이것이 고급 인덱싱을 고려하는 것이 얼마나 효율적인지 확실하지는 않지만 그렇게하는 한 가지 방법입니다.
import numpy as np
a = np.array([8, 18, 5, 15, 12])
b = a[:, None] - a
# Fill lower triangle with largest negative
b[np.tril_indices(len(a))] = np.iinfo(b.dtype).min # np.finfo for float
# Put diagonals as rows
s = 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 zero
c = np.r_[0, diags.max(1)]
print(c)
# [ 0 13 3 6 -4]
편집하다:
당신이 찾고있는 것이 아닐 수도있는 다른 대안은 다음과 같이 Numba를 사용하는 것입니다.
import numpy as np
import numba as nb
def 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]
이 방법들을 원본과 비교하기 :
import numpy as np
import numba as nb
def 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 out
def 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())
# True
print((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)
첫 번째 배열은 작은 배열에서는 조금 나을 수 있지만 큰 배열에서는 잘 작동하지 않습니다. 반면에 Numba는 모든 경우에 매우 좋습니다.
a?