最大回撤计算

import numpy as np

def MaxDrawdown1(Asset):
    i = np.argmax((np.maximum.accumulate(Asset) - Asset) / np.maximum.accumulate(Asset)) 
    if i == 0:
        return 0
    j = np.argmax(Asset[:i]) 
    return (Asset[j] - Asset[i]) / (Asset[j])

def MaxDrawdown2(Asset):
    flag = Asset[0]
    bottom = Asset[0]
    drawdown = [0]
    d = 0
    for v in Asset:
        if v>=flag:
            flag = v
            bottom = v
            drawdown.append(d)
        elif v<=bottom:
            bottom = v
            d = (bottom-flag)/flag
    		drawdown.append(d)
    maxdrawdown = -min(drawdown)
    return maxdrawdown