慕村9548890
为了有效,窗口算法必须链接两个重叠窗口的结果。在这里,与 :med0中位数,排序后的元素med中的中位数 x \ med0,xl之前的元素med和xg之后的元素,可以看作: medfuncX(x)<|x-med|> + med = [sum(xg) - sum(xl) - |med0-med|] / windowsize + med 因此,一个想法是维护一个表示已排序当前窗口的缓冲区,sum(xg)并且sum(xl). 使用 Numba 即时编译,这里会出现非常好的性能。首先是缓冲区管理:init对第一个窗口进行排序并计算 left( xls) 和 right( xgs) 总和。import numpy as npimport numbawindowsize = 51 #odd, >1halfsize = windowsize//2@numba.njitdef init(firstwindow): buffer = np.sort(firstwindow) xls = buffer[:halfsize].sum() xgs = buffer[-halfsize:].sum() return buffer,xls,xgsshift是线性部分。它更新缓冲区,保持它的排序。np.searchsorted计算 中的插入和删除位置O(log(windowsize))。这是技术性的xin<xout,因为xout<xin不是对称的情况。@numba.njitdef shift(buffer,xin,xout): i_in = np.searchsorted(buffer,xin) i_out = np.searchsorted(buffer,xout) if xin <= xout : buffer[i_in+1:i_out+1] = buffer[i_in:i_out] buffer[i_in] = xin else: buffer[i_out:i_in-1] = buffer[i_out+1:i_in] buffer[i_in-1] = xin return i_in, i_outupdate更新缓冲区和左右部分的总和。这是技术性的xin<xout,因为xout<xin不是对称的情况。@numba.njitdef update(buffer,xls,xgs,xin,xout): xl,x0,xg = buffer[halfsize-1:halfsize+2] i_in,i_out = shift(buffer,xin,xout) if i_out < halfsize: xls -= xout if i_in <= halfsize: xls += xin else: xls += x0 elif i_in < halfsize: xls += xin - xl if i_out > halfsize: xgs -= xout if i_in > halfsize: xgs += xin else: xgs += x0 elif i_in > halfsize+1: xgs += xin - xg return buffer, xls, xgsfunc相当于原来funcX的on buffer。O(1).@numba.njit def func(buffer,xls,xgs): med0 = buffer[halfsize] med = (buffer[halfsize-1] + buffer[halfsize+1])/2 if med0 > med: return (xgs-xls+med0-med) / windowsize + med else: return (xgs-xls+med-med0) / windowsize + med med是全局函数。O(data.size * windowsize).@numba.njitdef med(data): res = np.full_like(data, np.nan) state = init(data[:windowsize]) res[halfsize] = func(*state) for i in range(windowsize, data.size): xin,xout = data[i], data[i - windowsize] state = update(*state, xin, xout) res[i-halfsize] = func(*state) return res 表现 :import pandasdata=pandas.DataFrame(np.random.rand(10**5))%time res1=data[0].rolling(window = windowsize, center = True).apply(funcX, raw = True)Wall time: 10.8 sres2=med(data[0].values)np.allclose((res1-res2)[halfsize:-halfsize],0)Out[112]: True%timeit res2=med(data[0].values)40.4 ms ± 462 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)它快了 250 倍,窗口大小 = 51。一小时变成了 15 秒。