更新时间:2023-11-11 09:19:28
这样的事情有用吗?
tot_length=200
steps=0.1
list_no = np.arange(0.0, tot_length, steps)
x, y, z = np.meshgrid(*[list_no for _ in range(3)], sparse=True)
a = ((x>=y) & (y>=z)).nonzero()
这仍然会为中间布尔数组使用 8GB 的内存,但避免重复调用 np.append
,因为它们很慢.
This will still use 8GB of memory for the intermediate array of booleans, but avoids repeated calls to np.append
which are slow.