rapidclus.py 文件源码

def close_outer(itr, b):
    """
    For sorted iterable `itr` and numeric `b`, return the index of the value in
    itr that is closest to b.  Also return the value itself and the distance
    between itr and b.
    """
    i, length, first, last = bisect(itr, b), len(itr), itr[0], itr[-1]
    if i == length:
        return length - 1, last, b - last
    if i == 0:
        return 0, first, first - b
    vnext, vprev = itr[i], itr[i - 1]
    dnext, dprev = vnext - b, b - vprev
    if dnext < dprev:
        return i, vnext, dnext
    return i - 1, vprev, dprev