    out = np.zeros(len(a), b.dtype)    out[-1] = b[-1, 0]    for i in range(1, len(a) - 1):        out[i] = np.diag(b, -i).max()    return outdef max_window_diffs_jdehesa_np(a):    a = np.asarray(a)    b = a[:, None] - a    dtinf = np.iinfo(b.dtype) if np.issubdtype(b.dtype, np.integer) else np.finfo(b.dtype)    b[np.tril_indices(len(a))] = dtinf.min    s = b.strides[1]    diags = np.ndarray((len(a) - 1, len(a) - 1), b.dtype, b, offset=s, strides=(s, (len(a) + 1) * s))    return np.concatenate([[0], diags.max(1)])def max_window_diffs_jdehesa_nb(a):    a = np.asarray(a)    dtinf = np.iinfo(b.dtype) if np.issubdtype(b.dtype, np.integer) else np.finfo(b.dtype)    out = np.full_like(a, dtinf.min)    _pwise_diffs(a, out)    return out@nb.njit(parallel=True)def _pwise_diffs(a, out):    out[0] = 0    for w in nb.prange(1, len(a)):        for i in range(len(a) - w):            out[w] = max(a[i] - a[i + w], out[w])np.random.seed(0)a = np.random.randint(0, 100, size=100)r = max_window_diffs_orig(a)print((max_window_diffs_jdehesa_np(a) == r).all())# Trueprint((max_window_diffs_jdehesa_nb(a) == r).all())# True%timeit max_window_diffs_orig(a)# 348 µs ± 986 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)%timeit max_window_diffs_jdehesa_np(a)# 91.7 µs ± 1.3 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)%timeit max_window_diffs_jdehesa_nb(a)# 19.7 µs ± 88.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)np.random.seed(0)a = np.random.randint(0, 100, size=10000)%timeit max_window_diffs_orig(a)# 651 ms ± 26 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)%timeit max_window_diffs_jdehesa_np(a)# 1.61 s ± 6.19 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)%timeit max_window_diffs_jdehesa_nb(a)# 22 ms ± 967 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

The first one may be a bit better for smaller arrays, but doesn't work well for bigger ones. Numba on the other hand is pretty good in all cases.

You can use numpy.diagonal:

a = np.array([8, 18, 5,15,12])b = a - a[:, None]c = np.tril(b)for i in range(b.shape[0]):    print(max(c.diagonal(-i)))

Output:

01336-4