Find index given multiple values of array with Numpy

Question

I understand that Numpy can generate index of an array given the value that we are looking for with numpy.where. My question: Is there a function that can generate index given multiple values.

For example, with this array

a = np.array([1.,0.,0.,0.,1.,1.,0.,0.,0.,0.,...,1.,1.])

It will be great if I can just specify 4 zeros and the function can tell the index of it then I can replace those right away with another value. I have a function that can recognize the pattern but it is not efficient. Any pointers will be very helpful

Answer 1

Seems like I give this answer once a week or so. Fastest and most memory-efficient way is to use void views over as_strided views

def rolling_window(a, window):  #based on @senderle's answer: https://stackoverflow.com/q/7100242/2901002
    shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
    strides = a.strides + (a.strides[-1],)
    c = np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
    return c

def vview(a):  #based on @jaime's answer: https://stackoverflow.com/a/16973510/4427777
    return np.ascontiguousarray(a).view(np.dtype((np.void, a.dtype.itemsize * a.shape[1])))

def pattwhere_void(pattern, a): # Using @PaulPanzer's template form above
    k, n = map(len, (pattern, a))
    pattern = np.atleast_2d(pattern)
    a = np.asanyarray(a)
    if k>n:
        return np.empty([0], int)
    return np.flatnonzero(np.in1d(vview(rolling_window(a, k)), vview(pattern)))