我目前正在用python实现差分进化算法,当在较小尺寸下工作时一切都很好,但是,当我开始增加搜索空间的尺寸时,运行该算法所需的时间呈指数增长。经过一些分析后,我发现大部分时间都花在了突变功能上,如下所示,
def _mutate(self, candidate: int) -> np.ndarray:
# r0, r1, & r2 are np.ndarrays of shape (dimension,)
r0, r1, r2 = self._select_samples(candidate)
# mutant is an np.ndarray of shape (dimension,)
mutant = np.copy(self.population[candidate])
j_rand = int(np.random.uniform() * self.dimensions)
for j in range(self.dimensions):
if np.random.uniform() < self.cr or j == j_rand:
# bound the mutant to the search space
mutant[j] = np.clip(r0[j] + self.F * (r1[j] - r2[j]),
self.range[0], self.range[1])
现在,对于population size
的{{1}}和100
的{{1}},运行算法所需的总时间约为40秒,其中约20秒dimension
中花费了几秒钟。
现在,我已经取消了对该功能的优化,以使其比以前的版本缩短了约3秒。
20
但是显然,这还不够。我想知道mutate
中是否有技巧可以删除在def _mutate_2(self, candidate: int) -> np.ndarray:
r0, r1, r2 = self._select_samples(candidate)
mutant = np.copy(self.population[candidate])
j_rand = np.random.randint(self.dimensions)
cross_indxs = np.flatnonzero(np.random.rand(self.dimensions) < self.cr)
cross_indxs = np.append(
cross_indxs, [j_rand]) if j_rand not in cross_indxs else cross_indxs
for j in cross_indxs:
mutant[j] = np.clip(r0[j] + self.F * (r1[j] - r2[j]), self.range[0],
self.range[1])
return mutant
上应用元素操作的for循环。要注意的是,只能使用索引在numpy
中的元素。
答案 0 :(得分:0)
尝试一下:
mutant[cross_indxs] = (r0[cross_indxs] + self.F[cross_indxs] * (r1[cross_indxs] - r2[cross_indxs])).clip(self.range[0],self.range[1])