左循环 numpy 数组的最快方法(如弹出、推送队列) [英] Fastest way to left-cycle a numpy array (like pop, push for a queue)

问题描述

使用 numpy 数组，我想执行此操作:

• 移动 x[1],...,x[n-1] 到 x[0],...,x[n-2] (左移),
• 在最后一个索引中写入一个新值:x[n-1] = newvalue.

这类似于先进后出队列的pop(), push(newvalue)(仅反转).

一个简单的实现是:x[:-1] = x[1:];x[-1] = 新值.

另一个使用 np.concatenate 的实现速度较慢:np.concatenate((x[1:], np.array(newvalue).reshape(1,)), axis=0).

有没有最快的方法?

解决方案

经过一些实验，很明显:

• 需要复制，
• 对于nparray(numpy 数组)来说，最快和最简单的方法是切片和复制.

所以解决方案是:x[:-1] = x[1:];x[-1] = 新值.

这是一个小基准:

<预><代码>>>>x = np.random.randint(0, 1e6, 10**8);新值 = -100>>>%timeit x[:-1] = x[1:];x[-1] = 新值1000 个循环，最好的 3 个:每个循环 73.6 毫秒>>>%timeit np.concatenate((x[1:], np.array(newvalue).reshape(1,)), axis=0)1 个循环，最好的 3 个:每个循环 339 毫秒

但是如果您不需要快速访问数组中的所有值，而只需要快速访问第一个或最后一个，请使用 deque 更智能.

With numpy arrays, I want to perform this operation:

• move x[1],...,x[n-1] to x[0],...,x[n-2] (left shift),
• write a new value in the last index: x[n-1] = newvalue.

This is similar to a pop(), push(newvalue) for a first-in last-out queue (only inverted).

A naive implementation is: x[:-1] = x[1:]; x[-1] = newvalue.

Another implementation, using np.concatenate, is slower: np.concatenate((x[1:], np.array(newvalue).reshape(1,)), axis=0).

Is there a fastest way to do it?

解决方案

After some experiments, it is clear that:

• copying is required,
• and the fastest and simplest way to do that, for nparray (numpy arrays) is a slicing and copying.

So the solution is: x[:-1] = x[1:]; x[-1] = newvalue.

Here is a small benchmark:

>>> x = np.random.randint(0, 1e6, 10**8); newvalue = -100
>>> %timeit x[:-1] = x[1:]; x[-1] = newvalue
1000 loops, best of 3: 73.6 ms per loop
>>> %timeit np.concatenate((x[1:], np.array(newvalue).reshape(1,)), axis=0) 
1 loop, best of 3: 339 ms per loop

But if you don't need to have a fast access to all values in the array, but only the first or last ones, using a deque is smarter.

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