我目前正在python中实现差分进化算法，并且在较低维度上工作时一切都很好，但是，当我开始增加搜索空间的维度时，运行算法所需的时间呈指数增长。做了一点profiling后发现大部分时间都花在了mutation函数上，具体如下，
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 apopulation size和100a dimension，20运行算法的总时间约为 40 秒，其中约 20 秒花费在mutate.
现在，我已经对这个功能进行了优化，使其比以前的版本缩短了大约 3 秒。
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
但显然，这还不够。我想知道是否有一个技巧numpy可以删除 for 循环在r0, r1, r2, and mutant. 问题是只能cross_indxs使用索引为 in 的元素。