Fastest way to left-cycle a numpy array (like pop, push for a queue)

Question

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?

Answer 1

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.