如果您的数组很小,那么使用itertools.combinations 的Samy's pure-Python solution 应该没问题:
from itertools import combinations, chain
def all_products1(a):
p = (x * y for x, y in combinations(a, 2))
return list(chain(a, p))
但是,如果您的数组很大,那么您将通过使用 numpy.triu_indices 对计算进行完全矢量化来获得显着的加速,如下所示:
import numpy as np
def all_products2(a):
x, y = np.triu_indices(len(a), 1)
return np.r_[a, a[x] * a[y]]
让我们比较一下:
>>> data = np.random.uniform(0, 100, (10000,))
>>> timeit(lambda:all_products1(data), number=1)
53.745754408999346
>>> timeit(lambda:all_products2(data), number=1)
12.26144006299728
使用numpy.triu_indices的解决方案也适用于多维数据:
>>> np.random.uniform(0, 100, (3,2))
array([[ 63.75071196, 15.19461254],
[ 94.33972762, 50.76916376],
[ 88.24056878, 90.36136808]])
>>> all_products2(_)
array([[ 63.75071196, 15.19461254],
[ 94.33972762, 50.76916376],
[ 88.24056878, 90.36136808],
[ 6014.22480172, 771.41777239],
[ 5625.39908354, 1373.00597677],
[ 8324.59122432, 4587.57109368]])
如果要对列而不是行进行操作,请使用:
def all_products3(a):
x, y = np.triu_indices(a.shape[1], 1)
return np.c_[a, a[:,x] * a[:,y]]
例如:
>>> np.random.uniform(0, 100, (2,3))
array([[ 33.0062385 , 28.17575024, 20.42504351],
[ 40.84235995, 61.12417428, 58.74835028]])
>>> all_products3(_)
array([[ 33.0062385 , 28.17575024, 20.42504351, 929.97553238,
674.15385734, 575.4909246 ],
[ 40.84235995, 61.12417428, 58.74835028, 2496.45552756,
2399.42126888, 3590.94440122]])