The solution I came up with is to separately find the indices where the first occurrence of 1 is and the indices where the last occurrence of 1 is.

def get_indices2(arr):
    value_diff = np.diff(arr, prepend=0, append=0)
    start_idx = np.nonzero(value_diff == 1)[0]
    end_idx = np.nonzero(value_diff == -1)[0] - 1  # -1 to include endpoint
    idx = np.stack((start_idx, end_idx), axis=-1)
    return idx

Note that the result is not a list of tuples, but a 2D numpy array like the one below.

array([[0, 0],
       [3, 5],
       [7, 9]])

Here is benchmark:

import timeit

import numpy as np


def get_indices(array):
    indices = []
    xstart = None
    for x, col in enumerate(array):
        if col == 0 and xstart is not None:
            indices.append((xstart, x - 1))
            xstart = None
        elif col == 1 and xstart is None:
            xstart = x

    if xstart is not None:
        indices.append((xstart, x))

    return indices


def get_indices2(arr):
    value_diff = np.diff(arr, prepend=0, append=0)
    start_idx = np.nonzero(value_diff == 1)[0]
    end_idx = np.nonzero(value_diff == -1)[0] - 1  # -1 to include endpoint
    idx = np.stack((start_idx, end_idx), axis=-1)
    return idx


def benchmark():
    rng = np.random.default_rng(0)
    arr = rng.integers(0, 1, endpoint=True, size=20_000_000)
    expected = np.asarray(get_indices(arr))

    for f in [get_indices, get_indices2]:
        t = np.asarray(f(arr))
        assert expected.shape == t.shape and np.array_equal(expected, t), f.__name__
        elapsed = min(timeit.repeat(lambda: f(arr), repeat=10, number=1))
        print(f"{f.__name__:25}: {elapsed}")


benchmark()

Result:

get_indices              : 4.708652864210308
get_indices2             : 0.21052680909633636

One thing that concerns me is that on my PC, your function takes less than 5 seconds to process 20 million elements, while you mention that it takes 8 seconds to process 2 million elements. So I may be missing something.

Update

Matt provided an elegant solution using reshape in his answer. However, if performance is important, I would suggest optimizing the np.diff part first.

def custom_int8_diff(arr):
    out = np.empty(len(arr) + 1, dtype=np.int8)
    out[0] = arr[0]
    out[-1] = -arr[-1]
    np.subtract(arr[1:], arr[:-1], out=out[1:-1])
    return out


def get_indices2_custom_diff(arr):
    mask = custom_int8_diff(arr)  # Use custom diff. Others unchanged.
    start_idx = np.nonzero(mask == 1)[0]
    end_idx = np.nonzero(mask == -1)[0] - 1
    return np.stack((start_idx, end_idx), axis=-1)

For Matt’s reshape solution, we can use logical_xor, which is even faster.

def custom_bool_diff(arr):
    out = np.empty(len(arr) + 1, dtype=np.bool_)
    out[0] = arr[0]
    out[-1] = arr[-1]
    np.logical_xor(arr[1:], arr[:-1], out=out[1:-1])
    return out


def get_indices3_custom_diff(arr):
    value_diff = custom_bool_diff(arr)  # Use custom diff. Others unchanged.
    idx = np.nonzero(value_diff)[0]
    idx[1::2] -= 1
    return idx.reshape(-1, 2)

Benchmark (2 million elements):

get_indices              : 0.463582425378263
get_indices2             : 0.01675519533455372
get_indices3             : 0.01814895775169134
get_indices2_custom_diff : 0.010258681140840054
get_indices3_custom_diff : 0.006368924863636494

Benchmark (20 million elements):

get_indices              : 4.708652864210308
get_indices2             : 0.21052680909633636
get_indices3             : 0.19463363010436296
get_indices2_custom_diff : 0.14093663357198238
get_indices3_custom_diff : 0.08207075204700232

How about:

import numpy as np
r = np.array([1, 0, 0, 1, 1, 1, 0, 1, 1, 1])
s = np.diff(r, prepend=0, append=0)
t = np.where(s)[0]
t[1::2] -= 1
# don't do `tolist` if you don't need to, though
t.reshape(-1, 2).tolist() 
# [[0, 0], [3, 5], [7, 9]]

Update: I see there is an independent post of the same core solution at about the same time. I suspect the in-place operations + reshape here is slightly preferable to separating, creating a copy with the arithmetic, and stacking (which creates another copy). tolist takes almost all the time here, so just stop after the reshape if you don’t actually need a list.