Shift elements in a numpy array

Following-up from this question years ago, is there a canonical "shift" function in numpy? I don't see anything from the documentation.

Here's a simple version of what I'm looking for:

    def shift(xs, n):
        if n >= 0:
            return np.r_[np.full(n, np.nan), xs[:-n]]
        else:
            return np.r_[xs[-n:], np.full(-n, np.nan)]

Using this is like:

    In [76]: xs
    Out[76]: array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.])

    In [77]: shift(xs, 3)
    Out[77]: array([ nan,  nan,  nan,   0.,   1.,   2.,   3.,   4.,   5.,   6.])

    In [78]: shift(xs, -3)
    Out[78]: array([  3.,   4.,   5.,   6.,   7.,   8.,   9.,  nan,  nan,  nan])

This question came from my attempt to write a fast rolling_product yesterday. I needed a way to "shift" a cumulative product and all I could think of was to replicate the logic in np.roll().

So np.concatenate() is much faster than np.r_[]. This version of the function performs a lot better:

    def shift(xs, n):
        if n >= 0:
            return np.concatenate((np.full(n, np.nan), xs[:-n]))
        else:
            return np.concatenate((xs[-n:], np.full(-n, np.nan)))

An even faster version simply pre-allocates the array:

    def shift(xs, n):
        e = np.empty_like(xs)
        if n >= 0:
            e[:n] = np.nan
            e[n:] = xs[:-n]
        else:
            e[n:] = np.nan
            e[:n] = xs[-n:]
        return e

Not numpy but scipy provides exactly the shift functionality you want,

    import numpy as np
    from scipy.ndimage.interpolation import shift

    xs = np.array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.])

    shift(xs, 3, cval=np.NaN)

where default is to bring in a constant value from outside the array with value cval, set here to nan. This gives the desired output,

    array([ nan, nan, nan, 0., 1., 2., 3., 4., 5., 6.])

and the negative shift works similarly,

    shift(xs, -3, cval=np.NaN)

Provides output

    array([  3.,   4.,   5.,   6.,   7.,   8.,   9.,  nan,  nan,  nan])

From: stackoverflow.com/q/30399534